hdu 2069 Coin Change（完全背包）

最新推荐文章于 2020-03-17 01:18:21 发布

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最新推荐文章于 2020-03-17 01:18:21 发布

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文章标签： java

原文链接：http://www.cnblogs.com/pshw/p/5171223.html

本文解决了一个涉及使用五种不同面额硬币（50分、25分、10分、5分、1分）计算特定金额不同组合数量的问题。通过使用完全背包算法，文章详细解释了如何计算最多100个硬币组成的任意金额的不同方式。

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Coin Change

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16592 Accepted Submission(s): 5656

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

Sample Output

Author

Lily

Source

浙江工业大学网络选拔赛

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第一次写完全背包的题，这题很简单，不过要注意硬币数目不能大于100

题意：给你面值有1，5，10，25，50的币种，然后随意输入一个钱的数目，问用这些面值刚好凑成这个钱的方法有多少个（最多100个硬币）

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int dp[101][7500];
 6 int main()
 7 {
 8     int coin[5]= {1,5,10,25,50};
 9     int i,j,n,m,k;
10     memset(dp,0,sizeof(dp));
11     dp[0][0]=1;
12     for(i=0; i<5; i++)   //用的硬币
13     {
14         for(k=1; k<=100; k++)  //硬币数目
15         {
16             for(j=coin[i]; j<=7500; j++)
17             {
18                 dp[k][j]+=dp[k-1][j-coin[i]];
19             }
20         }
21 
22     }
23     while(~scanf("%d",&n))
24     {
25         int ans=0;
26         for(i=0; i<=100; i++)
27             ans+=dp[i][n];
28         printf("%d\n",ans);
29     }
30     return 0;
31 }