zoukankan      html  css  js  c++  java
  • hdu-5578 Friendship of Frog(暴力)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5578

    Friendship of Frog

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 556    Accepted Submission(s): 355


    Problem Description
    N frogs from different countries are standing in a line. Each country is represented by a lowercase letter. The distance between adjacent frogs (e.g. the 1st and the 2nd frog, the N1th and the Nth frog, etc) are exactly 1. Two frogs are friends if they come from the same country.

    The closest friends are a pair of friends with the minimum distance. Help us find that distance.
     
    Input
    First line contains an integer T, which indicates the number of test cases.

    Every test case only contains a string with length N, and the ith character of the string indicates the country of ith frogs.

     1T50.

     for 80% data, 1N100.

     for 100% data, 1N1000.

     the string only contains lowercase letters.
     
    Output
    For every test case, you should output "Case #x: y", where x indicates the case number and counts from 1 and y is the result. If there are no frogs in same country, output 1 instead.
     
    Sample Input
    2 abcecba abc
     
    Sample Output
    Case #1: 2
    Case #2: -1
    AC代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    char str[1009];
    int main()
    {
        int t;
        scanf("%d",&t);
        int cnt=1;
        while(t--)
        {
            scanf("%s",str);
            int len=strlen(str);
            int ans=20000,flag=0;
            for(int i=0;i<len;i++)
            {
                for(int j=i+1;j<len;j++)
                {
                    if(str[j]==str[i])
                    {
                        ans=min(ans,j-i);
                        flag=1;
                        break;
                    }
                }
            }
            if(!flag)ans=-1;
            printf("Case #%d: %d
    ",cnt++,ans);
        }
        return 0;
    }
  • 相关阅读:
    Mysql Explain 详解【转】
    Windows下Gradle安装与配置
    MYSQ创建联合索引,字段的先后顺序,对查询的影响分析
    给.Net 5 Api增加JwtBearer认证
    纸壳CMS 3.3.6发布升级.Net 5
    PL/SQL Developer登入时候报ORA-12638: 身份证明检索失败的解决办法
    Mybatis 日志工厂
    Mybatis 配置解析
    Mybatis 完成增删改查
    Mybatis 简介
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5260119.html
Copyright © 2011-2022 走看看