uva 357 - Let Me Count The Ways

最新推荐文章于 2018-04-23 19:12:56 发布

滑头鬼之亨

最新推荐文章于 2018-04-23 19:12:56 发布

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分类专栏：动态规划文章标签： combinations input output numbers each file

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本文介绍了一个有趣的问题：如何计算出给定金额下不同美国硬币组合的数量。通过一个具体的例子，详细展示了使用动态规划方法解决该问题的过程，并提供了一段C语言实现的代码。

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Let Me Count The Ways

After making a purchase at a large department store, Mel's change was 17 cents. He received 1 dime, 1 nickel, and 2 pennies. Later that day, he was shopping at a convenience store. Again his change was 17 cents. This time he received 2 nickels and 7 pennies. He began to wonder ' "How many stores can I shop in and receive 17 cents change in a different configuration of coins? After a suitable mental struggle, he decided the answer was 6. He then challenged you to consider the general problem.

Write a program which will determine the number of different combinations of US coins (penny: 1c, nickel: 5c, dime: 10c, quarter: 25c, half-dollar: 50c) which may be used to produce a given amount of money.

Input

The input will consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.

Output

The output will consist of the appropriate statement from the selection below on a single line in the output file for each input value. The number m is the number your program computes, n is the input value.

There are m ways to produce n cents change.

There is only 1 way to produce n cents change.

Sample input

17 
11
4

Sample output

There are 6 ways to produce 17 cents change. 
There are 4 ways to produce 11 cents change. 
There is only 1 way to produce 4 cents change.

以加入不同的面值来划分决策阶段，则种类数目等于之前的加上，加入此面之后增加的数目；

加入面值b[I]; 种类数a[I][J]=a[I][J-1]+a[I-b[I]][J];
  0 1 2 3 4 5 6 7 8 9 10  15 20 25
0    
1   1 1 1 1 1 1 1 1 1  1 
5   1 1 1 1 2 2 2 2 2  3  4  5  6
10                      
25
50

#include<stdio.h>
#include<string.h>
#define max 30000
long long i,j,n,a[max+1][6],b[6]={0,1,5,10,25,50};
int main()
{
 memset(a,0,sizeof(a));
 for (i=1;i<6;i++)
 {
  ++a[b[i]][i-1];//加入面值b[I];
  for (j=1;j<=max;j++)
  {
   if (j-b[i]>=0) a[j][i]=a[j][i-1]+a[j-b[i]][i];
            else a[j][i]=a[j][i-1];
  }
 }
 while (scanf("%lld",&n)!=EOF)
 if (a[n][5]==1||n==0) printf("There is only 1 way to produce %lld cents change.\n",n);//面值为0也要输出1种Y-Y
        else     printf("There are %lld ways to produce %lld cents change.\n",a[n][5],n);
 return 0;
}