1235 - Coin Change (IV)

【1235 - Coin Change (IV)】的更多相关文章

1235 - Coin Change (IV)

1235 - Coin Change (IV) PDF (English) Statistics Forum Time Limit: 1 second(s) Memory Limit: 32 MB Given n coins, values of them are A1, A2 ... An respectively, you have to find whether you can pay K using the coins. You can use any coin at most t…

LightOJ 1235 - Coin Change (IV) (折半枚举)

题目链接: http://www.lightoj.com/volume_showproblem.php?problem=1235 题目描述: 给出n个硬币,每种硬币最多使用两次,问能否组成K面值? 解题思路: 因为K草鸡大,尽管n很小,但是2^n很大,无法用背包做到O(nK)的复杂度.如果暴力枚举复杂度立马飙升到O(2^(n+1)).··········· 所以引进一种新的算法,折半查找:把所要枚举的值,先把前部分的值所有情况枚举出来,再把后部分的值所有情况枚举出来并排序,结合二分进行查找. 这…

Lightoj 1235 - Coin Change (IV) 【二分】

题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1235 题意: 有N个硬币(N<=18).问是否能在每一个硬币使用不超过两次的情况下支付正好K的面额. 思路 : dfs构造出用这些硬币用前一半能支付的全部费用和后一半能支付的全部费用. 之后排序,枚举前一半的每一个面值在第二个里面二分寻找就可以.(或者用set保存). 代码:(set) #include <stdio.h> #include <ctime>…

Coin Change (IV) （dfs）

Coin Change (IV) Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu [Submit] [Go Back] [Status] Description Given n coins, values of them are A1, A2 ... An respectively, you have to find whether you can pay K using the coin…

[LeetCode] Coin Change 硬币找零

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 of the coins,…

HDOJ 2069 Coin Change(母函数)

Coin Change Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10289 Accepted Submission(s): 3451 Problem Description Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1…

HDU 2069 Coin Change

Coin Change Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10924 Accepted Submission(s): 3669 Problem Description Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent…