HDU3792---Twin Prime Conjecture（树状数组）

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2950    Accepted Submission(s): 1016

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

 

题目大意：

题目的意思是，给定一个数字n。求小于n的孪生素数有多少对。孪生素数的概念就是说，相邻的两个素数之间差值为2。例如：5和7，17和19，这些都是。

解题思路：

先用筛法打了一个素数表，在素数表的基础上打出来一个直接保存结果的数组c[]（树状数组直接处理的数组）。这样就可以直接用树状数组来解决了。基础的BIT应用。

如果不懂树状数组的话，可以去看这篇博客：搞懂树状数组

源代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<deque>
#include<map>
#include<set>
#include<algorithm>
#include<string>
#include<iomanip>
#include<cstdlib>
#include<cmath>
#include<sstream>
#include<ctime>
using namespace std;

typedef long long ll;
#define MAXN 100005
int c[MAXN];//c[]从下标为1的地方开始

int lowbit(int x)
{
    return x&(-x);
}

void add(int x, int num)//x代表位置，num代表增量
{
    while(x<=MAXN)
    {
        c[x]+=num;
        x+=lowbit(x);
    }
}

int sum(int x)//代表求解1~x的和（x是下标）
{
    int res = 0;
    while(x>0)
    {
        res+=c[x];
        x-=lowbit(x);
    }
    return res;
}

int primer[MAXN];
int number[MAXN];

void init()
{
    memset(number,0,sizeof(number));
    int i,j;
    int cnt=0;
    for(i=2;i<=100000;i++)
    {
        if(number[i]==0)
        {
            cnt++;
            primer[cnt]=i;
            for(j=i*2;j<=100000;j+=i)
            {
                number[j]=1;
            }
        }
    }//素数表OK
    for(i=2;i<=cnt;i++)
    {
        if(primer[i]-primer[i-1]==2)
        {
            add(primer[i],1);//发现一对增加一个
        }
    }
}
int main()
{
    int n;
    init();
    while(scanf("%d",&n)!=EOF&&n>0)
    {
        printf("%d\n",sum(n));
    }
	return 0;
}