poj 2689 Prime Distance （素数二次筛法）

2689 -- Prime Distance

　　没怎么研究过数论，还是今天才知道有素数二次筛法这样的东西。

　　题意是，要求求出给定区间内相邻两个素数的最大和最小差。

　　二次筛法的意思其实就是先将1~sqrt(b)内的素数先筛出来，然后再对[a,b]区间用那些筛出来的素数再次线性筛。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 66666;
int prm[N >> 3], pn;
bool np[N];

void getbase() {
    np[pn = 0] = np[1] = true;
    prm[pn++] = 2;
    for (int i = 3; i < N; i++, i++) {
        if (!np[i]) prm[pn++] = i;
        for (int j = 0, tmp; j < pn; j++) {
            tmp = i * prm[j];
            if (tmp >= N) break;
            np[tmp] = true;
            if (i % prm[j] == 0) break;
        }
    }
//    cout << pn << endl;
//    for (int i = 0; i < 10; i++) cout << prm[i] << endl;
}

bool mk[1111111];

int main() {
//    freopen("in", "r", stdin);
    getbase();
    int T;
    long long a, b;
    while (~scanf("%lld%lld", &a, &b)) {
        a = max(2ll, a);
        memset(mk, 0, sizeof(mk));
        for (int i = 0; i < pn && prm[i] <= b; i++) {
            long long t = a / prm[i] * prm[i];
            if (t != a) t += prm[i];
            t = max(t, (long long) prm[i] << 1);
            for ( ; t <= b; t += prm[i]) mk[t - a] = true;
        }
        long long p = a;
        int mini = 1111111, maxi = -1111111, minp, maxp;
        while (p <= b && mk[p - a]) p++;
        for (long long i = p + 1; i <= b; i++) {
            if (mk[i - a]) continue;
//            cout << i << endl;
            if (maxi < i - p) maxi = i - p, maxp = p;
            if (mini > i - p) mini = i - p, minp = p;
            p = i;
        }
        if (maxi < 0) puts("There are no adjacent primes.");
        else printf("%d,%d are closest, %d,%d are most distant.
", minp, minp + mini, maxp, maxp + maxi);
    }
    return 0;
}

——written by Lyon