Primes on Interval

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 ≤ 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：给出三个正整数a、b、k，求最小的L（1?≤?L≤?b?-?a?+?1）满足对于[a, b-L+1]中的任意一个数X，在[X, X+L-1]这L个数中，至少有k个素数。如果不存在满足条件的L，输出-1.
解题思路：首先判断是否有解，即判断当L=b-a+1时是否有解，因此时L最大，若此时都无解，则肯定无解。
若有解，则1?≤?L≤?b?-?a?+?1，二分求最小的L即可。*/

#include
#include
using namespace std;
int su[1000100]={1,1};
int con[1000100];
void sushudabiao()
{
	int i,j;
	for(i=2;i<=1000010;i++)
	{
		if(su[i]==1)
			continue;
		for(j=2*i;j<=1000010;j+=i)
		{
			su[j]=1;
		}
	}
}

void shusu()
{
	sushudabiao();
	int i;
	con[0]=0;
	for(i=1;i<=1000010;i++)
	{
		if(!su[i]){
			con[i]=con[i-1]+1;
		}
		else{
			con[i]=con[i-1];
		}
	}
}

bool panduan(int a,int b,int k,int l)
{
	int i,r=b-l+1;
	for(i=a;i<=r;i++)
	{
		if(con[i+l-1]-con[i-1]>=k){
			continue;
		}
		else{
			return false;
		}
	}
	return true;
}

int qiumin(int a,int b,int k)
{
	int l=1,r=b-a+1;
	int mid;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(panduan(a,b,k,mid))
			r=mid-1;
		else
			l=mid+1;
	}
	return l;
}

int main()
{
	shusu();
	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",qiumin(a,b,k));
	}
	return 0;
}