HDU1405 The Last Practice

质数分解练习题解析

最新推荐文章于 2020-03-31 20:30:01 发布

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本文介绍了一道关于质数分解的编程题，任务是将输入的正整数分解为若干质数的乘积，并输出这些质数及其出现次数。文章提供了一个C++实现示例，包括质数判断函数及主程序流程。

1405

The Last Practice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11018 Accepted Submission(s): 2495

Problem 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

#include<stdio.h>
#include<vector>
using namespace std;
int isPrime(int n)
{
    int i;
    for (i = 2; i*i <= n; i++)
        if (n%i == 0)
            return false;
    return true;
}
int main()
{
    bool flag = false;
    vector<int>Prime;
    vector<int>::iterator Iter;
    int n, i, count,cns=1;
    for (i = 2; i <= 65536; i++)
    {
        if (isPrime(i))
            Prime.push_back(i);
    }
    while (scanf("%d", &n)&&(n>=0)) 
    {
        if(flag)printf("\n");
        flag = true;
        printf("Case %d.\n", cns++);
        for (Iter = Prime.begin(); Iter != Prime.end(); Iter++)
        {
            count = 0;
            while (n%(*Iter)==0)
            {
                count++;
                n /= *Iter;
            }
            if(count)
            printf("%d %d ", *Iter, count);
            if (n < 2)
            {
                printf("\n");
                break;
            }           
        }       
    }
    return 0;
}