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  • 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

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  • 原文地址:https://www.cnblogs.com/yym2013/p/3777551.html
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