Even? Odd? G

最新推荐文章于 2025-01-23 12:28:34 发布

原创最新推荐文章于 2025-01-23 12:28:34 发布 · 374 阅读

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Description

Bessie’s cruel second grade teacher has assigned a list of N (1 <= N <= 100) positive integers I (1 <= I <= 10^60) for which Bessie must determine their parity (explained in second grade as ‘Even… or odd?’). Bessie is overwhelmed by the size of the list and by the size of the numbers. After all, she only learned to count recently.

Write a program to read in the N integers and print ‘even’ on a single line for even numbers and likewise ‘odd’ for odd numbers.

POINTS: 25

Bessie那惨无人道的二年级老师搞了一个有 N 个正整数 I 的表叫Bessie去判断“奇偶性”（这个词语意思向二年级的学生解释，就是“这个数是单数，还是双数啊？”）。Bessie被那个表的长度深深地震惊到了，竟然跟栋栋的泛做表格一样多道题！！！毕竟她才刚刚学会数数啊。

写一个程序读入N个整数，如果是双数，那么在单立的一行内输出"even"，如果是单数则类似地输出"odd".

数据范围：每个正整数不超过 10601060。

Input

* Line 1: A single integer: N

* Lines 2…N+1: Line j+1 contains I_j, the j-th integer to determine even/odd

Output

* Lines 1…N: Line j contains the word ‘even’ or ‘odd’, depending on the parity of I_j

Sample 1

Inputcopy	Outputcopy
2 1024 5931	even odd

Hint

Two integers: 1024 and 5931

1024 is eminently divisible by 2; 5931 is not

思路：

本题由于输入的数很大，用字符串来输入，判断这个数是奇数还是偶数，只需要判断最后一个数是奇数还是偶数。

代码：

#include <iostream>

using namespace std;

int main()
{
int n;
cin >>n;
while(n--)
{
string s;
cin >>s;
if((s[s.size()-1] - '0') % 2 == 0) cout <<"even"<<endl;
else cout <<"odd"<<endl;
}
return 0;
}

本题比较简单，知道一个数是奇数还是偶数，只看最后一个数字就可以判断