对于面试来说,常见的面试题就是贪心法的找零问题。 一个经典的贪心算法的简单例子是找零钱问题(Coin Change Problem)。假设你是一名售货员,需要找零给客户,而你手头有不同面额的硬币。你的目标是找零的硬币数量尽可能少。 假设你有 1 元、5 元、10 元、20 元和 50 元的硬币,现在需要找零 36 元。 贪心算法...
http://comproguide.blogspot.com/2013/12/minimum-coin-change-problem.html http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/ 提前致谢。我访问的每个网站仅说明了该解决方案的工作原理,而不说明其他解决方案为何不起作用。 algorithm dynamic-programming coin-change zc9*_*917 lucky-day...
一种的多重背包背包——Coin Change http://acm.zjnu.edu.cn/CLanguage/showproblem?problem_id=1140题目链接我的暴力算法#include <iostream> #include <algorithm> using namespace std; int main() { int n; while (cin >> n){ int counter = 0; int sum = 0; for (int i = 0;i <= n;i ...
#include <iostream>#include<vector>#include<algorithm>#include<numeric>usingnamespacestd; typedef unsignedlonglongULL;intmain() { unsigned n, m; cin>> n >>m; vector<ULL>v(m);for(unsigned i =0; i < m; i++) cin>>v[i]; std::sort(v.begin(), v.end());//dp[val][ending coi...
HackerRank# The Coin Change Problem 原题地址 背包问题,没啥好说的,记得用long long,否则会爆 代码: 1#include <cmath>2#include <cstdio>3#include <vector>4#include <iostream>5#include <algorithm>6#include <cstring>7usingnamespacestd;89#defineMAX_N 25610#defineMAX_M 641112longlongf[MAX_N];...
Task Write a function to compute the minimum number of coins needed for change using Coin Change Algorithm. Acceptance Criteria All tests must pass. Summary of Changes Added a new function to solve the Coin Change problem using dynamic programming. The implementation efficiently calculates the minimum...
LightOJ 1232 - Coin Change (II) 【完全背包】 http:///volume_showproblem.php?problem=1232 题意:每个物品价值为val[i] (>=1),每个物品有k种,组成价值为k的方案数。完全背包。 解法:完全背包计数。 代码: AI检测代码解析 #include <stdio.h>...
HackerRank - "The Coin Change Problem" AI检测代码解析 #include <iostream>#include<vector>#include<algorithm>#include<numeric>usingnamespacestd; typedef unsignedlonglongULL;intmain() { unsigned n, m; cin>> n >>m; vector<ULL>v(m);for(unsigned i =0; i < m; i++)...
#include <iostream> #include <stdio.h> #include <algorithm> using namespace std; int money[5] = { 50, 25,10,5,1 }; int sum = 0; int n; int coin = 0; //硬币数量 void dfs(int x, int pos) { //递归出口 if (x == n) { sum++; return; } if (coin >= 100) { //硬...
Also, I would like to know generally how oneprovethat DP cannot solve some problem, if possible ? Thanks for reading You mean I will only need to add the condition x<=a[i] to while loop in this code which solve the problem when having inf. coins.Indeed. how stupid my question!