集训第六周 数学概念与方法 J题 数论,质因数分解

本文介绍了一种质因数分解算法的实现方法,并通过一个具体的C++代码示例展示了如何将一个正整数分解为若干质数的乘积。输入文件包含多个测试用例,每个用例是一个正整数,输出则是该整数的质因数分解结果。

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 

求质因数,使用循环求,数组记录就好
#include"iostream"
#include"cstring"
#include"cstdio"
using namespace std;
const int maxn=65536+10;
int vis[maxn];
int main()
{
    int n,ans,ca=1;
    while(cin>>n&&n>=0)
    {
     memset(vis,0,sizeof(vis));
     ans=n;
     int i=2;
     while(n!=1)
     {
         if(n%i==0)
         {
             vis[i]++;
             n/=i;
             i=2;
         }
         else
             i++;
     }
     if(ca!=1) cout<<endl;
     cout<<"Case "<<ca++<<"."<<endl;
     for(int j=0;j<=ans;j++)
     {
         if(vis[j]!=0) cout<<j<<" "<<vis[j]<<" ";
     }
     cout<<endl;
    }
    return 0;
}
View Code

 

 

转载于:https://www.cnblogs.com/zsyacm666666/p/4744614.html

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