hdu 3792-浙大计算机研究生保研复试上机考试-2011年

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 974    Accepted Submission(s): 286

Problem Description

If we define dn as: dn = p(n+1)-p(n), 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
   
   
   
   

 

Source

浙大计算机研究生保研复试上机考试-2011年

 

关键在求prime时的效率，类似找朋友那道题的做法，把x的倍数标记为非素数即可。

#include "iostream"
#include "stdio.h"
#include "math.h"
#include "vector"
#include "queue"
#include "memory.h"
#include "algorithm"
#include "string"
using namespace std;
#define N 100000
bool prime[N+10];
int twinp[N+10];

int main()
{
    memset(prime,true,sizeof(prime));
    prime[0]=prime[1]=false;
    int i,j,n;
    //initial prime array
    for(i=2;i<=(int)sqrt(1.0*N);i++)
    {
        for(j=2;i*j<=N;j++)
            prime[i*j]=false;
    }
    //calculate twin prime conjecture array
    twinp[0]=twinp[1]=twinp[2]=twinp[3]=twinp[4]=0;
    
    for(i=5;i<=N;i++)
        if(prime[i]&&prime[i-2])
            twinp[i]=twinp[i-1]+1;
        else twinp[i]=twinp[i-1];
    
    while(cin>>n&&n>=0)
        printf("%d\n",twinp[n]);
}