Simple String Problem Recently, you have found your interest in string theory. Here is an interesting question about strings. You are given a string S of length n consisting of the first k lowercase letters. You are required to find two non-empty sub…

Description 题目描述 Recently, you have found your interest in string theory. Here is an interesting question about strings. You are given a string S of length n consisting of the first k lowercase letters. You are required to find two non-empty substrin…

FZU - 2218Simple String Problem 题目大意:给一个长度为n含有k个不同字母的串,从中挑选出两个连续的子串,要求两个子串中含有不同的字符,问这样的两个子串长度乘积最大是多少? 根据题目所给的k<=16很自然的想到用状压dp来处理,但不知道该dp个什么,在观摩大佬的做法后才明白.我们用0000000000000000到1111111111111111表示pomnlkjhgfedcba的字符存不存在的状态,某个字符存在的话对应位就是1,反之就是0.一开始dp[x]就表示,…

A - Simple String Problem Time Limit:10000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu  Practice POJ 2236 Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network wi…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2791    Accepted Submission(s): 1659 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

Description 题目描述 You are given an polynomial of x consisting of only addition marks, multiplication marks, brackets, single digit numbers, and of course the letter x. For example, a valid polynomial would be: (1+x)*(1+x*x+x+5)+1*x*x. You are required…

Accept: 2    Submit: 16 Time Limit: 2000 mSec    Memory Limit : 32768 KB  Problem Description Recently, you have found your interest in string theory. Here is an interesting question about strings. You are given a string S of length n consisting of t…

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings: String Rank SKYLONG 1 KYLONGS 2 YLONGSK 3 LONGSKY 4 ONGSKYL 5 NGSKYLO 6 GSKYLON 7 and lexicograp…

题意:比如给你一个串,要求判断wyh是不是它的子序列,那么你只需要找一个w,找一个y,再找一个h,使得w在y前面,y在h前面即可.有一天小学生拿着一个串问他“wyh是不是这个串的子序列?”.但是wyh2000有重度近视眼,如果字符串中有一段连续的v(至少两个),那么他会把它看成一个w.例如,字符串vvv会被看成w,字符串vvwvvv会被看成www,字符串vwvv会被看成vww.请问wyh2000会怎么回答这个问题?输出yes和no. 思路:其实就是将两个以上的v换成w,再来看是否能有'wyh'子…

首先,定义S,表示前k个字符出现的集合,用二进制来压缩. 接下来,推出dp1[S],表示集合为S的子串的最长长度. 然后根据dp1[S]再推出dp2[S],表示集合为S或S的子集的子串的最长长度. 最后答案就是max(dp2[S]*dp2[补(S)]) #include<cstdio> #include<cstring> #include<algorithm> using namespace std; <<]; int main(){ int t,n,k;…