zoukankan      html  css  js  c++  java
  • HDUOJ 2069

    Coin Change

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 7167    Accepted Submission(s): 2385


    Problem Description
    Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.

    For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.

    Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.
     

    Input
    The input file contains any number of lines, each one consisting of a number ( ≤250 ) for the amount of money in cents.
     

    Output
    For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.
     

    Sample Input
    11 26
     

    Sample Output
    4 13
     



    #include <iostream>
    using namespace std;
    int main()
    {
    	int a,b,c,d,m,count;
    	while(scanf("%d",&m)!=EOF)
    	{
    		if(m==0) count=1;
    		else
    		{
    			count=0;
    			for(a=0;a<=m;a+=50)
    				for(b=0;a+b<=m;b+=25)
    					for(c=0;a+b+c<=m;c+=10)
    						for(d=0;a+b+c+d<=m;d+=5)
    						{
    							if(m-a-b-c-d<=100-a/50-b/25-c/10-d/5) 
    							{
    								count++;
    							}
    						}
    		}
    		printf("%d\n",count);
    	}
    	return 0;
    }


  • 相关阅读:
    域账户-配置文件
    创建任务计划
    查看系统和PowerShell版本
    查找数组中元素的索引位置
    更改计算机名称,修改密码
    生成GUID
    微服务架构 vs. SOA架构
    springMvc架构简介
    Spring Boot和Spring cloud
    Spring链接汇总
  • 原文地址:https://www.cnblogs.com/cszlg/p/2910543.html
Copyright © 2011-2022 走看看