题解 01背包板子 + 记录路径。这次的记录路径比较特殊，要从多组解中找到一组由尽量小价值的硬币组成的解。所以不能利用一维数组记录路径，path[目前重量] = 物品序号，因为这样最后只能记录一个可能符合或不符合要求解。所以应该利用二维数组记录路径，path[ 物品序号 ][ 目前重量 ]
You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin. Example 1: Input: amount = 5,
A. Collecting Coins time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Polycarp has three sisters: Alice, Barbara, and Cerene. They’re collecting coins. Currently, Alice has a coins, Barbara has b
9.Algorithm Gossip: 八枚银币 说明 现有八枚银币a b c d e f g h，已知其中一枚是假币，其重量不同于真币，但不知是较轻或较重，如何使用天平以最少的比较次数，决定出哪枚是假币，并得知假币比真币较轻或较重。 解法 单就求假币的问题是不难，但问题限制使用最少的比较次数，所以我们不能
链接： https://codeforces.com/contest/1282/problem/B2 题意： This is the hard version of this problem. The only difference is the constraint on k — the number of gifts in the offer. In this version: 2≤k≤n. Vasya came to the store to buy goods for his friends f
描述：代替递归求解 例如：斐波那契函数f(n)=f(n-1)+f(n-2)。计算f(n)需要计算f(n-1)和f(n-2)。当计算f(n-1)时要计算f(n-2)和f(n-3)。因此在计算f(n)中f(n-2)被计算了两次。 为了减少重复的递归调用，我们可以反过来计算。先计算f(2)，有了f(2)再计算f(3)，以此类推，计算到f(n)。在此过程
跳跃游戏 给定一个非负整数数组，你最初位于数组的第一个位置。 数组中的每个元素代表你在该位置可以跳跃的最大长度。 判断你是否能够到达最后一个位置。 思路 根据题目意思，最大跳跃距离，说明可以跳0--nums[i]的距离 可以把跳跃看成走nums[i]步，如果能走到下一位置则可以加油获取更多
Problem Statement You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination
题目：已知不同面值的钞票，求如 何用最少数量的钞票组成某个金额，求可 以使用的最少钞票数量。如果任意数量的已知面值钞票都无法组成该金额， 则返回-1。 示例： Input: coins = [1, 2, 5], amount = 11Output: 3 Explanation: 11 = 5 + 5 + 1Input: coins = [2],
Exercise 1: Pascal’s Triangle The following pattern of numbers is called Pascal’s triangle. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 ... The numbers at the edge of the triangle are all 1, and each number inside the triangle is the sum of the two numbers
问题 给定一个二叉树的root节点，二叉树中每个节点有node.val个coins，一种有N coins。 现在要求移动节点中的coins 使得二叉树最终每个节点的coins value都为1。每次移动，我们只能向相邻接点移动，也就是说coins 只能向父节点或者子节点移动，那么求最终需要移动多少步。 Leetcode原题链接
算法（三）--------扔鸡蛋问题和找零钱问题 扔鸡蛋问题描述：You are given two eggs, and access to a 100-storey building. The aim is to find out the highest floor from which an egg will not break when dropped out of a window from that floor. What strategy should y
反初始化（析构过程）（学习笔记） 环境Xcode 11.0 beta4 swift 5.1 反初始化（析构过程） 类实例销毁之前会立即调用析构器，用关键字 deinit 反初始化（析构过程）如何工作 Swift仍然使用ARC管理机制，大部分情况下实例销毁不用手动清理；但有些情况下需要自己做一些额外的操作在实例销毁前，此时
One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭，快来快来 数一数，二四六七八". And then the cashier put the counted coins back morosely and count again...
零钱兑换 给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回 -1。 示例 1: 输入: coins = [1, 2, 5], amount = 11 输出: 3 解释: 11 = 5 + 5 + 1 示例 2: 输入: coins
洛谷 UVA562 Dividing coins Description 给你一堆硬币，让你分成两堆，分别给A，B两个人，求两人得到的最小差。 Input 输入有多组测试样例。第一行一个数字n，代表有n组测试样例。接下来是n组测试样例，每个样例包括两行，第一行m代表有m个硬币，第二行输入m个硬币的面值。 Output 输出每组
time limit per test : 2 seconds memory limit per test : 256 megabytes Musicians of a popular band “Flayer” have announced that they are going to “make their exit” with a world tour. Of course, they will visit Berland as well. There are n cities in
Rikka with Coin Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 1722 Accepted Submission(s): 568 Problem Description Rikka hates coins, and she used to never carry any coins with her
1048 Find Coins (25 分) Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special require
Problem Description 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 an
思路 dp问题，空间换时间，递推公式(初始化+转移方程), F(S) = F(S-C) + 1 # S 代表总额(amount), F(S)代表最少兑换次数，C代表兑换的最后一个面值，其中 S为0时，F(S) = 0, 零钱数组为空时，F(S)=-1. 解法 暴力穷举，回溯 public int coinChange(int[] coins, int amount) { retu
题目原文 传送门 POJ P1013 Description（题目描述） Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silv
People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change
People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change
给出硬币面额及每种硬币的个数，求从1到m能凑出面额的个数。 简单的多重部分和问题，动态规划解决 递推式可以参考https://mp.csdn.net/postedit/98054811 #include<iostream> #include<cstdio> #include<fstream> #include<queue> #include<vector> #include<algorithm> #include
