质数相关

快速判断素数

bool prime(int n)  {      if(n==3 || n==2)      return true;      if(n%2==0 || n%3==0 || n==1)      return false;      int k = sqrt(n)+1;      for(int i = 5;i < k;i += 6)          if(n%i==0 || n%(i+2)==0)          return false;  return true;  }  

编译出来100w内的质数

#include <stdio.h>#include <stdlib.h>#include <string.h>#define PRIME_SIZE  100000int main(void){    int i,j=0,k;    int a[PRIME_SIZE];    char temp[100] = "";    FILE*fp=fopen("prime.txt","w");    a[0]=2;    j=1;    sprintf(temp, "%d\n", 2);    fwrite(temp,1,strlen(temp),fp);    for(i=3;i<1000000;i+=2)    {        for(k=0;k<j;k++)        {            if(i%a[k]==0)                break;        }        if(k >= j)        {            //prime number            a[j]=i;            j++;            sprintf(temp, "%d\n", i);            fwrite(temp,1,strlen(temp),fp);        }    }    fclose(fp);    printf("DONE!!\n");    return 0;}

求解第n个质数

内存开销小还跑得快

#include <bits/stdtr1c++.h>#define MAXN 100#define MAXM 10001#define MAXP 40000#define MAX 400000#define clr(ar) memset(ar, 0, sizeof(ar))#define read() freopen("lol.txt", "r", stdin)#define dbg(x) cout << #x << " = " << x << endl#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))using namespace std;namespace pcf{    long long dp[MAXN][MAXM];    unsigned int ar[(MAX >> 6) + 5] = {0};    int len = 0, primes[MAXP], counter[MAX];    void Sieve(){        setbit(ar, 0), setbit(ar, 1);        for (int i = 3; (i * i) < MAX; i++, i++){            if (!chkbit(ar, i)){                int k = i << 1;                for (int j = (i * i); j < MAX; j += k) setbit(ar, j);            }        }        for (int i = 1; i < MAX; i++){            counter[i] = counter[i - 1];            if (isprime(i)) primes[len++] = i, counter[i]++;        }    }    void init(){        Sieve();        for (int n = 0; n < MAXN; n++){            for (int m = 0; m < MAXM; m++){                if (!n) dp[n][m] = m;                else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];            }        }    }    long long phi(long long m, int n){        if (n == 0) return m;        if (primes[n - 1] >= m) return 1;        if (m < MAXM && n < MAXN) return dp[n][m];        return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);    }    long long Lehmer(long long m){        if (m < MAX) return counter[m];        long long w, res = 0;        int i, a, s, c, x, y;        s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);        a = counter[y], res = phi(m, a) + a - 1;        for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;        return res;    }}long long solve(long long n){    int i, j, k, l;    long long x, y, res = 0;    for (i = 0; i < pcf::len; i++){        x = pcf::primes[i], y = n / x;        if ((x * x) > n) break;        res += (pcf::Lehmer(y) - pcf::Lehmer(x));    }    for (i = 0; i < pcf::len; i++){        x = pcf::primes[i];        if ((x * x * x) > n) break;        res++;    }    return res;}int main(){    pcf::init();    long long n, res,L,R,M,ca=1;    while (scanf("%lld", &n) != EOF){        if(n==0) break;        L=2;R=1e8;        while(L<R){            M=(L+R)/2;            res=pcf::Lehmer(M);            if(res>=n) R=M;            else L=M+1;        }        printf("Case %lld: %lld\n",ca++,L);    }    return 0;}

内存开销大跑得快 Meisell-Lehmer算法+二分搜索

#include <bits/stdtr1c++.h>#define MAXN 100#define MAXM 10001#define MAXP 40000#define MAX 400000#define clr(ar) memset(ar, 0, sizeof(ar))#define read() freopen("lol.txt", "r", stdin)#define dbg(x) cout << #x << " = " << x << endl#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))using namespace std;namespace pcf{    long long dp[MAXN][MAXM];    unsigned int ar[(MAX >> 6) + 5] = {0};    int len = 0, primes[MAXP], counter[MAX];    void Sieve(){        setbit(ar, 0), setbit(ar, 1);        for (int i = 3; (i * i) < MAX; i++, i++){            if (!chkbit(ar, i)){                int k = i << 1;                for (int j = (i * i); j < MAX; j += k) setbit(ar, j);            }        }        for (int i = 1; i < MAX; i++){            counter[i] = counter[i - 1];            if (isprime(i)) primes[len++] = i, counter[i]++;        }    }    void init(){        Sieve();        for (int n = 0; n < MAXN; n++){            for (int m = 0; m < MAXM; m++){                if (!n) dp[n][m] = m;                else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];            }        }    }    long long phi(long long m, int n){        if (n == 0) return m;        if (primes[n - 1] >= m) return 1;        if (m < MAXM && n < MAXN) return dp[n][m];        return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);    }    long long Lehmer(long long m){        if (m < MAX) return counter[m];        long long w, res = 0;        int i, a, s, c, x, y;        s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);        a = counter[y], res = phi(m, a) + a - 1;        for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;        return res;    }}long long solve(long long n){    int i, j, k, l;    long long x, y, res = 0;    for (i = 0; i < pcf::len; i++){        x = pcf::primes[i], y = n / x;        if ((x * x) > n) break;        res += (pcf::Lehmer(y) - pcf::Lehmer(x));    }    for (i = 0; i < pcf::len; i++){        x = pcf::primes[i];        if ((x * x * x) > n) break;        res++;    }    return res;}int main(){    pcf::init();    long long n, res,L,R,M,ca=1;    while (scanf("%lld", &n) != EOF){        if(n==0) break;        L=2;R=1e8;        while(L<R){            M=(L+R)/2;            res=pcf::Lehmer(M);            if(res>=n) R=M;            else L=M+1;        }        printf("Case %lld: %lld\n",ca++,L);    }    return 0;}

Meisell-Lehmer算法模板（ 计算2~n中的素数个数

bool np[maxn];int prime[maxn],pi[maxn];int getprime(){    int cnt=0;    np[0]=np[1]=true;    pi[0]=pi[1]=0;    for(int i=2; i<maxn; ++i)    {        if(!np[i]) prime[++cnt]=i;        pi[i]=cnt;        for(int j=1; j<=cnt&&i*prime[j]<maxn; ++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 getphi(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<maxn)    {        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 getphi(x,s-1)-getphi(x/prime[s],s-1);}ll getpi(ll x){    if(x<maxn) return pi[x];    ll ans=getphi(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<maxn) 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=getphi(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;}