Think:
確實是很 EASY 的簽到題。。。。
判斷字符串長度 是奇數或者偶數 然后輸出就好了

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

#include<stdio.h> #include<string.h> int main() {char str[123456];int d;int T;int kase = 0;while(~scanf("%d",&T)){while(T --){scanf("%s", str);d = strlen(str);printf("Case %d: ", ++ kase);if (d % 2 == 0)printf("Even\n");elseprintf("Odd\n");memset(str, 0, sizeof(str));}} }

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