HDOJ 2069 Coin Change(母函数)

最新推荐文章于 2021-07-28 19:00:42 发布

转载最新推荐文章于 2021-07-28 19:00:42 发布 · 68 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/CKboss/p/3165222.html

文章标签：

#java

本文介绍了一个经典的动态规划问题——硬币找零问题。该问题的目标是计算出使用特定面额的硬币组合来构成给定金额的不同方式的数量。通过一段C++代码示例，展示了如何解决这个问题，并提供了完整的程序实现。

Coin Change

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10289 Accepted Submission(s): 3451

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

Author

Lily

Source

浙江工业大学网络选拔赛

Recommend

linle

 1 //加一维限制数量。。。。。。
 2 
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <cstring>
 6 
 7 using namespace std;
 8 
 9 const int value[6]={0,1,5,10,25,50};
10 
11 int c1[300][110],c2[300][110];
12 int n;
13 
14 int main()
15 {
16 while(scanf("%d",&n)!=EOF)
17 {
18    memset(c1,0,sizeof(c1));
19    memset(c2,0,sizeof(c2));
20 
21    for(int i=0;i<=min(n,100);i++)
22    {
23         c1[i][i]=1;
24    }
25 
26    for(int i=2;i<=5;i++)
27    {
28        for(int j=0;j<=n;j++)
29        {
30            for(int k=0;k+j<=n;k+=value[i])
31            {
32                for(int l=0;l<=100&&l+k/value[i]<=100;l++)
33                {
34                    c2[j+k][l+k/value[i]]+=c1[j][l];
35                }
36            }
37        }
38 
39        for(int j=0;j<=n;j++)
40        {
41            for(int k=0;k<=100;k++)
42            {
43                c1[j][k]=c2[j][k];
44                c2[j][k]=0;
45            }
46        }
47    }
48 
49    int ans=0;
50    for(int j=0;j<=100;j++)
51    {
52        ans+=c1[n][j];
53    }
54 
55    printf("%d\n",ans);
56 
57 }
58     return 0;
59 }