Coins in a Line

最新推荐文章于 2019-02-25 14:02:18 发布

原创最新推荐文章于 2019-02-25 14:02:18 发布 · 290 阅读

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本文介绍了一个简单的游戏博弈算法，通过递推的方式预判玩家的操作来决定胜负。算法使用一个布尔型数组记录不同回合下玩家是否能赢得游戏。

请判定第一个玩家是输还是赢？

n = 1, 返回 true.
n = 2, 返回 true.
n = 3, 返回 false.
n = 4, 返回 true.
n = 5, 返回 true.

var firstWillWin = function (n) {
    var dp = [];
    for (var i = 1; i <= n; i++) {
        if (i == 1 || i == 2) dp[i] = true;
        else dp[i] = !dp[i - 1] || !dp[i - 2];
    }
    return dp[n];
}

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专栏目录

PAT-A1068 Find More Coins 题目内容及题解

BourbonWhiskey 的博客

02-04

666

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,...

LeetCode题-coins in a line

qq_26919721的博客

08-15

693

There are n coins in a line. Two players take turns to take acoin from one of the ends of the line until there are no more coinsleft. The player with the larger amount of money wins. Assume thatyou

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【LintCode】Coins in a Line 硬币排成行

wutingyehe的专栏

07-15

1459

有 n 个硬币排成一条线。两个参赛者轮流从右边依次拿走 1 或 2 个硬币，直到没有硬币为止。拿到最后一枚硬币的人获胜。请判定第一个玩家是输还是赢？样例 n = 1, 返回 true. n = 2, 返回 true. n = 3, 返回 false. n = 4, 返回 true. n = 5, 返回 true.挑战 O(1) 时间复杂度且O(1) 存储。相似例子（引自http:

LintCode 394. Coins in a Line (博弈类DP 经典题)

roufoo的博客

02-25

500

Coins in a Line There are n coins in a line. Two players take turns to take one or two coins from right side until there are no more coins left. The player who take the last coin wins. Could you ple...

#394 Coins in a Line

易水居

08-19

500

题目描述： There are n coins in a line. Two players take turns to take one or two coins from right side until there are no more coins left. The player who take the last coin wins. Could you please

Monocarp is going to make a purchase with cost of exactly m m burles. He has two types of coins, in the following quantities: coins worth 1 1 burle: a 1 a 1 regular coins and infinitely many fancy coins; coins worth k k burles: a k a k regular coins and infinitely many fancy coins. Monocarp wants to make his purchase in such a way that there's no change — the total worth of provided coins is exactly m m. He can use both regular and fancy coins. However, he wants to spend as little fancy coins as possible. What's the smallest total number of fancy coins he can use to make a purchase? Input The first line contains a single integer t t ( 1 ≤ t ≤ 3 ⋅ 10 4 1≤t≤3⋅10 4 ) — the number of testcases. The only line of each testcase contains four integers m , k , a 1 m,k,a 1 and a k a k ( 1 ≤ m ≤ 10 8 1≤m≤10 8 ; 2 ≤ k ≤ 10 8 2≤k≤10 8 ; 0 ≤ a 1 , a k ≤ 10 8 0≤a 1 ,a k ≤10 8 ) — the cost of the purchase, the worth of the second type of coin and the amounts of regular coins of both types, respectively. Output For each testcase, print a single integer — the smallest total number of fancy coins Monocarp can use to make a purchase. Examples Inputcopy Outputcopy 4 11 3 0 0 11 3 20 20 11 3 6 1 100000000 2 0 0 5 0 1 50000000 Note In the first testcase, there are no regular coins of either type. Monocarp can use 2 2 fancy coins worth 1 1 burle and 3 3 fancy coins worth 3 3 (since k = 3 k=3) burles to get 11 11 total burles with 5 5 total fancy coins. In the second testcase, Monocarp has a lot of regular coins of both types. He can use 11 11 regular coins worth 1 1 burle, for example. Notice that Monocarp doesn't have to minimize the total number of used coins. That way he uses 0 0 fancy coins. In the third testcase, Monocarp can use 5 5 regular coins worth 1 1 burle and 1 1 regular coin worth 3 3 burles. That will get him to 8 8 total burles when he needs 11 11. So, 1 1 fancy coin worth 3 3 burles is enough.

03-12

a. s → 花式1的数量是0 → 总花式数= x-ak. 此时，为了让总花式数最小，x应该尽可能小。此时，x的最小可能值是ak+1。此时，总花式数是1。但必须满足s= m - (ak+1)*k >=0 → ak+1 → 即 (ak+1)*k → 需要m >=k*(ak+1...

Line 7: Char 29: warning: variable length arrays in C++ are a Clang extension [-Wvla-cxx-extension] 7 | int member[coins.size()+10][amount+10]; | ^~~~~~~~~ Line 7: Char 29: note: function parameter 'amount' with unknown value cannot be used in a constant expression Line 3: Char 42: note: declared here 3 | int coinChange( vector<int>& coins , int amount ) { | ^ Line 7: Char 12: warning: variable length arrays in C++ are a Clang extension [-Wvla-cxx-extension] 7 | int member[coins.size()+10][amount+10]; | ^~~~~~~~~~~~~~~ Line 7: Char 12: note: function parameter 'coins' with unknown value cannot be used in a constant expression Line 3: Char 30: note: declared here 3 | int coinChange( vector<int>& coins , int amount ) { | ^ Line 23: Char 15: error: use of undeclared identifier 'i' 23 | return member[i][j] ; | ^

03-22

### 解决 C++ 编译警告和错误在 C++ 中，变量长度数组（Variable Length Array, VLA）并不是标准的一部分。然而，在某些编译器（如 GCC 和 Clang）中支持作为扩展功能[^1]。当启用严格的标准模式时，这些扩展可能...

使用cpp Coin Change 时间限制: 1s 类别: 循环结构->竞赛问题描述 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 and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent. Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins. 输入说明 The first line of the input contains an integer T which means the number of test cases. Then T lines follow,each one consisting of a number ( ≤250 ) for the amount of money in cents. 输出说明 For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins. 输入范例 5 1 100 200 250 198 输出范例 1 292 2061 3830 1886

最新发布

12-30

const int MAX_COINS = 100; const std::vector<int> coinValues = {50, 25, 10, 5, 1}; // 递归函数，用于计算找零方式的数量 int countWays(int amount, int coinIndex, int usedCoins) { if (amount == 0 && ...

memory limit per test256 megabytes    You have n coins with denominations a1,a2,…,an and a natural number k . You also have a bag, which is initially empty, where you can place coins. You need to perform exactly k actions. In each action, you take one coin from those you have left and put it in your bag. After that, you can no longer take that coin. At the same time, you have a cat that loves even numbers, so every time the sum of the denominations of the coins in your bag becomes even, your cat empties the bag, meaning it takes all the coins to a place known only to it, and the bag is empty again. Note that the bag is emptied every time the sum becomes even during the process of adding coins, not just at the very last moment. Let your score be the sum of the denominations of the coins in the bag. Your task is to perform k actions such that your final score is maximized. Find the answer for all 1≤k≤n . DeepL 翻译    您有 n 枚面值为 a1,a2,…,an 的硬币和一个自然数 k 。您还有一个最初是空的袋子，可以用来放置硬币。您需要进行 k 次操作。在每次行动中，你都要从剩下的硬币中取出一枚硬币放进袋子里。之后，您就不能再拿这枚硬币了。同时，你有一只喜欢偶数的猫，所以每当你包里的硬币面值之和变成偶数时，你的猫就会清空包，也就是说它会把所有硬币带到一个只有它自己知道的地方，然后包里的硬币就又空了。请注意，在添加硬币的过程中，每次总和变为偶数时，袋子都会被清空，而不是在最后一刻才清空。让你的分数成为袋子里硬币面值的总和。你的任务是进行 k 次操作，使你的最终得分最大。找出所有 1≤k≤n 的答案。    Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104 ). The description of the test cases follows. The first line of each test case contains a single integer n (1≤n≤2⋅105 ) — the number of coins you have. The second line of each test case contains n natural numbers a1,a2,…,an (1≤ai≤109 ) — the denominations of the coins. It is guaranteed that the sum of n across all test cases does not exceed 2⋅105 . DeepL 翻译    输入每个测试包含多个测试用例。第一行包含测试用例的数量 t ( 1≤t≤104 )。测试用例说明如下。每个测试用例的第一行都包含一个整数 n ( 1≤n≤2⋅105 ) - 您拥有的硬币数量。每个测试用例的第二行包含 n 个自然数 a1,a2,…,an （ 1≤ai≤109 ）。( 1≤ai≤109 ) - 硬币的面值。保证所有测试用例中 n 的总和不超过 2⋅105 。    Output For each test case, output n numbers — the maximum possible score that can be achieved by performing exactly k actions for all k from 1 to n . DeepL 翻译    输出对于每个测试用例，输出 n 个数字，即从 1 到 n 的所有 k 中，执行 k 个操作所能获得的最大分数。 Example InputCopy 6 3 1 1 1 3 1 2 3 5 4 1 3 1 2 5 4 2 3 1 3 3 4 1 2 3 4 2 2 OutputCopy 1 0 1 3 5 0 3 7 9 7 9 3 7 9 7 9 1 5 7 0 0 0    Note In the first set of input data, you have coins with denominations [1,1,1 ]. k=1 : in this case, the sum of the denominations in the bag is 1 , regardless of the chosen coin. The final score is also 1 . k=2 : in this case, the sum of the denominations in the bag is initially 1 , regardless of the choice of coin. When choosing the second coin, the sum will be 2 , regardless of the choice of coin, and thus the bag will be emptied. The final score is 0 . k=3 : in this case, when choosing the first two coins, the sum will be 2 and the bag will be emptied, after which the sum will be 0 . When choosing the third coin, the sum will become 1 . In the second set of input data, you have coins with denominations [1,2,3 ]. k=1 : in this case, when choosing the coin with denomination 2 , the bag will be emptied and the final score will be 0 . When choosing the coin with denomination 1 or 3 , the final scores will be 1 and 3 , respectively. The maximum possible final score is 3 . k=2 : in this case, when choosing two coins 1 and 3 in any order, their sum will be 4 , after which the bag will be emptied and the final score will be 0 . However, when choosing 3 as the first coin, the score will be 3 . Then you can choose, for example, coin 2 , and the final score will become 5 . k=3 : in this case, the sum of all coins will be 6 , and since each time the bag is emptied it does not change the parity of the accumulated sum, the final score is 0 .

12-13

你有 $ n $ 枚硬币，面值为 $ a_1, a_2, \dots, a_n $。你要进行恰好 $ k $ 次操作（对每个 $ k = 1 $ 到 $ n $），每次将一枚未使用过的硬币放入袋中。**一旦袋中硬币的总和变为偶数，猫就会清空袋子（即总和归零）...

Coins in a Line II

leetcode 解题思路

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There are n coins with different value in a line. Two players take turns to take one or two coins fromleft sideuntil there are no more coins left. The player who take the coins with the most value w...

lintcode：Coins in a Line 硬币排成线

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题目硬币排成线有 n 个硬币排成一条线。两个参赛者轮流从右边依次拿走 1 或 2 个硬币，直到没有硬币为止。拿到最后一枚硬币的人获胜。请判定第一个玩家是输还是赢？样例 n = 1, 返回 true. n = 2, 返回 true. n = 3, 返回 false. n = 4, 返回 true. n = 5, 返回 true. 挑战 ...

lintcode coins-in-a-line 硬币排成线

stevewong的专栏

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问题描述笔记设置buff[i]为：有i个硬币的时候第一个玩家是赢还是输。如果buff[i-1]和buff[i-2]都是赢，那我一开始无论取1个还是2个，对方都可以赢。（我取完之后可以看做对方是先手）如果buff[i-1]是赢，buff[i-2]是输，那我一开始就取2个，对方剩下i-2个只能输。那就是我可以赢。如果buff[i-2]是赢，buff[i-1]是输，那我一开始就取1个，对方剩下i-1

[LintCode]Coins in a Line

Three_Thousand_World的博客

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//启发根源来自于这个题目version2。在version2的基础上，如何能够发现version3的类似解法。 //题目 //There are n coins with different value in a line. //Two players take turns to take one or two coins from left side until

#395 Coins in a Line II

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题目描述： There are n coins with different value in a line. Two players take turns to take one or two coins from left side until there are no more coins left. The player who take the coins with the

LintCode 394. Coins in a Line

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Description There are n coins in a line. Two players take turns to take one or two coins from right side until there are no more coins left. The player who take the last coin wins. Could you please ...

394.Coins in a Line-硬币排成线（中等题）

三色旗飞扬

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461

硬币排成线题目有 n 个硬币排成一条线。两个参赛者轮流从右边依次拿走 1 或 2 个硬币，直到没有硬币为止。拿到最后一枚硬币的人获胜。请判定第一个玩家是输还是赢？样例n = 1, 返回 true. n = 2, 返回 true. n = 3, 返回 false. n = 4, 返回 true. n = 5, 返回 true. 挑战O(1) 时间复杂度且O(1) 存储。题解从博弈论

LintCode 395. Coins in a Line II (博弈类DP经典题!!!)

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Coins in a Line II There are n coins with different value in a line. Two players take turns to take one or two coins from left side until there are no more coins left. The player who take the coins ...