【codeforce237C】Primes on Interval

最新推荐文章于 2019-08-04 00:37:15 发布

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==数学and思维==

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本文探讨了如何找到一个最小长度L，使得在给定区间[a,b]内的任意子区间[x,x+L-1]中都至少包含k个素数。通过预处理素数表并采用二分查找的方法来高效解决问题。

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 lintegers 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.

Example

Input

2 4 2

Output

Input

6 13 1

Output

Input

1 4 3

Output

-1

题意：给你一个闭区间[a,b]，求一个最小的L，使得在区间[a,b-L+1]内任取一个数x，可以满足在x,x+1,x+2,……,x+L-2,x+L-1内至少包含k个素数。(1<=a,b,k<=10^6)

题解：素数打表，统计出每个区间有多少个素数，对L二分查找

code:

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a,b,k;
int c[1000000+11];
int f[1000011];
int cheak(int mid){
for(int i=a;i<=b-mid+1;i++)
{
	if(f[i+mid-1]-f[i-1]<k)
return 0;
}	
return 1;
}
int main()
{
	for(int i=2;i<=1000011;i++)
	{
		f[i]=f[i-1];
		if(c[i]==1) continue;
		f[i]++;
		for(int j=2*i;j<=1000011;j+=i){
			c[j]=1;
		}
	}//素数打表
	while(~scanf("%d%d%d",&a,&b,&k)){
		if(f[b]-f[a-1]<k) {
			printf("-1\n");return 0;
		}
			int l=1;int r=b-a+1,ans;
			while(l<=r){
				int mid=(l+r)>>1;
				if(cheak(mid)) ans=mid,r=mid-1;
				else l=mid+1;
			}
			printf("%d\n",ans);
	       
		}
	
	return 0;
}