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最新推荐文章于 2020-10-29 11:24:19 发布

转载最新推荐文章于 2020-10-29 11:24:19 发布 · 216 阅读

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原文链接：http://www.cnblogs.com/naturepengchen/articles/3877701.html

本文介绍了一个关于新西兰货币的组合计数问题，并提供了一个使用完全背包算法解决该问题的C++程序示例。该程序可以计算出任意金额下，新西兰货币的所有可能组合数量。

2014-07-30 12:05:23

Dollars

New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1 20c, 2 10c, 10c+2 5c, and 4 5c.

Input

Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).

Output

Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.

Sample input

0.20
2.00
0.00

Sample output

  0.20                4
  2.00              293

思路：经典的完全背包。（1）定义币值 （2）dp【0】 = 1 （给好种子） （3）从前往后推的dp

 1 /*************************************************************************
 2 
 3     > File Name: j.cpp
 4     > Author: Nature
 5     > Mail: 564374850@qq.com 
 6     > Created Time: Tue 29 Jul 2014 07:49:33 PM CST
 7 ************************************************************************/
 8 
 9 #include <cstdio>
10 #include <cstring>
11 #include <cstdlib>
12 #include <cmath>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 #define ll long long
17 
18 int main(){
19     ll v[11] = {10000,5000,2000,1000,500,200,100,50,20,10,5};
20     ll dp[30005] = {1};
21     for(int i = 0; i < 11; ++i)
22         for(int j = 0; j + v[i] <= 30000; ++j)
23             dp[j + v[i]] += dp[j];
24     int n,m;
25     while(scanf("%d.%d",&n,&m) == 2 && (n || m)){
26         printf("%3d.%02d%17lld\n",n,m,dp[n * 100 + m]);
27     }
28     return 0;
29 }

转载于:https://www.cnblogs.com/naturepengchen/articles/3877701.html