hdu 3792 Twin Prime Conjecture n之内的孪生素数个数

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1425    Accepted Submission(s): 432

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

1520-2

 

Sample Output

014

 

Source

浙大计算机研究生保研复试上机考试-2011年

#include<stdio.h>int a[100100];int c[100100];void init(){    for(int i=2; i<=100100; i++)//打印素数表    {        if(a[i]==0)        {            for(int j=i+i; j<=100100; j+=i)                a[j]=1;        }    }    for(int i=5;i<=100100;i++)//    {        if(a[i]==0&&a[i-2]==0)//没有标记的就是素数 孪生素数的定义是相差是2 所以如果距离为2的没有标记就成立 +1        c[i]+=c[i-2]+1;        else        c[i]=c[i-1];    }}int main(){    int n;    init();    while(~scanf("%d",&n))    {        if(n<0) break;        printf("%d\n",c[n]);    }    return 0;}