题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3792点击打开链接
Twin Prime Conjecture
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3417 Accepted Submission(s): 1257
Problem Description
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
题意：让你计算0~n里面相差2的素数有几个 记忆化 树状数组 线性筛素数
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include<algorithm>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <queue>
#include <stack>
using namespace std;
int su[1111111];
int ans[111111];
int c[111111];
int lowbit(int x)
{return x&(-x);}
void add(int k,int num)
{
while(k<100000)
{
c[k]+=num;
k+=lowbit(k);
}
}
int read(int k)
{
int s=0;
while(k>0)
{
s+=c[k];
k-=lowbit(k);
}
return s;
}
int main()
{
for(int i=2;i<=100000;i++)
{
if(su[i]==0)
{
for(int j=i;j<=100000;j+=i)
{
if(j!=i)
su[j]=1;
}
}
}
su[0]=su[1]=1;
for(int i=1;i<=100000;i++)
{
if(su[i]==0&&su[i-2]==0)
add(i,1);
ans[i]=read(i);
}
int n=0;
while(scanf("%d",&n)&&n>=0)
printf("%d\n",ans[n]);
}
