UVA 674 Coin Change（dp）

UVA 674  Coin Change  解题报告

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/E

题目：

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

11
26

Sample Output

11
26

题目大意：

有5种硬币，分别是1，5,10,25,50。现有一定金额的钱由这5种硬币组成，求共有多少种组成方式。

分析：

完全背包，因为硬币数量不限。注意1+10和10+1是一样的，用递推的方式解，要将硬币从小到大排序。

要注意估算时间复杂度。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int maxn=7500;

int dp[maxn];//钱数为n时的最多组成方式
int coin[5]={1,5,10,25,50};//5种类型的硬币

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<5;i++)
            for(int j=0;j<=n;j++)
                dp[j+coin[i]]+=dp[j];//注意dp的转移顺序
        printf("%d\n",dp[n]);
    }
    return 0;
}