HDU2069 Coin Change(暴力)

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18240    Accepted Submission(s): 6279

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

1126

 

Sample Output

413

 

Author

Lily

 

Source

浙江工业大学网络选拔赛  

http://acm.split.hdu.edu.cn/showproblem.php?pid=2069

此题最坑的地方是没讲换成的硬币数量不能超过100

原题  https://www.bnuoj.com/v3/problem_show.php?pid=17813 是没要求100的  数据量也不一样 方法也不一样

给你这么多钱让你换 有五种(50,25,10,5,1)有多少种方法 1 5 和 5 1算一种

这题有太多的方法了 母函数(解决排列组合问题)  dp背包  记忆化搜索  

我这里要讲的是暴力  给的是250 明显让你暴力(优化一下)

还有的仁兄直接暴力打表  太可怕了 用搜索打表 

#include<bits/stdc++.h>using namespace std;int  main(){    int n,i,j,k,u,sum;    while(scanf("%d",&n)!=EOF)    {        sum=0;        for(i=0; i<=n/50; i++)        {            int g1=n-i*50;            for(j=0; j<=g1/25; j++)            {                int  g2=g1-j*25;                for(k=0; k<=g2/10; k++)                {                    int g3=g2-k*10;                    for(u=0; u<=g3/5; u++)                    {                        int g4=g3-u*5;                        if(i+j+k+u+g4<=100)                            sum++;                    }                }            }        }        printf("%d\n",sum);    }    return 0;}