题解 | Even? Odd?-牛客假日团队赛13D题
Even? Odd?
https://ac.nowcoder.com/acm/contest/1082/D
题目描述
Bessie's cruel second grade teacher has assigned a list of N (1 <= N <= 100) positive integers I (1 <= I <= 106010^{60}1060) 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
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
输入描述:
* Line 1: A single integer: N* Lines 2..N+1: Line j+1 contains IjI_jIj, the j-th integer to determine even/odd
输出描述:
* Lines 1..N: Line j contains the word 'even' or 'odd', depending on the parity of
示例1
输入
2
1024
5931
输出
even
odd
说明
Two integers: 1024 and 5931
1024 is eminently divisible by 2; 5931 is not
解答
思路：此题是判断一个数是不是偶数，但是这个数比较大（不大于10的6次方），可以选择用字符串读取该数字，然后判断最后一个字符的奇偶性就可以了；
代码：
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
char num[100];
scanf("%d", &n);
for(int i= 0; i < n; i++){
scanf("%s", num);
int len = strlen(num) - 1;
if((num[len]- '0')% 2 == 0)
printf("even\n");
else
printf("odd\n");
}
return 0;
} 来源：xiydang
，去找课程学习，在牛客上看看大家面试笔试都需要会什么，岗位有什么需求就去学什么，努力的人就一定会有收获，这句话从来都经得起考验，像我现在大三了啥也不会，被迫强行考研，炼狱难度开局，啥也不会
，找工作没希望了，考研有丝丝机会