MillerRabin与PollardRho

BZOJ 4802 欧拉函数

Description

已知N，求phi(N)
Input

正整数N。N<=10^18
Output

输出phi(N)
Sample Input

8
Sample Output

4

直接套板子

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<vector>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define vll vector<ll> #define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10LL+ch-'0';ch=getchar();}    return x*f;}inline ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);} inline ll abs2(ll x){if (x>=0) return x;return -x;} inline ll add(ll a,ll b,ll p){a+=b; return a>=p ? a - p : a;}inline ll mul(ll a,ll b,ll p){    ll rst = 0;    while (b) {        if (b&1) rst = add(rst, a, p);        b >>= 1, a = add(a, a, p);    }    return rst;}inline ll pow2(ll a,ll b,ll p)  //a^b mod p {      ll rst = 1;    while (b) {        if (b&1) rst = mul(rst, a, p);        b >>= 1, a = mul(a, a, p);    }    return rst;}inline bool witness2(ll a, ll x) {    ll m = x - 1, t = 0;    while (! (m&1)) {        m>>=1, ++t;    }    ll fac = pow2(a, m, x), lst = fac;    while (t--) {        fac = mul(fac, fac, x);        if (fac == 1 and lst != 1 and lst != x - 1) return true;        lst = fac;    } return lst != 1;}#define CAP (20) inline ll range(ll l,ll r){return l+rand()%(r-l+1); }inline bool MillerRabin(ll x) {    if (x < 2LL) return false;    if (x == 2LL) return true;    if (! (x&1LL)) return false;    for (int i = 1; i <= CAP; i++) {        ll a = range(2, x-1);        if (witness2(a, x)) return false;    } return true;}inline ll PollardRho(ll x, ll a) {    ll x1 = range(0, x - 1), x2 = x1;    while(1) {        x2 = add(mul(x2, x2, x), a, x);        x2 = add(mul(x2, x2, x), a, x);        ll d = gcd(abs2(x2 - x1), x);        if (d != 1 and d != x) return d;        x1 = add(mul(x1, x1, x), a, x);        if (x1 == x2) return x;    } return x; }vector<ll> ls;inline void findFac(ll x) {    if (MillerRabin(x))        return ls.push_back(x), void(0);    ll p = x;    while (p >= x) p = PollardRho(p, range(0, x - 1));    findFac(p), findFac(x / p  );}int main(){//  freopen("bzoj4802.in","r",stdin);//  freopen(".out","w",stdout);    srand(23333);    ll n=read();    if (n==1) puts("1");    else if (MillerRabin(n)) cout<<n-1<<endl;    else {        findFac(n);        sort(ALL(ls));        ls.erase(unique(ALL(ls)),ls.end());        ll phi=n;        vll::iterator it;        for(it=ls.begin();it!=ls.end();it++) {            phi=phi/(*it)*((*it)-1);        }        cout<<phi<<endl;    }    return 0;}

HDU 1164 Eddy’s research I

Problem Description

Eddy’s interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can’t write program, so Eddy has to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .

Input

The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.

Output

You have to print a line in the output for each entry with the answer to the previous question.

Sample Input

11

9412

Sample Output

11

2*2*13*181

Author

eddy

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<vector>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define vll vector<ll> #define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10LL+ch-'0';ch=getchar();}    return x*f;}inline ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);} inline ll abs2(ll x){if (x>=0) return x;return -x;} inline ll add(ll a,ll b,ll p){a+=b; return a>=p ? a - p : a;}inline ll mul(ll a,ll b,ll p){    ll rst = 0;    while (b) {        if (b&1) rst = add(rst, a, p);        b >>= 1, a = add(a, a, p);    }    return rst;}inline ll pow2(ll a,ll b,ll p)  //a^b mod p {      ll rst = 1;    while (b) {        if (b&1) rst = mul(rst, a, p);        b >>= 1, a = mul(a, a, p);    }    return rst;}inline bool witness2(ll a, ll x) {    ll m = x - 1, t = 0;    while (! (m&1)) {        m>>=1, ++t;    }    ll fac = pow2(a, m, x), lst = fac;    while (t--) {        fac = mul(fac, fac, x);        if (fac == 1 and lst != 1 and lst != x - 1) return true;        lst = fac;    } return lst != 1;}#define CAP (20) inline ll range(ll l,ll r){return l+rand()%(r-l+1); }inline bool MillerRabin(ll x) {    if (x < 2LL) return false;    if (x == 2LL) return true;    if (! (x&1LL)) return false;    for (int i = 1; i <= CAP; i++) {        ll a = range(2, x-1);        if (witness2(a, x)) return false;    } return true;}inline ll PollardRho(ll x, ll a) {    ll x1 = range(0, x - 1), x2 = x1;    while(1) {        x2 = add(mul(x2, x2, x), a, x);        x2 = add(mul(x2, x2, x), a, x);        ll d = gcd(abs2(x2 - x1), x);        if (d != 1 and d != x) return d;        x1 = add(mul(x1, x1, x), a, x);        if (x1 == x2) return x;    } return x; }// need clear ls// x>1 vector<ll> ls;inline void findFac(ll x) {    if (MillerRabin(x))        return ls.push_back(x), void(0);    ll p = x;    while (p >= x) p = PollardRho(p, range(0, x - 1));    findFac(p), findFac(x / p  );}int main(){//  freopen("hdu1164.in","r",stdin);//  freopen(".out","w",stdout);    srand(time(NULL));    int n;    while(scanf("%d",&n)!=EOF) {        int ans=0;        ls.erase(ALL(ls));        findFac(n);        sort(ALL(ls));        Rep(i,SI(ls)) {            if (i) putchar('*');            printf("%lld",ls[i]);        }        puts("");    }    return 0;}

HDU D_num

一个数是否有且只有4个因子。

分解质因数，2种情况

• 1,a,b,a∗b
• 1,a,a2,a3

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<vector>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define vll vector<ll> #define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10LL+ch-'0';ch=getchar();}    return x*f;}inline ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);} inline ll abs2(ll x){if (x>=0) return x;return -x;} inline ll add(ll a,ll b,ll p){a+=b; return a>=p ? a - p : a;}inline ll mul(ll a,ll b,ll p){    ll rst = 0;    while (b) {        if (b&1) rst = add(rst, a, p);        b >>= 1, a = add(a, a, p);    }    return rst;}inline ll pow2(ll a,ll b,ll p)  //a^b mod p {      ll rst = 1;    while (b) {        if (b&1) rst = mul(rst, a, p);        b >>= 1, a = mul(a, a, p);    }    return rst;}inline bool witness2(ll a, ll x) {    ll m = x - 1, t = 0;    while (! (m&1)) {        m>>=1, ++t;    }    ll fac = pow2(a, m, x), lst = fac;    while (t--) {        fac = mul(fac, fac, x);        if (fac == 1 and lst != 1 and lst != x - 1) return true;        lst = fac;    } return lst != 1;}#define CAP (20) inline ll range(ll l,ll r){return l+rand()%(r-l+1); }inline bool MillerRabin(ll x) {    if (x < 2LL) return false;    if (x == 2LL) return true;    if (! (x&1LL)) return false;    for (int i = 1; i <= CAP; i++) {        ll a = range(2, x-1);        if (witness2(a, x)) return false;    } return true;}inline ll PollardRho(ll x, ll a) {    ll x1 = range(0, x - 1), x2 = x1;    while(1) {        x2 = add(mul(x2, x2, x), a, x);        x2 = add(mul(x2, x2, x), a, x);        ll d = gcd(abs2(x2 - x1), x);        if (d != 1 and d != x) return d;        x1 = add(mul(x1, x1, x), a, x);        if (x1 == x2) return x;    } return x; }// need clear ls// x>1 vector<ll> ls;inline void findFac(ll x) {    if (MillerRabin(x))        return ls.push_back(x), void(0);    ll p = x;    while (p >= x) p = PollardRho(p, range(0, x - 1));    findFac(p), findFac(x / p  );}int main(){//  freopen("hdu3864.in","r",stdin);//  freopen(".out","w",stdout);    srand(time(NULL));    ll n;    while(scanf("%lld",&n)!=EOF) {        if (n==1) {            puts("is not a D_num");            continue;        }        int ans=0;        ls.erase(ALL(ls));        findFac(n);        sort(ALL(ls));        if(SI(ls)==2) {            if (ls[0]!=ls[1]) printf("%lld %lld %lld\n",ls[0],ls[1],n);            else puts("is not a D_num");        }        else if (SI(ls)==3) {            if (ls[0]==ls[2]) printf("%lld %lld %lld\n",ls[0],ls[0]*ls[0],n);            else puts("is not a D_num");        }         else puts("is not a D_num");    }    return 0;}

HDU 4910 Problem about GCD

题意：给定一个数n，求所有小于等于n且与n互素的数的乘积再mod n

打表，发现
f(n)={n−1,1,n=1,2,4 or n=pk,2∗pk,p is prime and p≠2 else

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<vector>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define vll vector<ll> #define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10LL+ch-'0';ch=getchar();}    return x*f;}inline ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);} inline ll abs2(ll x){if (x>=0) return x;return -x;} inline ll add(ll a,ll b,ll p){a+=b; return a>=p ? a - p : a;}inline ll mul(ll a,ll b,ll p){    ll rst = 0;    while (b) {        if (b&1) rst = add(rst, a, p);        b >>= 1, a = add(a, a, p);    }    return rst;}inline ll pow2(ll a,ll b,ll p)  //a^b mod p {      ll rst = 1;    while (b) {        if (b&1) rst = mul(rst, a, p);        b >>= 1, a = mul(a, a, p);    }    return rst;}inline bool witness2(ll a, ll x) {    ll m = x - 1, t = 0;    while (! (m&1)) {        m>>=1, ++t;    }    ll fac = pow2(a, m, x), lst = fac;    while (t--) {        fac = mul(fac, fac, x);        if (fac == 1 and lst != 1 and lst != x - 1) return true;        lst = fac;    } return lst != 1;}#define CAP (20) inline ll range(ll l,ll r){return l+rand()%(r-l+1); }inline bool MillerRabin(ll x) {    if (x < 2LL) return false;    if (x == 2LL) return true;    if (! (x&1LL)) return false;    for (int i = 1; i <= CAP; i++) {        ll a = range(2, x-1);        if (witness2(a, x)) return false;    } return true;}inline ll PollardRho(ll x, ll a) {    ll x1 = range(0, x - 1), x2 = x1;    while(1) {        x2 = add(mul(x2, x2, x), a, x);        x2 = add(mul(x2, x2, x), a, x);        ll d = gcd(abs2(x2 - x1), x);        if (d != 1 and d != x) return d;        x1 = add(mul(x1, x1, x), a, x);        if (x1 == x2) return x;    } return x; }// need clear ls// x>1 vector<ll> ls;inline void findFac(ll x) {    if (MillerRabin(x))        return ls.push_back(x), void(0);    ll p = x;    while (p >= x) p = PollardRho(p, range(0, x - 1));    findFac(p), findFac(x / p  );}int main(){//  freopen("hdu3864.in","r",stdin);//  freopen(".out","w",stdout);    srand(time(NULL));//  For(i,100) {//      int p=1;//      For(j,i) if (gcd(i,j)==1) p=p*j%i;//      cout<<i<<':'<<p<<endl;//  }    ll n;    while(scanf("%lld",&n)!=EOF&& n!=-1) {        ll ans;        if (n==1||n==2||n==4) ans=n-1;        else {            ls.erase(ALL(ls));            findFac(n);            sort(ALL(ls));            int sz=SI(ls);            if (ls[0]==ls[sz-1]) {                ans=(ls[0]==2) ?1: (n-1);            } else {                ans=1;                if (SI(ls)>=2) {                    if (ls[0]==2&&ls[0]!=ls[1]&&ls[1]==ls[sz-1]) ans=n-1;                }            }        }        cout<<ans<<endl;    }    return 0;}