FZU 2146 Easy Game

思路：水题

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);string s;int ca, t=1;scanf("%d", &ca); getchar();while(ca--){cin>>s;if(s.size()%2) printf("Case %d: Odd\n", t++);else printf("Case %d: Even\n", t++);}        return 0;}

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

4wellthisisthesimplestprobleminthiscontest

Sample Output

Case 1: EvenCase 2: OddCase 3: OddCase 4: Odd

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