Primes on Interval 【打表+二分】

A - Primes on Interval

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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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

代码：

#include<stdio.h>
#include<string.h>
int a,b,k;
int dp[1000000+10],c[1000000+10];
void dabiao()
{
	memset(c,0,sizeof(c));
	c[0]=1;
	c[1]=1;
	for(int i=2;i<=1000000;i++)
	{
		if(!c[i])
		{
			for(int j=i+i;j<=1000000;j+=i)
			{
				c[j]=1;
			} 
		}
	}
	dp[0]=0;
	for(int i=1;i<=1000000;i++)
	{
		dp[i]=dp[i-1];
		if(!c[i])
		dp[i]++;
	} 
}
int judge(int l)
{
	for(int i=a;i<=b-l+1;i++)
	{
		if(dp[i+l-1]-dp[i-1]<k)
		return 0;
	}
	return 1;
}
int main()
{
	dabiao();
	scanf("%d%d%d",&a,&b,&k);
	int l=1,r=b-a+1,mid,flag=0;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(judge(mid))
		{
			flag=1;
			r=mid-1;
			//ans=mid;
		}
		else
		{
			l=mid+1;
		}
	}
	if(flag)
	printf("%d\n",l);
	else
	printf("-1\n");
	return 0;
}