Codeforces 237C Primes on Interval

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 integerx (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 找到最小的l满足1 <= l <= b - a + 1，使得对于任意的x(a <= x <= b - l + 1)均有[x, x + l - 1]里面至少有k个质数

素数打表 标记前i个数中有多少个素数 先判断最大的l不符合直接输出-1 否则用二分查找最小的l

#include<cstdio>
#include<algorithm>
using namespace std;
int su[1000010];
int vis[1000010];//标记有多少素数 
void dabiao()//素数打表 
{
	for(int i=2;i<=1000010;i++)
	{
		if(su[i]==1)
		{
			continue;
		}
		for(int j=2*i;j<=1000010;j+=i)
		{
			su[j]=1;
		}
	}
	su[0]=1;
	su[1]=1;
}
void sushu()//标记前i个数有多少个素数 
{
	dabiao();
	vis[0]=0;
	for(int i=1;i<1000010;i++)
	{
		if(!su[i])
		{
			vis[i]=vis[i-1]+1;
		}
		else
		{
			vis[i]=vis[i-1];
		}
	}
	
}
bool panduan(int a,int b,int k,int l)
{
	  int s=b-l+1;
	  for(int i=a;i<=s;i++)
	  {
	  	if(vis[i+l-1]-vis[i-1]>=k)
	  	{
	  		continue;
		  }
		  else
		  {
		  	return 0;
		  }
	  }
	  return 1;
}
int erfen(int a,int b,int k)
{
	int mid, l=1,r=b-a+1;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(panduan(a,b,k,mid))
		{
			r=mid-1;
		}
		else
		{
			l=mid+1;
		}
	}
	return l;
 } 
int main()
{
	sushu();
    int a,b,k;
    while(scanf("%d%d%d",&a,&b,&k)!=EOF)
    {
    	if(!panduan(a,b,k,b-a+1))
    	{
    		printf("-1\n");
		}
		else
		{
			printf("%d\n",erfen(a,b,k));
		}
	}
	return 0;
 }