UV 1404-prime K-tuple (number theory)

Link to the question: Ultraviolet A 1404 1404-prime K-tuple

If K adjacent prime numbers are P1, P2 ,..., PK: Can it meet the PK? P1 = S. These prime numbers form a K-tuple of prime numbers whose distance is S. The number of K-tuples between S and A and B is determined.

Solution: filter the prime number method, first pre-process the prime number table of [1, SQRT (INF)], and then specify the range [, b] filter out prime numbers based on the pre-processed prime number table.

#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;}