【解题报告】uva674_Coin Change（硬币找零, dp, 完全背包）

674 - Coin Change

Time limit: 3.000 seconds

  Coin Change 

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 

1126

Sample Output 

413

题目大意：

有5种面值的硬币，面值分别为1、5、10、25、50。给定所需找零的金额数，求用上述硬币找零，共有多少种硬币的组合方式（不考虑硬币顺序，如[1,5]和[5,1]为一种组合方式）。

解题思路：

完全背包问题。定义状态d(i,j)，代表用前i种硬币组合出金额i时的组合方式。

状态转移方程：d(i,j) = ∑{ dp(i-1,j-c[i]*k) | 1<=k<=j/c[i], c[i]∈{1,5,10,25,50} }

由于不考虑硬币顺序，计算时先枚举硬币种类c[j]，其次枚举所有找零金额状态，确保每种硬币不会被重复计算。             

#include <cstdio>int c[5]={1,5,10,25,50};int d[7490];int main(){    d[0]=1;    for(int j=0;j<5;++j){        for(int i=c[j];i<=7489;++i){            d[i]+=d[i-c[j]];        }    }    int n;    while(~scanf("%d",&n)) printf("%d\n",d[n]);    return 0;}