HDU1405（质因数分解）

题目：HDU - 1405 

Tomorrow is contest day, Are you all ready? 
We have been training for 45 days, and all guys must be tired.But , you are so lucky comparing with many excellent boys who have no chance to attend the Province-Final. 

Now, your task is relaxing yourself and making the last practice. I guess that at least there are 2 problems which are easier than this problem. 
what does this problem describe? 
Give you a positive integer, please split it to some prime numbers, and you can got it through sample input and sample output. 

Input

Input file contains multiple test case, each case consists of a positive integer n(1

Output

For each test case you should output its factor as sample output (prime factor must come forth ascending ), there is a blank line between outputs.

Sample Input

60
12
-1

Sample Output

Case 1.
2 2 3 1 5 1

Case 2.
2 2 3 1

        
  

Hint

60=2^2*3^1*5^1

        
 

输出真的太坑爹了，一直PE

#include 
#include 
#include 
#include 

using namespace std;

const int maxn = 10005;
int prime[maxn];
void getprime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2;i <= maxn;i ++)
    {
        if(! prime[i]) prime[++ prime[0]] = i;
        for(int j = 1;j <= prime[0] && prime[j] <= maxn / i;j ++)
        {
            prime[prime[j] * i] = 1;
            if(i % prime[j] == 0) break;
        }
    }
}

int factor[100][2]; //factor[i][0]存底数，factor[i][1]存指数
int cnt;
void getFactors(int n) {
    cnt = 0;
    for(int i = 2; i <= n / 2; i++) {
        if(n % i == 0) {
            factor[cnt][0] = i;
            factor[cnt][1] = 0;
            while(n % i == 0) {
                factor[cnt][1]++;
                n /= i;
            }
            cnt++;
        }
    }
    if(n != 1) {
        factor[cnt][0] = n;
        factor[cnt][1] = 1;
        cnt++;
    }
}

int main()
{
    int n;
    int Case = 0;
    while(cin >> n && n > 0)
    {
        if(Case != 0) printf("\n");
        printf("Case %d.\n",++ Case);
        getFactors(n);
        for(int i = 0;i < cnt;i ++)
        {
            printf("%d %d ",factor[i][0],factor[i][1]);
        }
        printf("\n");
    }
    return 0;
}