Acdream 1111:LSS（水题，字符串处理）

LSS

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others)

Problem Description

Time flies, four years passed, colleage is over. When I am about to leave, a xuemei ask me an ACM problem, but I can't solve it, I am 功力尽失. Please help me so that I won't lose face in front of xuemei!

Give you a string , you should find the longest substring which is of the same character.

Input

First line there is a T , represents the test cases.

next T lines will be T strings.

the length of every string is less than 100

all the characters of the strings will be lowercase letters

Output

for each test case output a number

Sample Input

1

a

Sample Output

Source

*qi

Manager

*qi

　　水题，求最长连续相同字符子串长度。

　　就这么道破题纠结了我和旭旭一晚上，啊啊啊，脑子抽抽了，真不应该。无法很好的理解题意是大问题啊。这道题坑就坑在题目中提到的“字串”的含义应该是连续的，而我理解的字串应该是不连续的啊！像最长公共字串问题里面的字串不就是不连续的吗，有点凌乱了…… 总之卡在这么道题上，真给跪了，怨念无穷啊……

　　题意：求最长连续相同字符子串长度。

　　思路：从a[1]开始比较，每一个字符与它前面的字符比较如果相同，num加1，每次与sum比较，如果num>sum，令sum=num。最后输出sum。

　　代码：

 #include <iostream>

 using namespace std;

 int main()

 {

     int T,i;

     cin>>T;

     while(T--){

         char a[];

         cin>>a;

         //统计最长连续相同字符子串长度

         int num=,sum=;

         for(i=;a[i];i++){

             if(a[i]==a[i-])

                 num++;

             else

                 num=;

             if(num>sum)

                 sum=num;

         }

         cout<<sum<<endl;

     }

     return ;

 }

Freecode : www.cnblogs.com/yym2013

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