题意:求可重叠的k次最长重复子串的长度 链接:点我 和poj1743差不多 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; #define MOD 1000000007 const int I
Milk Patterns Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 16742 Accepted: 7390 Case Time Limit: 2000MS Description Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation,
Milk Patterns Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 7586 Accepted: 3448 Case Time Limit: 2000MS Description Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation,
#include "stdio.h" #define maxn 20010 int wa[maxn],wb[maxn],wv[maxn],ws[maxn]; int rank[maxn],height[maxn]; int r[maxn],sa[maxn],ans[maxn]; int n,res,k; int cmp(int *r,int a,int b,int l) { return r[a]==r[b]&&r[a+l]==r[b+l]; } void da(int
Life Forms Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have
http://acm.hdu.edu.cn/showproblem.php?pid=1217 Arbitrage Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9455 Accepted Submission(s): 4359 Problem Description Arbitrage is the use of discrepa
1这道题目就是给定两个字符串,然后求这两个字符串的最长公共子串的最大长度,假设我的f()方法是来求两个字符串的最大公共子串,从头开始逐一比较,如果相等,则 继续调用这个方法,使得递归的长度+1,如果不相等,则只要比较s1截掉一个和s2比较,和s2截掉和s1比较,两个中的最大者,如果s1或者s2中有一个长度为0,则最大公共长度就是0,return 2.代码示例: package zzl; public class 最长公共子串 { public static void main(String[]