Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11607    Accepted Submission(s): 5413Problem Description It is well known that AekdyCoin is good at string problems as well as number theory probl…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8845    Accepted Submission(s): 4104 Problem Description It is well known that AekdyCoin is good at string problems as well as n…

Count the string Problem Description It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:s: "abab"The prefi…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14096    Accepted Submission(s): 6462 Problem Description It is well known that AekdyCoin is good at string problems as well as…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4105    Accepted Submission(s): 1904 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3797    Accepted Submission(s): 1776 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:  s: "abab"  The prefixes are: "a", "ab&…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=3336 如果你是ACMer,那么请点击看下 题意:求每一个的前缀在母串中出现次数的总和. AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include…

[题意概述] 给定一个文本字符串,找出所有的前缀,并把他们在文本字符串中的出现次数相加,再mod10007,输出和. [题目分析] 利用kmp算法的next数组 再加上dp [存在疑惑] 在分析next数组和dp之间的关系,结论是 dp[i] = (dp[next[i]]+1); 搞不懂之间存在的联系 [AC] #include <bits/stdc++.h> ],next[]; ]; void getnext() { ,j=-; next[]=-; while(i<m) { ||s[i…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example: s: "abab" The prefixes are: "a", "ab&qu…

题目 以下不是KMP算法—— 以下是kiki告诉我的方法,好厉害的思维—— 就是巧用标记,先标记第一个出现的所有位置,然后一遍遍从标记的位置往下找. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { ],shunxu; ]; scanf("%d",&t); while(t--) { memset(xiabiao,…

题意: 求给定字符串,包含的其前缀的数量. 分析: 就是求所有前缀在字符串出现的次数的和,可以用KMP的性质,以j结尾的串包含的串的数量,就是next[j]结尾串包含前缀的数量再加上自身是前缀,dp[i]表示以i为结尾包含前缀的数量,则dp[i]=dp[next[i]]+1,最后求和即可. #include <map> #include <set> #include <list> #include <cmath> #include <queue>…

题意:统计前缀在串中出现的次数 思路:next数组,递推 #include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MaxSize 200005 #define Mod 10007 char str[MaxSize]; int _next[MaxSize]; int dp[MaxSize]; int len; void GetNext(char t[]){//…

题意:给一个字符串,问该字符串的所有前缀与该字符串的匹配数目总和是多少. 此题要用KMP的next和DP来做. next[i]的含义是当第i个字符失配时,匹配指针应该回溯到的字符位置. 下标从0开始. 设j=next[i],那么 如果j==0,回溯到起点说明该字符不匹配. 其他情况,说明字符串S[0,...j-1]与字符串S[0,..i-1]的某个后缀(准确的说是S[i-j,i-1])相同,这样的话,S[0,..i-1]的后缀(S[i-j,i-1])一定包含字符串S[0,..i-1]的后缀能够匹…

题解:利用next数组来保存前缀位置,递推求解. #include <cstdio> #include <cstring> char pat[200005]; int next[200005],M,f[200005]; const int MOD=10007; int getnext(){ int i=1,j=0;next[1]=0; while(i<M){ if(j==0||pat[j]==pat[i])next[++i]=++j; else j=next[j]; } }…

题意: 求一个字符串的所有前缀串的匹配次数之和. 思路: 首先仔细思考: 前缀串匹配. n个位置, 以每一个位置为结尾, 就可以得到对应的一个前缀串. 对于一个前缀串, 我们需要计算它的匹配次数. k = next [ j ] 表示前缀串 Sj 的范围内(可以视为较小规模的子问题), 前缀串 Sk 是最长的&能够匹配两次的前缀串. 这和我们需要的答案有什么关系呢? 题目是求所有前缀串的匹配次数之和, 那么可以先求前缀串 Si 在整个串中的匹配次数, 再加和. 到此, 用到了两个"分治&q…

dp[i]代表前i个字符组成的串中所有前缀出现的次数. dp[i] = dp[next[i]] + 1; 因为next函数的含义是str[1]~str[ next[i] ]等于str[ len-next[i]+1 ]~str[len],即串的前缀后缀中最长的公共长度. 对于串ababa,所有前缀为:a, ab,aba,abab, ababa, dp[3] = 3; 到达dp[5]的时候,next = 3, 它与前面的最长公共前缀为aba,因此dp[5]的凑法应该加上dp[3],再+1是加上aba…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6062    Accepted Submission(s): 2810 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

Problem Description It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example: s: "abab" The prefixes are: "…

一道字符串匹配的题目,仅仅借此题练习一下KMP 因为这道题目就是要求用从头开始的n个字符串去匹配原来的字符串,很明显与KMP中求next的过程很相似,所以只要把能够从头开始匹配一定个数的字符串的个数加起来就OK了(再此结果上还应该加上字符串的长度,因为每个从头开始的字符串本身也可以去匹配自己的),即将next中值不为-1和0的个数统计出来即可. 用GCC编译的,时间用了46MS. #include <stdio.h> #include <string.h> #define MAXL…

http://acm.hdu.edu.cn/showproblem.php?pid=3336 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10917    Accepted Submission(s): 5083 Problem Description It is well known that AekdyCoin is good a…

Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to another one according…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10513    Accepted Submission(s): 3671 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware…

参考连接: KMP+DP: http://www.cnblogs.com/yuelingzhi/archive/2011/08/03/2126346.html 另外给出一个没用dp做的:http://blog.sina.com.cn/s/blog_82061db90100usxw.html 题意: 给出一个字符串,求它的各个前缀在字符串中出现的次数总和. 思路:记 dp[i] 为前 i 个字符组成的前缀出现的次数则 dp[next[i]]+=dp[i] dp[i]表示长度为i的前缀出现的次数,初…

这道题本来想对了,可是因为hdu对pascal语言的限制是我认为自己想错了,结果一看题解发现自己对了…… 题意:给以字符串 计算出以前i个字符为前缀的字符中 在主串中出现的次数和 如: num(abab)=num(a)+num(ab)+num(aba)+num(abab)=2+2+1+1=6; 题解:next[i]记录的是 长度为i 不为自身的最大首尾重复子串长度  num[i]记录长度为next[i]的前缀所重复出现的次数 推介一篇博文,非常不错,和本代码解法不一样,但实质上是一样的. 附上代…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4738 题意:给定一个n个节点m条边的无向图(可能不连通.有重边),每条边有一个权值.判断其连通性,若双连通,输出-1:若非连通,输出0:否则,输出权值最小的桥的权值. 思路:进行双连通域分解,记下连通块的个数和所有桥的情况,对应输出结果即可. 注意对重边的处理.这里我按照上一道题学到的姿势如法炮制:先把所有边按“字典序”排序(u, v, w),这样重边聚集在一起了,然后扫描一遍,发现重边即在结构体E…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 题意:给出一棵n个节点的无根树,每条边有各自的权值.给出m个查询,对于每条查询返回节点u到v的最短路径的权值和,按查询顺序输出结果. 数据范围:n [2, 40000], m[1, 200] 思路:Tarjan算法:dfs遍历每个点,每遍历完 r 的一个孩子 c, 把 c 并入以 r 为祖先的集合,并处理 c 的所有查询 q:若qi的目标节点 v 已被遍历到,那么一定有lca(c, v) =…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3177 \(describe\): 有一个山洞,山洞的容积最大为\(v\).现在你有\(n\)个物品,这些物品在往山洞里搬和放在山洞所需要占用山洞的体积是两个不同的值\(B\),\(A\).你可以理解为在搬运这个物品进洞时需要的容积为一个\(B\),放下物品后的容积是一个\(A\).在任何时刻搬运物品都不允许超过山洞的最大容积.试求能不能把所有物品搬进去 题解: 这个题正解是贪心...没错... 题目…

题目: Given an input string, reverse the string word by word. For example,Given s = "the sky is blue",return "blue is sky the". 代码:oj在线测试通过 Runtime: 172 ms class Solution: # @param s, a string # @return a string def reverseWords(self, s)…

给出一个序列(长度>=2),问去掉一个元素后是否能成为单调不降序列或单调不增序列. 对任一序列,先假设其可改造为单调不降序列,若成立则输出YES,不成立再假设其可改造为单调不增序列,若成立则输出YES,不成立则输出NO. 由于持平不影响整体单调性,为了直观,我在以下把“不降”称为“递增/升序”,把“不增”称为“递减/降序”. 递增和递减是对称的,这里先考虑递增,递减改个符号和最值就好. 我们把为维护单调性而去掉的那个点称为“坏点”.由题目的要求,“可改造”可等价于“只存在一个坏点”. 对于“坏点…