Appoint description:
Description
You've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.
Consider positive integers a, a + 1, ..., b(a ≤ b). You want to find the minimum integer l(1 ≤ l ≤ b - a + 1) such that for any integer x(a ≤ x ≤ b - l + 1) among l integers x, x + 1, ..., x + l - 1 there are at least k prime numbers.
Find and print the required minimum l. If no value l meets the described limitations, print -1.
Input
A single line contains three space-separated integers a, b, k (1 ≤ a, b, k ≤ 106; a ≤ b).
Output
In a single line print a single integer — the required minimum l. If there's no solution, print -1.
Sample Input
Input
2 4 2
Output
3
Input
6 13 1
Output
4
Input
1 4 3
Output
-1
题意：
求最小的l使 x, x + 1, ..., x + l - 1 there are at least k prime numbers；
简单二分；
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; const int MAXN = 1e6 + 100; int dp[MAXN]; int vis[MAXN]; void db(){ memset(vis, 0, sizeof(vis)); vis[1] = 1; for(int i = 2; i <= sqrt(MAXN); i++){ if(!vis[i]){ for(int j = i * i; j < MAXN; j += i){ vis[j] = 1; } } } dp[0] = 0; for(int i = 1; i < MAXN; i++){ dp[i] = dp[i - 1]; if(!vis[i])dp[i]++; } } bool js(int l, int a, int b, int k){ for(int i = a; i <= b - l + 1; i++){ if(dp[i + l - 1] - dp[i - 1] < k)return false; } return true; } int erfen(int l, int r, int a, int b, int k){ int mid, ans = -1; while(l <= r){ mid = (l + r) >> 1; if(js(mid, a, b, k)){ ans = mid; r = mid - 1; } else l = mid + 1; } return ans; } int main(){ db(); int a, b, k; while(~scanf("%d%d%d", &a, &b, &k)){ printf("%d ", erfen(0, b - a + 1, a, b, k)); } return 0; }