uva 1404 - Prime k-tuple(数论)

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题目链接：uva 1404 1404 - Prime k-tuple

题目大意：如果k个相邻的素数p1,p2,…,pk，满足pk?p1=s，称这些素数组成一个距离为s的素数k元组，给定区间a，b，求有多少个距离s的k元组。

解题思路：筛选素数法，先预处理出[1, sqrt(inf)]的素数表，然后对给定区间[a,b]根据预处理出的素数表筛选出素数即可。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;
const int sqrt_inf = 46340;
const int maxn = 2 * 1e9;

int np, pri[sqrt_inf];
bool vis[maxn+5];
vector<int> vec;

void prime_table (int n) {
    np = 0;
    memset(vis, 0, sizeof(vis));

    for (int i = 2; i <= n; i++) {
        if (vis[i])
            continue;

        pri[np++] = i;
        for (int j = i * i; j <= n; j += i)
            vis[j] = 1;
    }
}

int solve () {
    int ret = 0;
    int a, b, s, k;
    vec.clear();
    memset(vis, 0, sizeof(vis));

    scanf("%d%d%d%d", &a, &b, &k, &s);

    for (int i = 0; i < np && pri[i] * pri[i] <= b; i++) {
        int u = pri[i], d = (u - a % u) % u;

        if (u == a + d)
            d += u;

        while (d <= b - a) {
            vis[d] = 1;
            d += u;
        }
    }

    for (int i = 0; i <= b-a; i++) {
        if (vis[i] == 0 && a + i > 1)
            vec.push_back(a+i);
    }


    for (int i = 0; i + k - 1 < vec.size(); i++) {
        if (vec[i+k-1] - vec[i] == s)
            ret++;
    }

    return ret;
}

int main () {
    prime_table(sqrt_inf);

    int cas;
    scanf("%d", &cas);

    while (cas--) {
        printf("%d\n", solve());
    }
    return 0;
}

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uva 1404 - Prime k-tuple(数论)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/38639525