AC Diary--Character token Openjudge 1.7 30

30: Character Token

Total time limit:

1000ms

Memory Limit:

65536kB

Describe

There are two rings composed of characters. Please write a program that calculates the length of the longest contiguous public string on the token of the two characters. For example, the strings "ABCEFAGADEGKABUVKLM" are joined together to form a ring; the string "Madjkluvkl" is joined together to form a second ring; "Uvklma" is a contiguous public string of these two rings.

Input

A row that contains two strings that correspond to one character token. The two strings are separated by a single space. The string length does not exceed 255, and it does not contain whitespace such as spaces.

Output

Outputs an integer that represents the length of the longest public string on the token of the two words.

Sample input

ABCEFAGADEGKABUVKLM MADJKLUVKL

Sample output

6

Ideas:

Dynamic return;

Come on, on the code:

#include <cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespacestd;intlen_1,len_2,ans,cur_ans=0, dp[520][520];BOOLci[520][520];stringword_1,word_2;intMain () {CIN>>word_1>>word_2; Len_1=word_1.length (), len_2=word_2.length (); Ans=min (len_1,len_2); //for (int i=0;i<len_1;i++) word_1[i+len_1]=word_1[i]; //for (int i=0;i<len_2;i++) word_2[i+len_2]=word_2[i];word_1+=word_1,word_2+=word_2; Len_2*=2, len_1*=2;  for(intI=0; i<len_1;i++)    {         for(intj=0; j<=len_2;j++)        {            if(Word_1[i]==word_2[j]) ci[i+1][j+1]=true, dp[i+1][j+1]++; }    }     for(intI=1; i<=len_1;i++)    {         for(intj=1; j<=len_2;j++)        {            if(Ci[i][j]) {if(ci[i-1][j-1]) Dp[i][j]=max (dp[i][j],dp[i-1][j-1]+1); Cur_ans=Max (cur_ans,dp[i][j]); }}} ans=min (Ans,cur_ans); printf ("%d\n", ans); return 0;}

AC Diary--Character token Openjudge 1.7 30