HDOJ-2069Coin Change（母函数加强）

最新推荐文章于 2017-10-11 19:28:47 发布

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最新推荐文章于 2017-10-11 19:28:47 发布

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文章标签： java

本文介绍了一个解决硬币兑换问题的算法实现，通过多种硬币组合凑出特定金额的方法，并限制了硬币总数不超过100个。算法通过动态规划方法计算不同组合方式的数量。

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Coin Change

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9939 Accepted Submission(s): 3343

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

题解：

题目出来不仅要凑出输入的钱数，而且所用硬币的数量之和不能大于100，

c1[i][j]:
表示 i分钱由 j个硬币组成的方案数, 然后在循环中增加一个循环:用于保存在当前硬币数量 k上增加
1...n 个硬币时(k+n <= 100)方案数。

 1 #include <cstdio>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 int sum[251];
 7 int c1[251][101], c2[251][101];
 8 int val[6] = {0, 1, 5, 10, 25, 50};
 9 
10 void init()
11 {
12     c1[0][0] = 1;
13     for(int i = 1; i <= 5; ++i)
14     {
15         for(int j = 0; j <= 250; ++j)
16         {
17             for(int k = 0; k*val[i]+j <= 250; ++k)
18                 for(int p = 0; p+k <= 100; ++p)   // 增加的循环 
19                     c2[j+k*val[i]][p+k] += c1[j][p];
20         }
21         for(int j = 0; j <= 250; ++j)
22         {
23             for(int k = 0; k <= 100; ++k)
24             {
25                 c1[j][k] = c2[j][k];
26                 c2[j][k] = 0;
27             }
28         }
29     }
30     for(int i = 0; i <= 250; ++i)
31         for(int k = 0; k <= 100; ++k)
32             sum[i] += c1[i][k];
33 }
34 
35 int main()
36 {
37     int tot;
38     init();
39     while(~scanf("%d", &tot))
40     {
41         printf("%d\n", sum[tot]);
42     }
43     return 0;
44 }