UVa 674 Coin Change (完全背包&最优解的方案数)

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本文介绍了一个经典的动态规划问题——硬币找零问题，通过编程实现寻找给定金额下不同找零方式的数量。使用C++语言进行实现，并考虑了最多7489美分的特殊情况。

674 - Coin Change

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=615

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.

11
26

Sample Output

4
13

完整代码：

/*0.042s*/

#include<cstdio>
#include<cstring>
const int maxn = 7490;
const int coin[5] = {1, 5, 10, 25, 50};

int dp[maxn];

int main(void)
{
	int n;
	memset(dp, 0, sizeof(dp));
	dp[0] = 1;
	for (int k = 0; k < 5; ++k)
		for (int i = coin[k]; i < maxn; ++i)
			dp[i] += dp[i - coin[k]]; ///方案数叠加起来
	while (~scanf("%d", &n))
		printf("%d\n", dp[n]);
	return 0;
}