Bestcoder round 18---A题（素数筛+素数打表+找三个素数其和==n）

Primes Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12    Accepted Submission(s): 11

Problem Description

Given a number n, please count how many tuple(p1, p2, p3) satisfied that p1<=p2<=p3, p1,p2,p3 are primes and p1 + p2 + p3 = n.

 

Input

Multiple test cases(less than 100), for each test case, the only line indicates the positive integer n(n≤10000).

 

Output

For each test case, print the number of ways.

 

Sample Input

3 9

 

Sample Output

0 2

 

Accepted 的代码：

 

#include <string>
#include <iostream>
#include <cstdio>
#include <math.h>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

int f[10001];

void sushu()
{
    int i, j;
    memset(f, 0, sizeof(f));
    f[1]=1;
    i=2;
    while(i<=200)
    {
        for(j=i*2; j<=10000; j+=i)
        {
            f[j]=1;
        }
        i++;
        while(f[i]==1)
        {
            i++;
        }
    }
}

int s[10000], e;

int main()
{
    int n;
    int i, j, k;
    int cnt;
    sushu();
     e=0;
    for(i=2; i<=10000; i++)
    {
        if(f[i]==0)
        {
            s[e++]=i;
        }
    }
    while(scanf("%d", &n)!=EOF)
    {
        if(n<6)
        {
            cout<<'0'<<endl;
            continue;
        }
        cnt=0;
        int flag=0;
        for(i=0; i<=n; i++)
        {
            if(s[i]>=n)
              break;
            for(j=i; j<=n; j++)
            {
                if( (s[i]+s[j])>=n )
                {
                    flag=1;
                    break;
                }
                else
                {
                    int dd=n-s[i]-s[j];
                    if(f[dd]==0 && dd>=s[i] && dd>=s[j] )
                    {
                        cnt++;
                    }

                }
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}

 

 

这个代码是超时的（3层循环太 耗时）：

#include <string>
#include <iostream>
#include <cstdio>
#include <math.h>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

int f[10001];

void sushu()
{
    int i, j;
    memset(f, 0, sizeof(f));
    f[1]=1;
    i=2;
    while(i<=200)
    {
        for(j=i*2; j<=10000; j+=i)
        {
            f[j]=1;
        }
        i++;
        while(f[i]==1)
        {
            i++;
        }
    }
}

int s[10000], e;

int main()
{
    int n;
    int i, j, k;
    int cnt;
    sushu();
     e=0;
    for(i=2; i<=10000; i++)
    {
        if(f[i]==0)
        {
            s[e++]=i;
        }
    }
    while(scanf("%d", &n)!=EOF)
    {
        if(n<6)
        {
            cout<<'0'<<endl;
            continue;
        }
        cnt=0;
        for(i=0; i<=n; i++)
        {

            for(j=i; j<=n; j++)
            {
                for(k=j; k<=n; k++)
                {
                    if((s[i]+s[j]+s[k])==n)
                    {
                        cnt++;
                    }
                }
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}