问题 A: THE FOOL

最新推荐文章于 2022-07-28 20:20:02 发布

原创最新推荐文章于 2022-07-28 20:20:02 发布 · 508 阅读

0 ·

CC 4.0 BY-SA版权

小小思维题专栏收录该内容

1 篇文章

订阅专栏

博客围绕一个数值奇偶判断问题展开，给出时间、内存限制及提交、解决情况。描述了题目，包括输入输出要求，最后给出解题思路，即通过打表观察端点值，归纳出判断数值开根的奇偶来确定原数奇偶。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

时间限制: 1 Sec 内存限制: 128 MB
提交: 322 解决: 148
[提交] [状态] [命题人:admin]

题目描述

The Fool is numbered 0 the number of unlimited potential -and therefore does not have a specific place in the sequence of the Tarot cards . The Fool can be placed either at the beginning of the Marjor Arcam or at the end. The Major Arcana is often considered as the Fool's journey through life and as such, he is ever present and therefore needs no number.

Given n∈N+, print the parity of

where

输入

The first line of the input contains one integer T≤100, denoting the number of testcases. Then T testcases follow.
In each of the T testcases, there is a positive number n≤109

输出

For each testcase, print a single line starting with “Case i:”( i indicates the case number) and then "even" or “odd’', separated with a single space.

样例输入

复制样例数据

样例输出

Case 1: odd
Case 2: even
Case 3: even

思路：直接打表，看出0为偶，1-3为奇，4-8为偶，9-15为奇......观察端点值，归纳出判断数值开跟的奇偶，即可判原数奇偶；

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        int n;
        scanf("%d",&n);
        int tmp=sqrt(n);
        if(tmp%2==1) printf("Case %d: odd\n",i);
        else printf("Case %d: even\n",i);
    }
    return 0;
}