UVA674 Coin Change

最新推荐文章于 2020-03-18 15:48:04 发布

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最新推荐文章于 2020-03-18 15:48:04 发布

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分类专栏：动态规划

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本文介绍了一种使用动态规划解决硬币找零问题的方法。针对五种不同面额的硬币，通过编程实现求解任意金额下的找零组合数量。采用完全背包思想，通过状态转移方程递推得出所有可能的找零方案。

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Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to makechanges 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-centcoin, 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 thatthere 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 ofmoney 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 moneyin cents.
Output
For each input line, output a line containing the number of different ways of making changes with theabove 5 types of coins.
Sample Input
11
26
Sample Output
4
13

题意：给定五种硬币，1，5，10，25，50；输入钱币面值，由这五种硬币组合而成，问有多少种组合；五种硬币无限次取，完全背包，
状态方程：dp[i][j] +=dp[i-1][j-coin[i]] 前j元能由i枚硬币组成的方案数

//货币组成DP

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>

using namespace std;

int coin[5]={1,5,10,25,50};
int dp[8000];

int main()
{
    int n;

    while(cin >> n){
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<5;i++){
            for(int j=coin[i];j<=8000;j++){
                dp[j] +=dp[j-coin[i]];
            }
        }
        cout << dp[n]<< endl;
    }
    return 0;
}