CodeForces 237C Primes on Interval(数学加二分）

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本文介绍了一个算法挑战，即找出在给定区间内含有至少k个质数的最短连续子区间长度。通过预处理所有小于等于100万的整数来判断其是否为质数，并使用滑动窗口的方法来解决这个问题。

Primes on Interval

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

Submit Status Practice CodeForces 237C

Appoint description:

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

Input

6 13 1

Output

Input

1 4 3

Output

-1

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int f[1000010],sum[1000010],a,b,k;
void init()
{
	f[0]=1;
	f[1]=1;
	for(int i=2;i<=1000000;i++)
	{
		if(f[i]==0)
		{
			for(int j=i+i;j<=1000000;j+=i)
			f[j]=1;
			
		}
	}
	sum[0]=sum[1]=0;
	for(int i=2;i<=1000000;i++)
	{
		if(f[i]==0)
		sum[i]=sum[i-1]+1;
		else
		sum[i]=sum[i-1];
	
		
    }
}
int fun(int x)  
{  
      
    for(int i=a;i<=b-x+1;i++)  
    if(sum[i+x-1]-sum[i-1]<k)  
    {  
        return 0;  
    }   
    return 1;  
}	
int main()
{     init();
   scanf("%d %d %d",&a,&b,&k);
    if(sum[b]-sum[a-1]<k)
   printf("-1\n");
    else
{
	int start=1,end=b-a+1,ans=0;  
            while(start <= end)  
            {  
                int mid = (start + end) / 2;  
                if(fun(mid))  
                {  ans=mid;
                    end = mid-1;  
                    
                }  
                else  
                start = mid + 1;  
            }  
            if(ans == 0)
			 puts("-1");  
            else
			 printf("%d\n",ans);  
        } 
	}