zoukankan      html  css  js  c++  java
  • hdoj 1405 The Last Practice

    The Last Practice

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 9037    Accepted Submission(s): 1905


    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<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
     
    题意:输入一个n,分解出其质因子,并输出其中有多少个相同的质因子
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    #include<math.h>
    #define MAX 1100
    using namespace std;
    int a[MAX],b[MAX];
    int main()
    {
    	int k,n,m,j,i;
    	k=1;
    	while(scanf("%d",&n)&&n>=0)
    	{
    		if(k>1)
    		printf("
    ");
    		memset(a,0,sizeof(a));
    		memset(b,0,sizeof(b));
    		printf("Case %d.
    ",k++);
    		m=n;j=0;
    		for(i=2;i<=n;i++)
    		{
    			int sum=0;
    			if(m%i==0)
    			{
    				while(m%i==0)
    				{
    					m/=i;
    					sum++;
    				}
    				a[j]=i;
    				b[j++]=sum;
    			}
    			if(m==1)
    			    break;
    		}
    		for(i=0;i<j;i++)
    		    printf("%d %d ",a[i],b[i]);
    		printf("
    ");
    	}
    	return 0;
    } 
    

      

  • 相关阅读:
    unit3d 4.6 document open solution
    Unity3dBug
    linq to xml
    A const field of a reference type other than string can only be initialized with null Error [duplicate]
    Redis数据类型
    redis快照与AOF
    redis实现高并发下的抢购/秒杀功能
    xss攻击怎么防止
    四种常见的索引类型
    什么是sql 注入及如何预防 sql 注入
  • 原文地址:https://www.cnblogs.com/tonghao/p/4994342.html
Copyright © 2011-2022 走看看