UVA - 1404 Prime k-tuple （素数筛选）

Description

{p₁,..., p_k : p₁ <p₂ <...< p_k} is called a prime k -tuple of distance s if p₁, p₂,..., p_k are consecutive prime numbers andp_k - p₁ = s . For example, withk = 4 , s = 8 ,{11, 13, 17, 19} is a prime 4-tuple of distance 8.

Given an interval [a, b] , k , and s , your task is to write a program to find the number of primek -tuples of distance s in the interval[a, b] .

Input 

The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.

For each data set, there is only one line containing 4 numbers, a , b , k ands(a, b < 2* 10⁹, k < 10, s < 40) .

Output 

For each test case, write in one line the numbers of prime k -tuples of distance s .

Sample Input 

1100 200 4 8

Sample Output 

2题意：如果有k个相邻的素数，满足pk-p1=s,称这些素数组成一个距离为s的素数k元组，输出区间[a, b]内距离为s的k元组思路：首先筛选出素数，然后在区间[a,b]内查找满足的个数，还有我决定了以后刷选用long long 
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>typedef long long ll;using namespace std;const int maxn = 100005;int prime[maxn], vis[maxn], cnt;void init() {memset(vis, 0, sizeof(0));vis[1] = vis[0] = 1;cnt = 0;for (ll i = 2; i < maxn; i++) if (!vis[i]) {prime[cnt++] = i;for (ll j = i*i; j < maxn; j += i) vis[j] = 1;}}int check(int n) {if (n < maxn)return vis[n] == 0;for (int i = 0; i < cnt && prime[i]*prime[i] <= n; i++)if (n % prime[i] == 0)return 0;return 1;}int main() {init();int t, a, b, k, s;scanf("%d", &t);while (t--) {scanf("%d%d%d%d", &a, &b, &k, &s);int tmp = 0, ans = 0;vector<int> num;for (int i = a; i <= b; i++)if (check(i))num.push_back(i);int size = num.size();for (int i = 0; i+k-1 < size; i++)if (num[i+k-1] - num[i] == s)ans++;printf("%d\n", ans);}return 0;}