2011年浙大:Twin Prime Conjecture

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2659    Accepted Submission(s): 906

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

来源:http://acm.hdu.edu.cn/showproblem.php?pid=3792

求区间素数,注意打表

#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=100005;
bool isPrime[MAXN];
int cnt[MAXN];
int main()
{
    int n;
    memset(isPrime,true,sizeof(isPrime));
    for(int i=2;i<MAXN;i++)
    {
        if(isPrime[i])
        {    
            for(int j=i+i;j<MAXN;j+=i)
                isPrime[j]=false;
        }
    }
    cnt[5]=1;
    int pre=5;
    int num=1;
    for(int i=6;i<MAXN;i++)
    {
        if(isPrime[i])
        {
            if(i-pre==2)
            {
                num++;
            }
            pre=i;
        }
        cnt[i]=num;
    }
    while(scanf("%d",&n)!=EOF&&n>=0)
    {    
        printf("%d
",cnt[n]);
    }
    return 0;
}