zoukankan      html  css  js  c++  java
  • fzu 2146 Easy Game

    http://acm.fzu.edu.cn/problem.php?pid=2146

     Problem 2146 Easy Game

    Accept: 661    Submit: 915
    Time Limit: 1000 mSec    Memory Limit : 32768 KB

     Problem Description

    Fat brother and Maze are playing a kind of special (hentai) game on a string S. Now they would like to count the length of this string. But as both Fat brother and Maze are programmers, they can recognize only two numbers 0 and 1. So instead of judging the length of this string, they decide to judge weather this number is even.

     Input

    The first line of the date is an integer T, which is the number of the text cases.

    Then T cases follow, each case contains a line describe the string S in the treasure map. Not that S only contains lower case letters.

    1 <= T <= 100, the length of the string is less than 10086

     Output

    For each case, output the case number first, and then output “Odd” if the length of S is odd, otherwise just output “Even”.

     Sample Input

    4
    well
    thisisthesimplest
    problem
    inthiscontest

     Sample Output

    Case 1: Even
    Case 2: Odd
    Case 3: Odd
    Case 4: Odd
     
    分析:
    就是字符串数组长度奇偶判断,一个strlen()函数解决。
    AC代码:
     1 #include <stdio.h>
     2 #include <algorithm>
     3 #include <iostream>
     4 #include <string.h>
     5 #include <string>
     6 #include <math.h>
     7 #include <stdlib.h>
     8 #include <queue>
     9 #include <stack>
    10 #include <set>
    11 #include <map>
    12 #include <list>
    13 #include <iomanip>
    14 #include <vector>
    15 #pragma comment(linker, "/STACK:1024000000,1024000000")
    16 #pragma warning(disable:4786)
    17 
    18 using namespace std;
    19 
    20 const int INF = 0x3f3f3f3f;
    21 const int MAX = 10086 + 10;
    22 const double eps = 1e-8;
    23 const double PI = acos(-1.0);
    24 char str[MAX];
    25 
    26 int main()
    27 {
    28     int T ;
    29     while(~scanf("%d",&T))
    30     {
    31         int first = 1;
    32         getchar();
    33         while(T --)
    34         {
    35             scanf("%s",str);
    36             int len = strlen(str);
    37             printf("Case %d: ", first ++);
    38             if(len % 2)
    39                 printf("Odd
    ");
    40             else
    41                 printf("Even
    ");
    42         }
    43     }
    44     return 0;
    45 }
    View Code
     
  • 相关阅读:
    刨根问底 | Elasticsearch 5.X集群多节点角色配置深入详解【转】
    ElasticSearch 内存那点事【转】
    Zookeeper之Zookeeper的Client的分析【转】
    Zookeeper之Zookeeper底层客户端架构实现原理(转载)
    elasticsearch 性能调优
    ElasticSearch性能优化策略【转】
    elasticsearch中 refresh 和flush区别【转】
    我理解的朴素贝叶斯模型【转】
    (转)Intellij IDEA 快捷键整理
    使用Mongo dump 将数据导入到hive
  • 原文地址:https://www.cnblogs.com/jeff-wgc/p/4449343.html
Copyright © 2011-2022 走看看