zoukankan      html  css  js  c++  java
  • 数学概念——J

    J - 数论,质因数分解
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
    Submit Status

    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 

    解题思路:
    这个题目很简单,但是要注意一个陷井,就是输入一个数的时候,它的结束条件是负数,一定要注意
    程序代码:
    #include <cstdio>
    #include <cstring>
    using namespace std;
    int b[66550];
    int n;
    void p( )
    {
        int m=n;
        while(1)
        {
            for(int i=2;i<=n;i++ )
            if(m%i==0)
            {
                b[i]++;
                m/=i;
                break;
            }
            if(m==1)
                break;
        }
    
    }
    int main()
    {
        int Case=0;
        while(scanf("%d",&n)==1&&n>0)
        {
            if(Case>0) printf("
    ");
    
            memset(b,0,sizeof(b));
            p();
            printf("Case %d.
    ",++Case);
            for(int i=2;i<=n;i++)
            if(b[i]!=0)
                printf("%d %d ",i,b[i]);
            printf("
    ");
        }
    }
    View Code


    版权声明:此代码归属博主, 请务侵权!
  • 相关阅读:
    JAVA编程规则【转自java编程思想】
    诊断 Java 代码: 轻松掌握 Java 泛型
    Linux开启telnet远程登录服务全攻略
    TCP详解
    UNIX环境高级编程文件描述符浅析
    DHCP与BOOTP有什么区别
    Linux 多播(组播)例程
    你所不知道的传输层
    虚电路方式,数据报方式
    java foreach 使用
  • 原文地址:https://www.cnblogs.com/www-cnxcy-com/p/4750476.html
Copyright © 2011-2022 走看看