模板 - 质数筛法（线性筛法、二次筛法）

整理的算法模板合集： ACM模板

---

试除法判定质数

bool is_prime(int x)
{
    if (x < 2) return false;
    for (int i = 2; i <= x / i; i ++ )
        if (x % i == 0)
            return false;
    return true;
}

线性筛法

int primes[N], cnt;
bool vis[N];

void init(int x){
    vis[1] = vis[0] = 1;
    for(int i = 2; i <= x; ++ i){
        if(!vis[i])primes[cnt ++ ] = i;
        for(int j = 0; primes[j] * i <= x; ++ j){
            vis[primes[j] * i] = true;
            if(i % primes[j] == 0)break;
        }
    }
}

二次筛法

/*AcWing 196. 质数距离
整数数列1~n,给出区间[L,R] (L,R<=2^31,R-L<=1e6) 求区间内相邻质数距离的最大值和最小值*/
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1000010;

int primes[N], cnt;
bool st[N];

void init(int n)
{
    memset(st, 0, sizeof st);
    cnt = 0;
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    int l, r;
    while (cin >> l >> r)
    {
        init(50000);

        memset(st, 0, sizeof st);
        for (int i = 0; i < cnt; i ++ )
        {
            LL p = primes[i];
            for (LL j = max(p * 2, (l + p - 1) / p * p); j <= r; j += p)
                st[j - l] = true;
        }

        cnt = 0;
        for (int i = 0; i <= r - l; i ++ )
            if (!st[i] && i + l >= 2)
                primes[cnt ++ ] = i + l;

        if (cnt < 2) puts("There are no adjacent primes.");
        else
        {
            int minp = 0, maxp = 0;
            for (int i = 0; i + 1 < cnt; i ++ )
            {
                int d = primes[i + 1] - primes[i];
                if (d < primes[minp + 1] - primes[minp]) minp = i;
                if (d > primes[maxp + 1] - primes[maxp]) maxp = i;
            }

            printf("%d,%d are closest, %d,%d are most distant.\n",
                primes[minp], primes[minp + 1],
                primes[maxp], primes[maxp + 1]);
        }
    }

    return 0;
}