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accepted1
- 解决时钟问题，acm竞赛题 A weird clock marked from 0 to 59 has only a minute hand. It won t move until a special coin is thrown into its box. There are different kinds of coins as your options. However once you make your choice, you cannot use any other kind.
设计一个用最少硬币找钱m的方法
- 设有n种不同面值的硬币，各硬币的面值存于数组T[1:n]中。现要用这些面值的硬币来找钱。可以使用的各种面值的硬币个数存于数组Coins[1:n]中。对任意钱数0≤m≤20001，设计一个用最少硬币找钱m的方法。-With n different coins, the face value of the coins stored in the array T [1: n] in the. Now use these coins to find the money. You can use the
FakeCoin
- 有一大堆金币其中有一枚假币在一堆金币中找出一个价币来。-Have a lot of these, a counterfeit gold coin in the pile of coins to find a price.
lesscoins
- 设有n 种不同面值的硬币，各硬币的面值存于数组T〔1:n 〕中。现要用这些面值的硬币来找钱。可以使用的各种面值的硬币个数存于数组Coins〔1:n 〕中。 对任意钱数0≤m≤20001，设计一个用最少硬币找钱m 的方法。-With n different coins, the face value of the coins stored in the array of T 〔1: n〕 in. Now use these coins to find the money. You can us
4.3
- 硬币找钱问题的解决方案，时间复杂度小，方法十分方便-Coin find money solution to the problem
At-least-the-issue-of-coins
- 有n种不同面值的硬币，各硬币面值存于数组T[1:n] 现用这些面值的钱来找钱；各面值的个数存在数组Coins[1:n]中。 对于给定的1<=n<=10，硬币面值数组、各面值的个数及钱数m，0<=m<=2001，编程计算找钱m的最少硬币数。 -There are n different denominations of coins, each coin denomination stored in the array T [1: n] current face v
yinbiwenti
- 在n个银币中有一个是不合格的，不合格的银币比合格银币要轻。 现用天平秤银币，找出不合格的银币，且在最坏情况下秤银币的次数最少 -In n silver in one is unqualified, not qualified silver qualified to light than silver. Now in the balance scale, and find out the unqualified silver coin, and in the worst c
sfdfsa
- 解决8枚硬币问题。在8枚硬币中有一枚是假硬币，把这个假硬币找出来。-Solve the eight coins problem. In the eight coins in a coin is fake, fake coins to find this out.
324
- 算法实现题3-1 最少硬币问题 问题描述：设有n种不同面值的硬币，各硬币的面值存于数组T[1:n]中。现要用这些面值的硬币来找钱。可以使用的各种面值的硬币个数存于数组Coins[1:n]中。对任意钱数0≤m≤20001，设计一个用最少硬币找钱m的方法。 算法设计：对于给定的1≤n≤10，硬币面值数组T和可以使用的各种面值的硬币个数数组Coins，以及钱数m，0≤m≤20001，计算找钱m的最少硬币数。 数据输入：由文件input.txt 提供输入数据，文件的第一行中只有1 个整数给出
CoinChange
- 硬币找零算法（动态规划）：有一堆不同面值的硬币，需要找到一个硬币数最少的组合来组成某一个数值。-Coin change algorithm (dynamic programming): There are a bunch of different denominations of coins, need to find a combination of the number of coins to form at least one value.