HDU 1405 The Last Practice - 解题心得

原题:

Description

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<n<65536), one per line. a negative terminates the input, and it should not to be processed.

 

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 

 

 

 

分析：

这题是裸的质因数分解，有小伙伴觉得自己结果对，可是就是wrong answer   那就是这题的格式，最后一个数后也有空格。

 

 

代码：

#include <stdio.h>
#include <string.h>
#include<iostream>
using namespace std;

int prime[1000000];

void sushu(void)
{
    fill(prime, prime + 1000000, 1);
    int i, j;
    for (i = 2; i <= 500000; i++)
    {
        if (prime[i])
        {
            for (j = i + i; j<1000000; j += i)
            {
                prime[j] = 0;
            }
        }
    }
}

int main()
{
    int n, cas = 1, k = 0;
    sushu();
    while (~scanf("%d", &n) && n >= 0)
    {
        int cnt = 0, i;
        if (k)
            printf("\n");
        printf("Case %d.\n", cas++);
        k++;
        if (prime[n])
        {
            printf("%d 1 \n", n);
            continue;
        }
        else
        {
            for (i = 2; i <= n; i += 1)
            {
                cnt = 0;
                if (prime[i] && n%i == 0)
                {
                    while (n%i == 0)
                    {
                        n /= i;
                        cnt++;
                    }
                    if (cnt)
                    {
                        printf("%d %d ", i, cnt);
                    }
                }
            }
            printf("\n");
        }
    }

    return 0;
}