CodeForces - 237C Primes on Interval （素数打表&二分）

CodeForces - 237C

Primes on Interval

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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

Find and print the required minimum l. If no value l

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

Source

Codeforces Round #147 (Div. 2)

//题意：
给你一个a,b,k,让你找出在a<=x<=b-l+1之间的每个x在[x,x+1,.....x+l-1]之间都至少有k个素数
//思路：
先将素数打表，在用二分查找就行了

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1000010
using namespace std;
int a,b,k;
int sum[N],p[N];
void f()
{
	p[1]=1;
	for(int i=2;i<=N;i++)
	{
		if(!p[i])
		{
			for(int j=i+i;j<=N;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<N;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;
}