标签:color family pack lib ++ size 硬币 问题 charge
找零问题:需找零金额为W,硬币面值有(d1, d2, d3,…,dm),最少需要多少枚硬币。
问题:需找零金额为8,硬币面值有(1, 3, 2, 5),最少需要多少枚硬币。
设F(j)表示总金额为j时最少的零钱数,F(0) = 0,W表示找零金额,有零钱一堆{d1, d2, d3,…,dm}。同样根据之前的经验,要达到为j,那么必然是j – di(1 <= i <= m)面值的硬币数再加1个di面值的硬币,当然j >= di,即F(j) = F(j - di) + 1, j >= di。
Java
1 package com.algorithm.dynamicprogramming; 2 3 import java.util.Arrays; 4 5 /** 6 * 找零问题 7 * Created by yulinfeng on 7/5/17. 8 */ 9 public class Money { 10 public static void main(String[] args) { 11 int[] money = {1, 3, 2, 5}; 12 int sum = 8; 13 int count = money(money, sum); 14 System.out.println(count); 15 } 16 17 private static int money(int[] money, int sum) { 18 int[] count = new int[sum + 1]; 19 count[0] = 0; 20 for (int j = 1; j < sum + 1; j++) { //总金额数,1,2,3,……,sum 21 int minCoins = j; 22 for (int i = 0; i < money.length; i++) { //遍历硬币的面值 23 if (j - money[i] >= 0) { 24 int temp = count[j - money[i]] + 1; //当前所需硬币数 25 if (temp < minCoins) { 26 minCoins = temp; 27 } 28 } 29 } 30 31 count[j] = minCoins; 32 } 33 System.out.println(Arrays.toString(count)); 34 return count[sum]; 35 } 36 }
Python3
1 #coding=utf-8 2 def charge_making(money, num): 3 ‘‘‘ 4 找零问题 5 ‘‘‘ 6 count = [0] * (num + 1) 7 count[0] = 0 8 for j in range(1, num + 1): 9 minCoins = j 10 for i in range(len(money)): 11 if j - money[i] >= 0: 12 temp = count[j - money[i]] + 1 13 if temp < minCoins: 14 minCoins = temp 15 16 count[j] = minCoins 17 18 return count[num] 19 20 money = [1, 3, 2, 5] 21 num = 8 22 count = charge_making(money, num) 23 print(count)
tag
标签:color family pack lib ++ size 硬币 问题 charge
原文地址:http://www.cnblogs.com/yulinfeng/p/7124122.html