Codeforces 237c:primes on Interval (prime number + two points)

C. Primes on Interval time limit per test 1 second memory limit per test, megabytes input standard input output Standar D Output

You ' ve decided to carry out a survey in the theory of prime numbers. Let us remind you the A prime number is a positive an integer that has exactly a 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 this for any integer x (a≤x≤b-l + 1) among L integer s x, x + 1, ..., x + l-1 There is 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. Examples input

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

Title: Give you the number A, B, K, let you ask for a * min * of the number L, so that in a, B interval of any length of the sub-continuous interval of l at least K primes.

Problem-Solving ideas: The number of characters to play the table, + a two points.

The code is as follows:

#include <cstdio>
int flag[1000010];
int sum[1000010];
int a,b,k,mid;
void Dabiao ()//Prime table, Sum[i] indicates the number of primes (including i) {flag[0]=flag[1]=1) in front of I
	;
	for (int i=2;i<1000010;i++)
	{
		if (flag[i]==1)
		continue;
		for (int j=i*2;j<1000010;j=j+i)
		{
			flag[j]=1;
		}
	}
	sum[0]=0;
	for (int i=1;i<1000010;i++)
	{
		if (flag[i]==0)
		{
			sum[i]=sum[i-1]+1;
		}
		else
		{
			sum[i]=sum[i-1];
}}}
	the range for 
	(int i=a;i<=size;i++)
	{
		if (sum[i+o-1]-) in the title of the bool judge (int o) {int size=b-o+1;// SUM[I-1]<K)//i+o-1 is the number of digits from the point of I, i-1 explanation: The number of characters between A~b =sum[b]-sum[a-1],a-1 is equivalent to i-1 
		{
			return false;
		}
	}
	return true;
}
int main ()
{
	dabiao ();
	scanf ("%d%d%d", &a,&b,&k);
	int l=1,r=b-a+1;//Test instructions set the left and right interval 
	int ans=-1;//cannot be found when output-1, so initialized to 1 while 
	(l<=r)//binary 
	{
		mid= (l+r) /2;
		if (judge (mid))
		{
			ans=mid;
			r=mid-1;
		}
		else
		{
			l=mid+1;
		}
	}
	printf ("%d\n", ans);
	return 0;