Problem Description You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.   Input The first line of the input…

题意:求n个串的字典序最小的最长公共子串. 解法:枚举第一个串的子串,与剩下的n-1个串KMP匹配,判断是否有这样的公共子串.从大长度开始枚举,找到了就break挺快的.而且KMP的作用就是匹配子串,近乎O(n)的速度,很快. P.S.对于字符串要仔细!!! #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; ; int n;…

Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14113   Accepted: 6260 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousa…

题意:给你多个字符串,问你这几个字符串的最长公共子串是哪个,如果有多个,输出字典序最大的那个,如果最长的公共子串长度小于3,输出一个奇怪的东西: 解题思路:首先看数据,数据不大,开始简单快乐的暴力之路,因为是公共的子串,所以我们随便找一个原串,把它的子串全部循环一遍,然后所有的子串都去和剩余的原串跑一遍kmp,如果匹配没啥问题,那么这个子串是剩余所有原串的公共子串,然后更新答案: 代码: #include<iostream> #include<algorithm> #include…

题目链接:http://poj.org/problem?id=2912 Time Limit: 5000MS Memory Limit: 65536K Description N children are playing Rochambeau (scissors-rock-cloth) game with you. One of them is the judge. The rest children are divided into three groups (it is possible t…

//截取字符串 ch 的 st~en 这一段子串返回子串的首地址 //注意用完需要根据需要最后free()掉 char* substring(char* ch,int st,int en) { ; char* pch=ch; ); pch=pch+st; ;i<length;i++) subch[i]=*(pch++); subch[length]='\0'; return subch; } 字符串截取 POJ 3080 Blue Jeans 题意 : 给出 n 个包含 60 个字符的字符串,问…

Blue Jeans  Time Limit: 1000MS        Memory Limit: 65536K Total Submissions: 21078        Accepted: 9340 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundred…

题目链接:hdu_2328_Corporate Identity 题意: 给你n个串,让你找这n个串的最大公共子串 题解: 串比较小,暴力枚举第一个的子串,然后KMP判断是否可行 #include<cstdio> #include<cstring> #define F(i,a,b) for(int i=a;i<=b;i++) ; ],ans,l,r,cnt; ][N]; int KMP(int n,char*a,int m,char*b){ int i,j; ]=j=-,i=…

完全想不到啊,同余模定理没学过啊,想起上学期期末考试我问好多同学'≡'这个符号什么意思,都说不知道,你们不是上了离散可的吗?不过看了别人的解法我现在会了,同余模定理介绍及运用点这里点击打开链接 简单说一下同余模定理:如果(a - b) / m = 0,说明a%m等于b%m,那么对于本题应该如何运用呢?  已知a % n = m,那么(a * 10 + x) % n = a * 10 % n + x % n = (a % n * 10 + x ) % n = (m *10 + x ) % n,有了…

题意:给你n个字符串,问你这n个串的最长公共子串 解题思路:暴力枚举任意一个字符串的所有子串,然后暴力匹配,和hdu1238差不多的思路吧,这里用string解决的: 代码: #include<iostream> #include<string> #include<cstdio> #include<algorithm> using namespace std; string t; int main() { int n; string a[4050]; ios…