zoukankan      html  css  js  c++  java
  • POJ3913

    此题比较简单!!不多说!!



                                                                                            Gnome Sequencing
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 1526   Accepted: 1022

    Description

    In the book All Creatures of Mythology, gnomes are kind, bearded creatures, while goblins tend to be bossy and simple-minded. The goblins like to harass the gnomes by making them line up in groups of three, ordered by the length of their beards. The gnomes, being of different physical heights, vary their arrangements to confuse the goblins. Therefore, the goblins must actually measure the beards in centimeters to see if everyone is lined up in order. Your task is to write a program to assist the goblins in determining whether or not the gnomes are lined up properly, either from shortest to longest beard or from longest to shortest.

    Input

    The input starts with line containing a single integer N, 0 < N < 30, which is the number of groups to process. Following this are N lines, each containing three distinct positive integers less than 100.

    Output

    There is a title line, then one line per set of beard lengths. See the sample output for capitalization and punctuation.

    Sample Input

    3
    40 62 77
    88 62 77
    91 33 18

    Sample Output

    Gnomes:
    Ordered
    Unordered
    Ordered

    Source

     
     
     
     
    1. #include <stdio.h>
    2. bool is_ordered(int a, int b, int c)
    3. {   
    4.   if (a < b && b < c)    
    5.      return true;   
    6.   if (a > b && b > c)      
    7.    return true;
    8.     return false;
    9. }
    10. int main()
    11. {   
    12.   int n, a, b, c;   
    13.   scanf("%d", &n);   
    14.   printf("Gnomes:\n");   
    15.   while (n--)    
    16. {        
    17. scanf("%d %d %d", &a, &b, &c);      
    18.    if (is_ordered(a, b, c))       
    19.       printf("Ordered\n");       
    20.   else            
    21. printf("Unordered\n");
    22.     }
    23.     return 0;
    24. }
     
  • 相关阅读:
    Diffbot:开发者工具 将web内容转换成应用
    算法之道—形而之上谓之道
    css三个ppt
    SMB的NTLM认证过程与NTLM挑战的编程实现
    c++计算圆周率
    SVN总结
    struts2技术实现用户名唯一的验证处理详解
    数字常量
    二叉树的创建和遍历
    php变量的定义和作用域
  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/2859488.html
Copyright © 2011-2022 走看看