HDU_3792_(素数筛+树状数组)

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3280    Accepted Submission(s): 1162

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：输入n，求小于n的孪生素数对有多少个。

 

素数筛+树状数组

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define N 100005

int s[N];
int prime[N],cntp;
bool isNot[N];

void getPrime()
{
    for(int i=2;i<=N;i++)
    {
        if(isNot[i]==0)
        {
            prime[cntp++]=i;
            for(int j=2;i*j<=N;j++)
                isNot[i*j]=1;
        }
    }
}
int lowbit(int x)
{
    return x&(-x);
}

void add(int x,int num)
{
    while(x<=N)
    {
        s[x]+=num;
        x+=lowbit(x);
    }
}

int sum(int x)
{
    int res=0;
    while(x>0)
    {
        res+=s[x];
        x-=lowbit(x);
    }
    return res;
}

int main()
{
    cntp=1;
    getPrime();
    for(int i=2;i<cntp;i++)
        if(prime[i]-prime[i-1]==2)
            add(prime[i],1);
    int n;
    while(scanf("%d",&n)!=EOF&&n>0)
    {
        printf("%d
",sum(n));
    }
    return 0;
}