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
1
Source
wuyiqi
Manager
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：
1 #include <iostream>
2
3 using namespace std;
4
5 int main()
6 {
7 int T,i;
8 cin>>T;
9 while(T--){
10 char a[101];
11 cin>>a;
12 //统计最长连续相同字符子串长度
13 int num=1,sum=1;
14 for(i=1;a[i];i++){
15 if(a[i]==a[i-1])
16 num++;
17 else
18 num=1;
19 if(num>sum)
20 sum=num;
21 }
22 cout<<sum<<endl;
23 }
24 return 0;
25 }
Freecode : www.cnblogs.com/yym2013
