hdu2069 母函数应用变形之总硬币数有限制

最新推荐文章于 2021-07-29 11:48:10 发布

最新推荐文章于 2021-07-29 11:48:10 发布 · 184 阅读

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#java

本文介绍了一种解决硬币找零问题的算法实现，通过动态规划的方法计算出给定金额下不同硬币组合的数量。该算法能够处理多达100枚硬币的情况，并提供了完整的C语言代码示例。

Coin Change

Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7813Accepted Submission(s): 2592

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

题意：有50 25 10 5 1 的硬币一共最多有100枚输入n输出有多少种表示方法

#include<stdio.h>
#include<string.h>
int c1[251][101],c2[251][101];//次数顶多为250 前一个括号是指系数后一个指硬币个数
int main()
{
int i,j,k,n,l,sum;
int val[6]={0,1,5,10,25,50};
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
for(i=0;i<=100;i++)
{
c1[i][i]=1;
}
n=250;
for(i=2;i<=5;i++)
{
for(j=0;j<=n;j++)
for(k=0;k+j<=n;k+=val[i])//次方最大为k+j所以要控制它小于n
for(l=0;l+k/val[i]<=100;l++)
c2[k+j][l+k/val[i]]+=c1[j][l];

for(k=0;k<=n;k++)
for(l=0;l<=100;l++)
{
c1[k][l]=c2[k][l];
c2[k][l]=0;
}
}
while(scanf("%d",&n)!=EOF)
{
sum=0;
for(i=0;i<=100;i++)
sum+=c1[n][i];
printf("%d\n",sum);
}
// }
return 0;
}

此题给了一个思路就是如何去求一个n拆分成各种硬币一共用了多少个硬币