If we define dn as: dn = pn+1-pn, where pi is the i-th prime. It is easy to see that d1 = 1 and dn=even for n>1. Twin Prime Conjecture states that "There are infinite consecutive primes differing by 2".
Now given any positive integer N (< 10^5), you are supposed to count the number of twin primes which are no greater than N.

Input

Your program must read test cases from standard input.
The input file consists of several test cases. Each case occupies a line which contains one integer N. The input is finished by a negative N.

Output

For each test case, your program must output to standard output. Print in one line the number of twin primes which are no greater than N.

Sample Input
1
5
20
-2
Sample Output
0
1
4

思路一：不采用前缀和，时间超时！暴力查找不可；

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
using namespace std;
const int maxa=1e5+10;
bool a[maxa];
int m;
bool prime(int n){
	for(int i=2;i<=sqrt(n);i++)
		if(n%i==0)	return false;
	return true;
} 
int main(){
	memset(a,false,sizeof a);
	for(int i=2;i<=maxa;i++)
		if(prime(i))	a[i]=!a[i];
	while(scanf("%d",&m)!=EOF&&m>=0){
		int ans=0;
		for(int i=5;i<=m;i++)
			if(a[i]&&a[i-2])
				ans++;
		cout<<ans<<endl;
	}
}

思路二：1.素数筛法标记出所有素数；

2.一个素数i,判断i-2是否是素数,是的话就把a[i]赋值为1;

3.最后求出所有前缀和;

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
using namespace std;
const int maxa=1e5+10;
bool a[maxa];
int ans[maxa],ANS[maxa];
int m;
bool prime(int n){
	for(int i=2;i<=sqrt(n);i++)
		if(n%i==0)	return false;
	return true;
} 
int main(){
	memset(a,false,sizeof a);
	memset(ans,0,sizeof ans);
	for(int i=2;i<=maxa;i++)
		if(prime(i)){
			a[i]=!a[i];
			if(a[i-2])	ans[i]=1; 
	} 
	ANS[0]=ans[0];
	for(int i=1;i<=maxa;i++)
		ANS[i]=ANS[i-1]+ans[i];//统计前缀和 
	while(scanf("%d",&m)!=EOF&&m>=0){
		printf("%d\n",ANS[m]);
	}
}