Codeforces--237C--Primes on Interval(二分水题)


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 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 ≤ 106a ≤ 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:在a--b之间找一个长度l使得a--b之间每一个数字x都有x--(x+l-1)之间有k个素数,二分水题,枚举最长长度就行
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a,b,k;
int sum[2000000+100],p[2000000+100];
void f()
{
	p[1]=1;
	for(int i=2;i<=2000100;i++)
	{
		if(!p[i])
		{
			for(int j=i+i;j<=2000100;j+=i)
			p[j]=1;
		}
	}
}
bool judge(int l)
{
	for(int i=a;i<=b-l+1;i++)
	{
		int r=i+l-1;
		if(sum[r]-sum[i-1]<k)
		return false;
	}
	return true;
}
int main()
{
	f();
	memset(sum,0,sizeof(sum));
	for(int i=1;i<2000100;i++)
	if(!p[i])
		sum[i]=sum[i-1]+1;
	else
		sum[i]=sum[i-1];
	while(scanf("%d%d%d",&a,&b,&k)!=EOF)
	{
		int ans=-1,l=1,r=b-a+1;
		while(l<=r)
		{
			int mid=(l+r)/2;
			if(judge(mid))
			{
				ans=mid;
				r=mid-1;
			}
			else
			l=mid+1;
		}
		printf("%d\n",ans);
	}
	return 0;
}

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