HDU 3792 Twin Prime Conjecture（素数筛法 + 树状数组）

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本文介绍了一种求解指定范围内的孪生素数数量的方法，通过素数筛法标记素数，并利用树状数组优化区间求和过程。

Twin Prime Conjecture

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

Sample Output

0
1
4

【思路分析】
求n的范围内孪生素数的个数。所谓的孪生素数就是相差为2的一对素数。
首先可以根据素数筛法用visited数组标记数字i是否为素数，并且可以判断数字i和数字i - 2是否均为素数，如果是，就将数组num[i]标记为1，最后再用一个数组记录num的每一段区间的和即可。求区间和的时候用树状数组优化了一下。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int n;
int num[maxn],c[maxn];
int ans[maxn];
int visited[maxn];//标记素数
int lowbit(int x)
{
    return x & (-x);
}
int sum(int i)
{
    int ans = 0;
    while(i > 0)
    {
        ans += c[i];
        i -= lowbit(i);
    }
    return ans;
}
void add(int i,int w)
{
    while(i <= maxn)
    {
        c[i] += w;
        i += lowbit(i);
    }
}
void init()
{
    memset(visited,0,sizeof(visited));
    memset(num,0,sizeof(num));
    memset(c,0,sizeof(c));
    visited[0] = visited[1] = 1;//即0和1
    for(int i = 2;i < maxn;i++)//i即表示数字i
    {
        if(!visited[i])
        {
            if(!visited[i - 2])
            {
                num[i] = 1;//标记i和i-2是否为孪生素数
                add(i,1);
            }
        }
        for(int j = i << 1;j < maxn;j += i)//筛素数
        {
            visited[j] = 1;
        }
    }
    /*
    ans[0] = 0;
    for(int i = 1;i < maxn;i++)//区间求和
    {
        ans[i] = ans[i - 1] + num[i];
    }
    */

}

int main()
{
    init();
    while(scanf("%d",&n) && n >= 0)
    {
        printf("%d\n",sum(n));
    }
    return 0;
}