CQUOJ 9711 Primes on Interval

 

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

/*
2016年4月21日23:49:25
题意：给出三个正整数a、b、k，求最小的L（1 ≤ L≤ b - a + 1）满足对于[a, b-L+1]中的任意一个数X，在[X, X+L-1]这L个数中，至少有k个素数。
　　　如果不存在满足条件的L，输出-1.
解题思路：首先判断是否有解，即判断当L=b-a+1时是否有解，因此时L最大，若此时都无解，则肯定无解。
       若有解，则1 ≤ L≤ b - a + 1，二分求最小的L即可。

*/

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
# include <vector>
# include <cmath>
# define INF 0x3f3f3f3f
using namespace std;
const int N = 1e6 + 5;
int vis[N], cnt[N];

//  标记素数 
void Sign()
{
    int i, j;
    int m = (int)sqrt(N);  // 可能是标记时 这个数不需要太大 后面的数可能在前面判断时都被标记过了 
    memset(vis, 0, sizeof(vis));  // 如果不开根号  好像会有问题... 
    for (i = 2; i <= m; i++){
        if (vis[i]) continue;
        for (j = i; j*i <= N; j++){
            if (!vis[j*i])    
                vis[j*i] = 1;    
        }
    }
}
//  cnt[i] 表示 从1到 i一共有多少个素数 
void Count()
{
    Sign();
    cnt[0] = cnt[1] = 0;
    for (int i = 2; i <= N; i++){
        if (!vis[i]) cnt[i] = cnt[i-1] + 1;
        else cnt[i] = cnt[i-1];
    }
}
// 判断当前的l是否满足题目要求  
int Check(int a, int b, int k, int l)
{
    int r = b - l + 1;
    for (int i = a; i <= r; i++){
        if (cnt[i + l - 1] - cnt[i - 1] >= k)
            continue;
        return 0;
    }
    return 1;
}
//  二分求 l的最小值 
int Solve(int a, int b, int k)
{
    int L = 1, R = b-a+1;
    while (L <= R){
        int mid = (L + R) / 2;
        if (Check(a, b, k, mid))
            R = mid - 1;
        else L = mid + 1;
    }
    return L;
}

int main(void)
{
    int a, b, k, ans;
    Count();
    while (~scanf("%d %d %d", &a, &b, &k)){
        if (!Check(a, b, k, b - a + 1)) //此处判断 l的最大值是否符合 
            printf("-1
");
        else {
            ans = Solve(a, b, k);
            printf("%d
", ans);
        }
    }    
    
    return 0;    
}