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 and p_k - p₁ = s . For example, with k = 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 prime k -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 and s(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 

1
100 200 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;
}