Fibonacci String Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5008    Accepted Submission(s): 1690 Problem Description After little Jim learned Fibonacci Number in the class , he was very int…

Fibonacci String Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4568    Accepted Submission(s): 1540 Problem Description After little Jim learned Fibonacci Number in the class , he was very int…

Fibonacci String Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5012 Accepted Submission(s): 1693 Problem Description After little Jim learned Fibonacci Number in the class , he was very interest…

Problem Description After little Jim learned Fibonacci Number in the class , he was very interest in it. Now he is thinking about a new thing – Fibonacci String . He defines : str[n] = str[n-1] + str[n-2] ( n > 1 ) He is so crazying that if someone g…

String HDU 5672(双指针) 传送门 题意:一个字符串中找到所有拥有不少于k个不同的字符的子串. import java.io.*; import java.util.*; public class Main { static final int N = 1000005; static final int inf= 0x3f3f3f3f; static final double eps= 1e-5; static int vis[]=new int[27]; static char…

Combine String HDU - 5707 题目大意:给你三个串a,b,c,问a和b是不是恰好能组成c,也就是a,b是不是c的两个互补的子序列. 根据题意就可以知道对于c的第一个就应该是a第一个或者b的第一个,如果第一个是a的第一个,那么c的第二个就应该是a的第二个或者是b的第一个,反之也是一样的.那么我们定义dp[i][j]就是a串匹配到i位置,b串匹配到j位置的合理性,那么dp[i][j]就由dp[i-1][j]和dp[i][j-1]推过来,a串匹配到i位置,b串匹配到j位置,那么c…

http://acm.hdu.edu.cn/showproblem.php?pid=3336 Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6672    Accepted Submission(s): 3089 Problem Description It is well known that Aek…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=5311 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1462    Accepted Submission(s): 521 Problem Description Today is the 1st anniversary of BestC…

思路 :二位数组维护数目. #include<iostream> #include<stdio.h> #include<stdlib.h> #include<memory.h> #include<string.h> #include<algorithm> #include<cmath> #include<stack> ; ; using namespace std; typedef long long ll;…

题目大意:给你一个串求出来这个串所有的前缀串并且与前缀串相等的数量,比如: ababa 前缀串{"a", "ab", "aba", "abab", "ababa"}: 每个前缀串出现的次数{3, 2, 2, 1, 1},那么结果就是 9.   分析:我们可以用dp[i],表示前i长度的串的结果,那么就可以得到下面的转移方程 dp[i] = dp[next[i]] + 1. 代码如下. ===========…