Turne的博客

TunnelsTime Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1859    Accepted Submission(s): 546Problem DescriptionBob is travelling i

题目描述：在计算机世界，我们一直追求用最小的资源产生最大的价值。现在，假设你可以支配m个0和n个1。同时有一些只有0和1组成的字符串。你的任务是用这些0和1去组成这些字符串，输出最多能组成多少个字符串。每个0和1只能被使用一次。样例输入样例一输入: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n =

不要62Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29687    Accepted Submission(s): 10460Problem Description杭州人称那些傻乎乎粘嗒嗒的人为62（音

N - 导弹拦截Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%lld & %lluSubmit Status Practice UESTC 251Appoint description: System Crawler  (2016-03-21)Description某国

dp[i][j]表示a[1]~a[i]和b[1]~b[j]并以b[j]结尾的最长公共上升子序列,如果a[i]不等于b[j]时，很明显dp[i][j]的值就等于dp[i-1][j]；如果a[i]等于b[j]时，就在b[1]~b[j]中寻找b[k]使得b[j]>b[k]而且dp[i][k]是最大的。即状态转移方程为：dp[i][j] = dp[i-1][j](a[i] != b[j]),dp[i][j

AlignmentTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 15323 Accepted: 4974DescriptionIn the army, a platoon is composed by n soldiers. During the mor

The Cow LexiconTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9199 Accepted: 4369DescriptionFew know that the cows have their own dictionary with W (

Cash MachineTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 31407 Accepted: 11288DescriptionA Bank plans to install a machine for cash withdrawal. The

KingdomsTime Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %lluSubmit StatusDescription题意：在钱数为k的限制下，求建与1直接或者间接相连的桥，使得人数最多。思路：初始化只有1的状态为0，其余为

BalanceTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 12659 Accepted: 7935DescriptionGigel has a strange "balance" and he wants to poise it. Actually,

D - 男神的约会Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%lld & %lluSubmit Status Practice UESTC 1134Appoint description: System Crawler  (2016-03-02)Description

题目描述给出一个字符串 s，对这个字符串进行切分，每个子串都必须是一个回文字符串，问最少要切多少次。Example：给出 s = “aab”，返回最少切分次数 1。因为只需要一次切分就能把 s 切分成符合要求的两个子串[“aa”, “b”] 。暴力法：枚举起点和终点去求该区间字符串的最长回文字串（时间复杂度O（n^2））,找到后就去递归剩下的字串（时间复杂度