Uva 674 Coin Change

完全背包问题解决方案数

最新推荐文章于 2020-05-03 15:31:32 发布

原创最新推荐文章于 2020-05-03 15:31:32 发布 · 734 阅读

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本文介绍了一种使用完全背包算法解决硬币兑换问题的方法，包括状态转移方程的使用和代码实现，详细解释了如何计算给定金额下不同兑换方案的数量。

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, two 5-cent coins and one 1-cent coin, 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 7489 cents.

Input

The input file contains any number of lines, each one consisting of a number 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

完全背包求解方案数问题

F[i]表示可达且可达的方案数！

状态转移方程：f[i]=f[i]+f[i-c[k]];

#include<stdio.h>
#include<string.h>

int f[8000];
int c[5]={1,5,10,25,50};

int main(void)
{
	int v,k,i;
	f[0]=1;
	for(k=0;k<5;k++)
		for(i=1;i<=7489;i++)
		{
			if(i>=c[k])								//这里不能忽略，不然指针访问出问题
				f[i]+=f[i-c[k]];
					
		}
	while(scanf("%d",&v)==1)
	{
		printf("%d\n",f[v]);
	}
	return 0;
}