筛素数——HDU 3792 Twin Prime Conjecture

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2653    Accepted Submission(s): 900

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

 

题意：输入一个数N（N<10⁵)，问这个数以内有多少对孪生素数（两个素数相差2）。

题解：筛出所需的素数，然后用一个数组来存到第i个素数时有多少对孪生素数。最后找到不大于N的最大素数，只要输出到这个素数有多少对孪生素数。

/*HDU 3792*/
/*筛素数*/
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=1e5+100;
bool isprime[maxn];
int prime[maxn];
int cnt=0;
int a[maxn];
void getprime()
{
    memset(isprime,1,sizeof(isprime));
    memset(prime,0,sizeof(prime));
    isprime[0]=isprime[1]=0;
    for(int i=2;i<maxn;i++)
    {
        if(isprime[i])
            for(int j=2*i;j<maxn;j+=i)
                isprime[j]=false;
    }

    for(int i=0;i<maxn;i++)
    {
        if(isprime[i])
            prime[cnt++]=i;
    }
    memset(a,0,sizeof(a));
    int t=0;
    for(int i=1;i<cnt;i++)
    {
        if((prime[i]-prime[i-1])==2)
        {
            a[i]=++t;/*当找到一对孪生素数时，在后一个+1*/
            //printf("%d %d
",prime[i],prime[i-1]);/**/
        }
        else
            a[i]=t;
    }
}

int find(int n,int l,int r)/*开始想用二分查找，可是不知道错哪了*/
{

    int mid=(l+r)/2;
    if(prime[mid]>n)
        return find(n,l,mid);
    else if(prime[mid]==n)
        return a[mid];
    else 
        return find(n,mid+1,r);
}
int main()
{
    int n;
    getprime();
    while(scanf("%d",&n)!=EOF)
    {
        if(n<0)    break;    
        int ans=0;
        int i;
        for(i=0;i<cnt-1;i++)
            if(prime[i]<=n&&prime[i+1]>n)
                break;
        ans=a[i];
        if(n==1)    ans=0;
        printf("%d
",ans);
    }
    return 0;
}