B. Obsessive String   Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following proble…

[CF-div.1 B]Obsessive String 题目大意 两个字符串\(S,T\),求划分方案数使得一个集合中两两划分不相交且划分都包含字符串\(T\) 试题分析 kmp先求出那个位置匹配. 然后怎么办呢?显然\(f_i\)表示考虑到第i个字符的答案. 传统思想:考虑这个地方加不加. 不加:\(f_{i-1}\). 加的话分两种情况讨论,一种是加入之前的某一个集合,就是\(\sum_{j=1}^{pre_i-1} f_j\),还有就是自己创建一个集合:\(pre_{i-1}\) \(p…

http://www.codeforces.com/problemset/problem/494/B 题意:给出两个串S,T,求有几种将S分成若干个子串,满足T都是这若干个子串的子串. 思路:f[n]代表前n个字符来划分,有多少种划分方式,sum[i]代表1到i的f的和,转移就是:对于i这个位置,可以不选,因此首先 f[i]=f[i-1],然后若是选了i,那一定至少是在有匹配位置的左边开始匹配,假设L为小于i的最右边的匹配位置,则 F[i]+=ΣF[j] (1<=j<=L-1),然后也有可能自…

这题的题意就很晦涩.题意是:问有多少种方法,把字符串s划分成不重叠的子串(可以不使用完s的所有字符,但是这些子串必须不重叠),使得t串是所有这些新串的子串.譬如第一个样例,"ababa"和"aba",共有5种方法:{aba}(前3个),{aba}(后3个),{abab},{baba},{ababa}. 先设s的长度为lena,t的长度为lenb. 做法是:用dp[i]表示到i为止,有几种方案数,所以最终答案是dp[lena].然后考虑转移.首先dp[i]至少等于dp…

题意: 给两个串S,T,问能找出多少的S的(a1,b1)(a2,b2)..(ak,bk),使Sa1---Sb1,...Sak---Sbk都包含子串T,其中k>=1,且(a1,b1)...(ak,bk)互不相交. 比如S = "abacaba",T="aba", 当k=1时,(0,6)满足,还有其他只包含一个aba串的也满足,k-2时,(0,2)(3,6)满足,(0,2)(4,6)也满足,(0,3)(4,6)也满足,所以总共有12种. 解法:dp.先用kmp找出…

题目传送门 /* 题意:找到一个字符串p,使得它和s,t的不同的总个数相同 贪心:假设p与s相同,奇偶变换赋值,当是偶数,则有答案 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <iostream> using namespace std; ; const int INF = 0x3f3f3f3f; int p[MAXN…

[codeforces494B]Obsessive String 试题描述 Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the…

题目连接:http://codeforces.com/contest/676/problem/C 题意:一串字符串,最多改变k次,求最大的相同子串 题解:很明显直接尺取法 #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<set> #include<map> #include<vector> #include<qu…

B. Equidistant String Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/545/problem/B Description Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings conta…

[Link]:http://codeforces.com/contest/828/problem/C [Description] 让你猜一个字符串原来是什么; 你知道这个字符串的n个子串; 且知道第i个字符t[i],在k[i]个位置出现过; 且告诉你这k[i]个位置在哪里; 数据不会产生矛盾; 让你输出最终的字符串,输出字典序最小的那个; [Solution] 对于输入的n个子串; 对于每个位置; 看看那个位置有没有子串之前出现过,没有的话,就放在那个位置; 否则,如果当前这个子串ti的长度比原…