数论板子(持续更新

bool issqr(__int128_t x)//开方{    __int128_t y=(__int128_t)ceil(sqrt((long double)x));    for(;y*y<=x;++y);        for(--y;y*y>x;--y);    return y*y==x;}bool iscub(__int128_t x)//三次方{    __int128_t y=(__int128_t)ceil(pow((long double)x,1.0/3));    for(;y*y*y<=x;++y);        for(--y;y*y*y>x;--y);    return y*y*y==x;}

int tot,pr[maxx],vis[maxx];void init()//素数表  O(nsqrt(n)){    for(int i=2;i<maxx;i++)    {        if(!vis[i])pr[tot++]=i;        for(int j=0;j<tot&&i*pr[j]<maxx;j++)        {            vis[i*pr[j]]=1;            if(i%pr[j]==0)break;        }    }} 

int d[maxx],pr[maxx],tot;void init(){    for(int i = 2; i < maxx; ++i)    {        if(!d[i])            pr[tot++] = d[i] = i;        for(int j = 0, k; (k = i * pr[j]) < maxx; ++j)        {            d[k] = pr[j];            if(d[i] == pr[j])                break;        }    }}

const int s=10;//Miller-Rabininline LL mul_mod(LL a, LL b, LL m)  O(n^(1/4)){    LL c = a*b-(LL)((long double)a*b/m+0.5)*m;    return c<0 ? c+m : c;}LL fast_exp(LL a,LL x,LL m){    LL b = 1;    while (x)    {        if (x & 1)            b = mul_mod(b, a, m);        a = mul_mod(a, a, m);        x >>= 1;    }    return b;}bool MR(LL n){    if (!(n&1))        return n == 2;    LL t = 0, u;    for (u = n-1; !(u&1); u >>= 1)        ++t;    for (int i = 0; i < s; ++i)    {        LL a = rand()%(n-2)+2, x = fast_exp(a, u, n);        for (LL j = 0, y; x != 1 && j < t; ++j, x = y)        {            y = mul_mod(x, x, n);            if (y == 1 && x != n-1)                return false;        }        if (x != 1)            return false;    }    return true;}

Miller-Rabin Pollard-rho

#include <cstdio>#include <cstdlib>#include <ctime>#include <algorithm>typedef long long ll;const int s = 10, MAX_F = 70;ll cnt, f[MAX_F];inline ll mul_mod(ll a, ll b, ll m){ll c = a*b-(ll)((long double)a*b/m+0.5)*m;return c<0 ? c+m : c;}ll fast_exp(ll a, ll x, ll m){ll b = 1;while (x){if (x & 1)b = mul_mod(b, a, m);a = mul_mod(a, a, m);x >>= 1;}return b;}bool MR(ll n){if (!(n&1))return n == 2;ll t = 0, u;for (u = n-1; !(u&1); u >>= 1)++t;for (int i = 0; i < s; ++i){ll a = rand()%(n-2)+2, x = fast_exp(a, u, n);for (ll j = 0, y; x != 1 && j < t; ++j, x = y){y = mul_mod(x, x, n);if (y == 1 && x != n-1)return false;}if (x != 1)return false;}return true;}inline ll abs(ll x){return x<0 ? -x : x;}ll gcd(ll a, ll b){return b ? gcd(b, a%b) : a;}ll PR(ll n, ll a){ll x = rand()%n, y = x, k = 1, i = 0, d = 1;while (d == 1){if ((x = (mul_mod(x, x, n)+a)%n) == y)return n;d = gcd(n, abs(y-x));if (++i == k){k <<= 1;y = x;}}return d;}void decomp(ll n){if (n == 1)return;if (MR(n)){f[cnt++] = n;return;}ll d = n, c = n-1;while (d == n)d = PR(n, c--);do{n /= d;}while (!(n%d));decomp(d);decomp(n);}int main(){srand(time(0));ll n;while (scanf("%lld", &n), n){cnt = 0;decomp(n);std::sort(f, f+cnt);cnt = std::unique(f, f+cnt)-f;ll ans = n;for (int i = 0; i < cnt; ++i){// printf("%lld\n", f[i]);ans = ans/f[i]*(f[i]-1);}printf("%lld\n", ans);}return 0;}

快速乘

LL fun(LL a,LL b){LL sum=0;W(b){if(b&1)sum=(sum+a)%p;b/=2;a=a*2%p;}return sum;}

快速幂

int _pow(LL a,LL n)  {      LL ret=1;      while(n)      {          if(n&1)  ret=ret*a%mod;          a=a*a%mod;          n>>=1;      }      return ret;  }  int inv(int x)  {      return _pow(x,mod-2);  }  

拓展gcd

void exgcd(ll a,ll b,ll &g,ll &x,ll &y){    if(!b) {g=a;x=1;y=0;}    else {exgcd(b,a%b,g,y,x);y-=x*(a/b);}}void work(ll a , ll b , ll d ,ll& g , ll& x , ll& y){    exgcd(a,b,g,x,y);      //此处的x，y接收 ax+by=gcd(a,b)的值    x *= d/g;             //求ax+by=c 的解x    int t = b/g;    x = (x%t + t) % t;    y = abs( (a*x - d) / b);}ll a,b,g,x,y,d;void solve(){    S_3(a,b,d);    work(a,b,d,g,x,y);   /* if(x==0)    {        for(int i=1;x<=0;i++)        {            x=x+b/g*i;            y=y-a/g*i;        }    }*/}

逆元

void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d)  {      if (!b) {d = a, x = 1, y = 0;}      else{          ex_gcd(b, a % b, y, x, d);          y -= x * (a / b);      }  }  LL inv2(LL t, LL p)  {//如果不存在，返回-1      LL d, x, y;      ex_gcd(t, p, x, y, d);      return d == 1 ? (x % p + p) % p : -1;  } 

错排 排列组合

const int mod=1e9+7;  const int maxn=1e4+7;  LL D[maxn], C[maxn][105];  int p[1000050];  void init()  {      D[0]=1;D[1]=0;      for(int i=1;i<maxn;i++)//错排          D[i]=((i-1)*(D[i-1]+D[i-2])+mod)%mod;      int i;      for (p[0]=i=1;i<=200000;i++) p[i]=1ll*p[i-1]*i%mod;      for (int i = 0; i < maxn; i++)//排列组合      {          for (int j = 0; j <= min(105, i); j++)          {              if (j == 0) C[i][j] = 1;              else                  C[i][j] = (C[i-1][j] + C[i-1][j-1]) % mod;          }      }  }

C(n,m)

先init 再comb(n,m) n在下

namespace COMB{int F[N<<1], Finv[N<<1], inv[N<<1];void init(){inv[1] = 1;for (int i = 2; i < N<<1; i++){inv[i] = (LL)(MOD - MOD/i) * inv[MOD%i] % MOD;}F[0] = Finv[0] = 1;for (int i = 1; i < N<<1; i++){F[i] = (LL)F[i-1] * i % MOD;Finv[i] = (LL)Finv[i-1] * inv[i] % MOD;}}int comb(int n, int m){if (m < 0 || m > n) return 0;return (LL)F[n] * Finv[n-m] % MOD * Finv[m] % MOD;}}using namespace COMB;

C（n,m） 不mod

double ln[3100000];int n,p;void init(){    int maxc=3e6;    for(int i=1;i<=maxc;i++)    {        ln[i]=log(i);    }}db comb(int x,int y){    db ans=0;    for(int i=1;i<=y;i++)    {        ans+=ln[x-i+1]-ln[i];    }    return ans;}db calc(int x){    return x*exp(comb(n,p-1)-comb(n+x,p));}

C(n,m) mod p (p为素数

//C(m,n) m在上 n在下,p为素数  LL n,m,p;    LL quick_mod(LL a, LL b)  {      LL ans = 1;      a %= p;      while(b)      {          if(b & 1)          {              ans = ans * a % p;              b--;          }          b >>= 1;          a = a * a % p;      }      return ans;  }    LL C(LL n, LL m)  {      if(m > n) return 0;      LL ans = 1;      for(int i=1; i<=m; i++)      {          LL a = (n + i - m) % p;          LL b = i % p;          ans = ans * (a * quick_mod(b, p-2) % p) % p;      }      return ans;  }    LL Lucas(LL n, LL m)  {      if(m == 0) return 1;      return C(n % p, m % p) * Lucas(n / p, m / p) % p;  }    int main()  {      int T;      scan_d(T);      while(T--)      {          scan_d(n),scan_d(m),scan_d(p);          print(Lucas(n,m));      }      return 0;  }  

C(n,m) mod p (p可能为合数

  const int N = 200005;    bool prime[N];  int p[N];  int cnt;    void isprime()  {      cnt = 0;      memset(prime,true,sizeof(prime));      for(int i=2; i<N; i++)      {          if(prime[i])          {              p[cnt++] = i;              for(int j=i+i; j<N; j+=i)                  prime[j] = false;          }      }  }    LL quick_mod(LL a,LL b,LL m)  {      LL ans = 1;      a %= m;      while(b)      {          if(b & 1)          {              ans = ans * a % m;              b--;          }          b >>= 1;          a = a * a % m;      }      return ans;  }    LL Work(LL n,LL p)  {      LL ans = 0;      while(n)      {          ans += n / p;          n /= p;      }      return ans;  }    LL Solve(LL n,LL m,LL P)  {      LL ans = 1;      for(int i=0; i<cnt && p[i]<=n; i++)      {          LL x = Work(n, p[i]);          LL y = Work(n - m, p[i]);          LL z = Work(m, p[i]);          x -= (y + z);          ans *= quick_mod(p[i],x,P);          ans %= P;      }      return ans;  }    int main()  {      int T;      isprime();      cin>>T;      while(T--)      {          LL n,m,P;          cin>>n>>m>>P;          n += m - 2;          m--;          cout<<Solve(n,m,P)<<endl;      }      return 0;  }  

FFT：

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>#include <complex>using namespace std;#define pi acos(-1)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long longtypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e3+10;const int maxx=1<<17;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;typedef complex<long double>Complex; /* * 进行FFT和IFFT前的反转变换。 * 位置i和 （i二进制反转后位置）互换 * len必须去2的幂 */void reverse(Complex y[],int len){    int i,j,k;    for(i = 1, j = len/2;i < len-1; i++)    {        if(i < j)swap(y[i],y[j]);        //交换互为小标反转的元素，i<j保证交换一次        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的        k = len/2;        while( j >= k)        {            j -= k;            k /= 2;        }        if(j < k) j += k;    }}/* * 做FFT * len必须为2^k形式， * on==1时是DFT，on==-1时是IDFT*/void fft(Complex a[],int n,int on){    reverse(a,n);    for(int i=1;i<n;i<<=1)    {        Complex wn=Complex(cos(PI/i),on*sin(PI/i));        for(int j=0;j<n;j+=(i<<1))        {            Complex w=Complex(1,0);            for(int k=0;k<i;k++,w=w*wn)            {                Complex x=a[j+k],y=w*a[j+k+i];                a[j+k]=x+y;                a[j+k+i]=x-y;            }        }    }    if(on==-1)        for(int i=0;i<n;i++)            a[i].real()/=n;}struct node{    int c[maxx];}A,B,C;Complex a[maxx],b[maxx],c[maxx];int main(){    int n,x;    W(scan_d(n))    {        FOr(0,n,i)        {            scan_d(x);            x+=20000;            A.c[x]++;            B.c[x+x]++;            C.c[x+x+x]++;        }        FOR(0,maxx,i)        {            a[i]=A.c[i],b[i]=B.c[i];        }        fft(a,maxx,1);        fft(b,maxx,1);        FOR(0,maxx,i)        {            c[i]=a[i]*(a[i]*a[i]-3.0l*b[i]);        }        fft(c,maxx,-1);        FOR(0,maxx,i)        {            LL ans=((LL)(c[i].real()+0.5)+2*C.c[i])/6;            if(ans>0)                printf("%d : %lld\n",i-60000,ans);        }    }}

因子分解--3：

LL F(LL a,LL i){LL l=1,r=a,b;W(l<r){LL mid=(l+r+1)/2;b=a;FOR(1,i,j) b/=mid;if(b) l=mid;else r=mid-1;}return l;}void solve(){FOR(1,i,j){LL root=F(n,j);}}

一个数因子数和：

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <map>#include <algorithm>using namespace std;#define pi acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=1e18+7;const int dx[]= {-1,0,1,0,1,-1,-1,1};const int dy[]= {0,1,0,-1,-1,1,-1,1};const int maxn=2005;const int maxx=1e5+100;const double EPS=1e-7;//const int mod=998244353;template<class T>inline T min(T a,T b,T c){return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c){return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d){return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d){return max(max(a,b),max(c,d));}inline LL Scan(){LL Res=0,ch,Flag=0;if((ch=getchar())=='-')Flag=1;else if(ch>='0' && ch<='9')Res=ch-'0';while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0';return Flag ? -Res : Res;}LL a,b,mod=9901;LL ans=1;LL tot,pr[maxx],vis[maxx];  void init()//素数表  O(nsqrt(n))  {      for(int i=2;i<maxx;i++)      {          if(!vis[i])pr[tot++]=i;          for(int j=0;j<tot&&i*pr[j]<maxx;j++)          {              vis[i*pr[j]]=1;              if(i%pr[j]==0)break;          }      }  }  LL _pow(LL a,LL n)  {      LL ret=1;      while(n)      {          if(n&1)  ret=ret*a%mod;          a=a*a%mod;          n>>=1;      }      return ret;  }  LL inv(LL x)  {      return _pow(x,mod-2);  }  void solve(){for(int i=0;i<tot;i++){int c=0;if(a%pr[i]==0){while(a%pr[i]==0){c++; a/=pr[i];//cout<<c<<" "<<pr[i]<<endl;}if(pr[i]%mod==0) continue;else if(pr[i]%mod==1) {ans=(ans*(c*b+1))%mod;}else {LL ret=(_pow(pr[i],(c*b+1))-1+mod)%mod;ans=(ans*ret*inv(pr[i]-1))%mod;}}}if(a>1){if(a%mod==0)return ;else if(a%mod==1){ans=(ans*(b+1))%mod;}else {LL ret=(_pow(a,(b+1))-1+mod)%mod;ans=(ans*ret*inv(a-1))%mod;}}}int main(){init();while(~scanf("%lld%lld",&a,&b)){ans=1;solve();cout<<ans<<endl;}}

baby-step A^x=B(mod c)

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define LL __int64#define N 1000000using namespace std;struct Node{int idx;int val;} baby[N];bool cmp(Node n1,Node n2){return n1.val!=n2.val?n1.val<n2.val:n1.idx<n2.idx;}int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int extend_gcd(int a,int b,int &x,int &y){if(b==0){x=1;y=0;return a;}int gcd=extend_gcd(b,a%b,x,y);int t=x;x=y;y=t-a/b*y;return gcd;}int inval(int a,int b,int n){int e,x,y;extend_gcd(a,n,x,y);e=((LL)x*b)%n;return e<0?e+n:e;}int PowMod(int a,int b,int MOD){LL ret=1%MOD,t=a%MOD;while(b){if(b&1)ret=((LL)ret*t)%MOD;t=((LL)t*t)%MOD;b>>=1;}return (int)ret;}int BinSearch(int num,int m){int low=0,high=m-1,mid;while(low<=high){mid=(low+high)>>1;if(baby[mid].val==num)return baby[mid].idx;if(baby[mid].val<num)low=mid+1;elsehigh=mid-1;}return -1;}int BabyStep(int A,int B,int C){LL tmp,D=1%C;int temp;for(int i=0,tmp=1%C; i<100; i++,tmp=((LL)tmp*A)%C)if(tmp==B)return i;int d=0;while((temp=gcd(A,C))!=1){if(B%temp) return -1;d++;C/=temp;B/=temp;D=((A/temp)*D)%C;}int m=(int)ceil(sqrt((double)C));for(int i=0,tmp=1%C; i<=m; i++,tmp=((LL)tmp*A)%C){baby[i].idx=i;baby[i].val=tmp;}sort(baby,baby+m+1,cmp);int cnt=1;for(int i=1; i<=m; i++)if(baby[i].val!=baby[cnt-1].val)baby[cnt++]=baby[i];int am=PowMod(A,m,C);for(int i=0; i<=m; i++,D=((LL)(D*am))%C){int tmp=inval(D,B,C);if(tmp>=0){int pos=BinSearch(tmp,cnt);if(pos!=-1)return i*m+pos+d;}}return -1;}int main(){int A,B,C;while(scanf("%d%d%d",&A,&C,&B)!=EOF){if(B>=C){puts("Orz,I can’t find D!");continue;}int ans=BabyStep(A,B,C);if(ans==-1)puts("Orz,I can’t find D!");elseprintf("%d\n",ans);}return 0;}

莫比乌斯：

int tot,p[maxx],miu[maxx],v[maxx];void Mobius(int n){    int i,j;    for(miu[1]=1,i=2;i<=n;i++)    {        if(!v[i]) p[tot++]=i,miu[i]=-1;        for(j=0;j<tot&&i*p[j]<=n;j++)        {            v[i*p[j]]=1;            if(i%p[j]) miu[i*p[j]]=-miu[i];            else break;        }    }}

递推式

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <map>#include <set>#include <cassert>using namespace std;#define rep(i,a,n) for (long long i=a;i<n;i++)#define per(i,a,n) for (long long i=n-1;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define fi first#define se second#define SZ(x) ((long long)(x).size())typedef vector<long long> VI;typedef long long ll;typedef pair<long long,long long> PII;const ll mod=1e9+7;ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}// headlong long _,n;namespace linear_seq {    const long long N=10010;    ll res[N],base[N],_c[N],_md[N];    vector<long long> Md;    void mul(ll *a,ll *b,long long k)     {        rep(i,0,k+k) _c[i]=0;        rep(i,0,k) if (a[i]) rep(j,0,k)             _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;        for (long long i=k+k-1;i>=k;i--) if (_c[i])            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;        rep(i,0,k) a[i]=_c[i];    }    long long solve(ll n,VI a,VI b)     { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...//        printf("%d\n",SZ(b));        ll ans=0,pnt=0;        long long k=SZ(a);        assert(SZ(a)==SZ(b));        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;        Md.clear();        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);        rep(i,0,k) res[i]=base[i]=0;        res[0]=1;        while ((1ll<<pnt)<=n) pnt++;        for (long long p=pnt;p>=0;p--)         {            mul(res,res,k);            if ((n>>p)&1)             {                for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;            }        }        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;        if (ans<0) ans+=mod;        return ans;    }    VI BM(VI s)     {        VI C(1,1),B(1,1);        long long L=0,m=1,b=1;        rep(n,0,SZ(s))         {            ll d=0;            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;            if (d==0) ++m;            else if (2*L<=n)             {                VI T=C;                ll c=mod-d*powmod(b,mod-2)%mod;                while (SZ(C)<SZ(B)+m) C.pb(0);                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;                L=n+1-L; B=T; b=d; m=1;            }            else             {                ll c=mod-d*powmod(b,mod-2)%mod;                while (SZ(C)<SZ(B)+m) C.pb(0);                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;                ++m;            }        }        return C;    }    long long gao(VI a,ll n)     {        VI c=BM(a);        c.erase(c.begin());        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));    }};int main() {    for (scanf("%I64d",&_);_;_--)     {        scanf("%I64d",&n);        printf("%I64d\n",linear_seq::gao(VI{31,197,1255,7997},n-2));    }}

求一个数的所有因子和

#include <cstdio>  const int maxn = 500000+2;  int sum[maxn];  int main()  {      for(int i=1; i<maxn; i++)          for(int j=1; i*j<maxn; j++)              sum[i*j] += i;      int a;      while(scanf("%d",&a)!=EOF)          printf("%d\n",sum[a]);      return 0;  }  

1-n 素数个数

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>#include <complex>using namespace std;//#define pi acos(-1)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long longtypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e3+10;const int maxx=1e5+10;const double EPS=1e-8;const double eps=1e-8;const LL mod=1e6+3;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;const int N = 5e6 + 2;bool np[N];int prime[N], pi[N];int getprime(){    int cnt = 0;    np[0] = np[1] = true;    pi[0] = pi[1] = 0;    for(int i = 2; i < N; i++)    {        if(!np[i]) prime[++cnt] = i;        pi[i] = cnt;        for(int j = 1; j <= cnt && i * prime[j] < N; j++)        {            np[i * prime[j]] = true;            if(i % prime[j] == 0)   break;        }    }    return cnt;}const int M = 7;const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;int phi[PM + 1][M + 1], sz[M + 1];void init(){    getprime();    sz[0] = 1;    for(int i = 0; i <= PM; i++)        phi[i][0] = i;    for(int i = 1; i <= M; i++)    {        sz[i] = prime[i] * sz[i - 1];        for(int j = 1; j <= PM; j++)            phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];    }}int sqrt2(LL x){    LL r = (LL)sqrt(x - 0.1);    while(r * r <= x)   ++r;    return int(r - 1);}int sqrt3(LL x){    LL r = (LL)cbrt(x - 0.1);    while(r * r * r <= x)   ++r;    return int(r - 1);}LL get_phi(LL x, int s){    if(s == 0)  return x;    if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];    if(x <= prime[s]*prime[s])   return pi[x] - s + 1;    if(x <= prime[s]*prime[s]*prime[s] && x < N)    {        int s2x = pi[sqrt2(x)];        LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;        for(int i = s + 1; i <= s2x; i++) ans += pi[x / prime[i]];        return ans;    }    return get_phi(x, s - 1) - get_phi(x / prime[s], s - 1);}LL getpi(LL x){    if(x < N)   return pi[x];    LL ans = get_phi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;    for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; i++)        ans -= getpi(x / prime[i]) - i + 1;    return ans;}LL lehmer_pi(LL x){    if(x < N)   return pi[x];    int a = (int)lehmer_pi(sqrt2(sqrt2(x)));    int b = (int)lehmer_pi(sqrt2(x));    int c = (int)lehmer_pi(sqrt3(x));    LL sum = get_phi(x, a) + (LL)(b + a - 2) * (b - a + 1) / 2;    for (int i = a + 1; i <= b; i++)    {        LL w = x / prime[i];        sum -= lehmer_pi(w);        if (i > c) continue;        LL lim = lehmer_pi(sqrt2(w));        for (int j = i; j <= lim; j++)            sum -= lehmer_pi(w / prime[j]) - (j - 1);    }    return sum;}int main(){    init();    LL n;    while(~scanf("%I64d",&n))    {        print(lehmer_pi(n));    }    return 0;}

大数：

#include <iostream>#include <cstring>#include <stdio.h>using namespace std;#define DIGIT   4      //四位隔开,即万进制#define DEPTH   10000        //万进制#define MAX     25100    //题目最大位数/4，要不大直接设为最大位数也行typedef int bignum_t[MAX+1];int read(bignum_t a,istream&is=cin){    char buf[MAX*DIGIT+1],ch ;    int i,j ;    memset((void*)a,0,sizeof(bignum_t));    if(!(is>>buf))return 0 ;    for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)    ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;    for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0');    for(i=1;i<=a[0];i++)    for(a[i]=0,j=0;j<DIGIT;j++)    a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ;    for(;!a[a[0]]&&a[0]>1;a[0]--);    return 1 ;}void write(const bignum_t a,ostream&os=cout){    int i,j ;    for(os<<a[i=a[0]],i--;i;i--)    for(j=DEPTH/10;j;j/=10)    os<<a[i]/j%10 ;}int comp(const bignum_t a,const bignum_t b){    int i ;    if(a[0]!=b[0])    return a[0]-b[0];    for(i=a[0];i;i--)    if(a[i]!=b[i])    return a[i]-b[i];    return 0 ;}int comp(const bignum_t a,const int b){    int c[12]=    {        1    }    ;    for(c[1]=b;c[c[0]]>=DEPTH;c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++);    return comp(a,c);}int comp(const bignum_t a,const int c,const int d,const bignum_t b){    int i,t=0,O=-DEPTH*2 ;    if(b[0]-a[0]<d&&c)    return 1 ;    for(i=b[0];i>d;i--)    {        t=t*DEPTH+a[i-d]*c-b[i];        if(t>0)return 1 ;        if(t<O)return 0 ;    }    for(i=d;i;i--)    {        t=t*DEPTH-b[i];        if(t>0)return 1 ;        if(t<O)return 0 ;    }    return t>0 ;}void add(bignum_t a,const bignum_t b){    int i ;    for(i=1;i<=b[0];i++)    if((a[i]+=b[i])>=DEPTH)    a[i]-=DEPTH,a[i+1]++;    if(b[0]>=a[0])    a[0]=b[0];    else    for(;a[i]>=DEPTH&&i<a[0];a[i]-=DEPTH,i++,a[i]++);    a[0]+=(a[a[0]+1]>0);}void add(bignum_t a,const int b){    int i=1 ;    for(a[1]+=b;a[i]>=DEPTH&&i<a[0];a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++);    for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);}void sub(bignum_t a,const bignum_t b){    int i ;    for(i=1;i<=b[0];i++)    if((a[i]-=b[i])<0)    a[i+1]--,a[i]+=DEPTH ;    for(;a[i]<0;a[i]+=DEPTH,i++,a[i]--);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void sub(bignum_t a,const int b){    int i=1 ;    for(a[1]-=b;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void sub(bignum_t a,const bignum_t b,const int c,const int d){    int i,O=b[0]+d ;    for(i=1+d;i<=O;i++)    if((a[i]-=b[i-d]*c)<0)    a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH ;    for(;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void mul(bignum_t c,const bignum_t a,const bignum_t b){    int i,j ;    memset((void*)c,0,sizeof(bignum_t));    for(c[0]=a[0]+b[0]-1,i=1;i<=a[0];i++)    for(j=1;j<=b[0];j++)    if((c[i+j-1]+=a[i]*b[j])>=DEPTH)    c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH ;    for(c[0]+=(c[c[0]+1]>0);!c[c[0]]&&c[0]>1;c[0]--);}void mul(bignum_t a,const int b){    int i ;    for(a[1]*=b,i=2;i<=a[0];i++)    {        a[i]*=b ;        if(a[i-1]>=DEPTH)        a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH ;    }    for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void mul(bignum_t b,const bignum_t a,const int c,const int d){    int i ;    memset((void*)b,0,sizeof(bignum_t));    for(b[0]=a[0]+d,i=d+1;i<=b[0];i++)    if((b[i]+=a[i-d]*c)>=DEPTH)    b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH ;    for(;b[b[0]+1];b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH);    for(;!b[b[0]]&&b[0]>1;b[0]--);}void div(bignum_t c,bignum_t a,const bignum_t b){    int h,l,m,i ;    memset((void*)c,0,sizeof(bignum_t));    c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1 ;    for(i=c[0];i;sub(a,b,c[i]=m,i-1),i--)    for(h=DEPTH-1,l=0,m=(h+l+1)>>1;h>l;m=(h+l+1)>>1)    if(comp(b,m,i-1,a))h=m-1 ;    else l=m ;    for(;!c[c[0]]&&c[0]>1;c[0]--);    c[0]=c[0]>1?c[0]:1 ;}void div(bignum_t a,const int b,int&c){    int i ;    for(c=0,i=a[0];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void sqrt(bignum_t b,bignum_t a){    int h,l,m,i ;    memset((void*)b,0,sizeof(bignum_t));    for(i=b[0]=(a[0]+1)>>1;i;sub(a,b,m,i-1),b[i]+=m,i--)    for(h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1)    if(comp(b,m,i-1,a))h=m-1 ;    else l=m ;    for(;!b[b[0]]&&b[0]>1;b[0]--);    for(i=1;i<=b[0];b[i++]>>=1);}int digit(const bignum_t a,const int b){    int i,ret ;    for(ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT;i;ret/=10,i--);    return ret%10 ;}#define SGN(x) ((x)>0?1:((x)<0?-1:0))#define ABS(x) ((x)>0?(x):-(x))int read(bignum_t a,int&sgn,istream&is=cin){    char str[MAX*DIGIT+2],ch,*buf ;    int i,j ;    memset((void*)a,0,sizeof(bignum_t));    if(!(is>>str))return 0 ;    buf=str,sgn=1 ;    if(*buf=='-')sgn=-1,buf++;    for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)    ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;    for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0');    for(i=1;i<=a[0];i++)    for(a[i]=0,j=0;j<DIGIT;j++)    a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ;    for(;!a[a[0]]&&a[0]>1;a[0]--);    if(a[0]==1&&!a[1])sgn=0 ;    return 1 ;}struct bignum{    bignum_t num ;    int sgn ;    public :    inline bignum()    {        memset(num,0,sizeof(bignum_t));        num[0]=1 ;        sgn=0 ;    }    inline int operator!()    {        return num[0]==1&&!num[1];    }    inline bignum&operator=(const bignum&a)    {        memcpy(num,a.num,sizeof(bignum_t));        sgn=a.sgn ;        return*this ;    }    inline bignum&operator=(const int a)    {        memset(num,0,sizeof(bignum_t));        num[0]=1 ;        sgn=SGN (a);        add(num,sgn*a);        return*this ;    }    ;    inline bignum&operator+=(const bignum&a)    {        if(sgn==a.sgn)add(num,a.num);        else if        (sgn&&a.sgn)        {            int ret=comp(num,a.num);            if(ret>0)sub(num,a.num);            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memcpy(num,a.num,sizeof(bignum_t));                sub (num,t);                sgn=a.sgn ;            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if(!sgn)            memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ;        return*this ;    }    inline bignum&operator+=(const int a)    {        if(sgn*a>0)add(num,ABS(a));        else if(sgn&&a)        {            int  ret=comp(num,ABS(a));            if(ret>0)sub(num,ABS(a));            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memset(num,0,sizeof(bignum_t));                num[0]=1 ;                add(num,ABS (a));                sgn=-sgn ;                sub(num,t);            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if            (!sgn)sgn=SGN(a),add(num,ABS(a));        return*this ;    }    inline bignum operator+(const bignum&a)    {        bignum ret ;        memcpy(ret.num,num,sizeof (bignum_t));        ret.sgn=sgn ;        ret+=a ;        return ret ;    }    inline bignum operator+(const int a)    {        bignum ret ;        memcpy(ret.num,num,sizeof (bignum_t));        ret.sgn=sgn ;        ret+=a ;        return ret ;    }    inline bignum&operator-=(const bignum&a)    {        if(sgn*a.sgn<0)add(num,a.num);        else if        (sgn&&a.sgn)        {            int ret=comp(num,a.num);            if(ret>0)sub(num,a.num);            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memcpy(num,a.num,sizeof(bignum_t));                sub(num,t);                sgn=-sgn ;            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if(!sgn)add (num,a.num),sgn=-a.sgn ;        return*this ;    }    inline bignum&operator-=(const int a)    {        if(sgn*a<0)add(num,ABS(a));        else if(sgn&&a)        {            int  ret=comp(num,ABS(a));            if(ret>0)sub(num,ABS(a));            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memset(num,0,sizeof(bignum_t));                num[0]=1 ;                add(num,ABS(a));                sub(num,t);                sgn=-sgn ;            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if            (!sgn)sgn=-SGN(a),add(num,ABS(a));        return*this ;    }    inline bignum operator-(const bignum&a)    {        bignum ret ;        memcpy(ret.num,num,sizeof(bignum_t));        ret.sgn=sgn ;        ret-=a ;        return ret ;    }    inline bignum operator-(const int a)    {        bignum ret ;        memcpy(ret.num,num,sizeof(bignum_t));        ret.sgn=sgn ;        ret-=a ;        return ret ;    }    inline bignum&operator*=(const bignum&a)    {        bignum_t t ;        mul(t,num,a.num);        memcpy(num,t,sizeof(bignum_t));        sgn*=a.sgn ;        return*this ;    }    inline bignum&operator*=(const int a)    {        mul(num,ABS(a));        sgn*=SGN(a);        return*this ;    }    inline bignum operator*(const bignum&a)    {        bignum ret ;        mul(ret.num,num,a.num);        ret.sgn=sgn*a.sgn ;        return ret ;    }    inline bignum operator*(const int a)    {        bignum ret ;        memcpy(ret.num,num,sizeof (bignum_t));        mul(ret.num,ABS(a));        ret.sgn=sgn*SGN(a);        return ret ;    }    inline bignum&operator/=(const bignum&a)    {        bignum_t t ;        div(t,num,a.num);        memcpy (num,t,sizeof(bignum_t));        sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn ;        return*this ;    }    inline bignum&operator/=(const int a)    {        int t ;        div(num,ABS(a),t);        sgn=(num[0]==1&&!num [1])?0:sgn*SGN(a);        return*this ;    }    inline bignum operator/(const bignum&a)    {        bignum ret ;        bignum_t t ;        memcpy(t,num,sizeof(bignum_t));        div(ret.num,t,a.num);        ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn ;        return ret ;    }    inline bignum operator/(const int a)    {        bignum ret ;        int t ;        memcpy(ret.num,num,sizeof(bignum_t));        div(ret.num,ABS(a),t);        ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a);        return ret ;    }    inline bignum&operator%=(const bignum&a)    {        bignum_t t ;        div(t,num,a.num);        if(num[0]==1&&!num[1])sgn=0 ;        return*this ;    }    inline int operator%=(const int a)    {        int t ;        div(num,ABS(a),t);        memset(num,0,sizeof (bignum_t));        num[0]=1 ;        add(num,t);        return t ;    }    inline bignum operator%(const bignum&a)    {        bignum ret ;        bignum_t t ;        memcpy(ret.num,num,sizeof(bignum_t));        div(t,ret.num,a.num);        ret.sgn=(ret.num[0]==1&&!ret.num [1])?0:sgn ;        return ret ;    }    inline int operator%(const int a)    {        bignum ret ;        int t ;        memcpy(ret.num,num,sizeof(bignum_t));        div(ret.num,ABS(a),t);        memset(ret.num,0,sizeof(bignum_t));        ret.num[0]=1 ;        add(ret.num,t);        return t ;    }    inline bignum&operator++()    {        *this+=1 ;        return*this ;    }    inline bignum&operator--()    {        *this-=1 ;        return*this ;    }    ;    inline int operator>(const bignum&a)    {        return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0);    }    inline int operator>(const int a)    {        return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0);    }    inline int operator>=(const bignum&a)    {        return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0);    }    inline int operator>=(const int a)    {        return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0);    }    inline int operator<(const bignum&a)    {        return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0);    }    inline int operator<(const int a)    {        return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0);    }    inline int operator<=(const bignum&a)    {        return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0);    }    inline int operator<=(const int a)    {        return sgn<0?(a<0?comp(num,-a)>=0:1):        (sgn>0?(a>0?comp(num,a)<=0:0):a>=0);    }    inline int operator==(const bignum&a)    {        return(sgn==a.sgn)?!comp(num,a.num):0 ;    }    inline int operator==(const int a)    {        return(sgn*a>=0)?!comp(num,ABS(a)):0 ;    }    inline int operator!=(const bignum&a)    {        return(sgn==a.sgn)?comp(num,a.num):1 ;    }    inline int operator!=(const int a)    {        return(sgn*a>=0)?comp(num,ABS(a)):1 ;    }    inline int operator[](const int a)    {        return digit(num,a);    }    friend inline istream&operator>>(istream&is,bignum&a)    {        read(a.num,a.sgn,is);        return  is ;    }    friend inline ostream&operator<<(ostream&os,const bignum&a)    {        if(a.sgn<0)            os<<'-' ;        write(a.num,os);        return os ;    }};int main(){    int t;    cin>>t;    while(t--)    {        int n;        cin>>n;        bignum jie,ans;        jie+=1;        for(int i=1;i<=n;i++)        {            jie*=i;        }        for(int i=1;i<=n;i++)        {            ans=ans+jie/i;        }        cout<<ans<<".0"<<endl;    }}