codeforces979A Pizza, Pizza, Pizza!!

Shiro在她的生日派对上需要将一个圆形比萨均匀地切成n+1份,以便她和她的朋友们都能享用到相同大小和形状的比萨片。本篇介绍了如何根据参与派对的朋友数量确定最少的切割次数。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.

Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.

She has ordered a very big round pizza, in order to serve her many friends. Exactly n

of Shiro's friends are here. That's why she has to divide the pizza into n+1

slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.

Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.

As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?

Input

A single line contains one non-negative integer n

(0n1018) — the number of Shiro's friends. The circular pizza has to be sliced into n+1

pieces.

Output

A single integer — the number of straight cuts Shiro needs.

Examples
Input
Copy
3
Output
Copy
2
Input
Copy
4
Output
Copy
5
Note

To cut the round pizza into quarters one has to make two cuts through the center with angle 90

between them.

To cut the round pizza into five equal parts one has to make five cuts.



情况一:没有朋友,需要1块,切0刀。

情况二:朋友为奇数,需要偶数n块,切n/2刀。

情况三:朋友为偶数,需要奇数n块,切n刀。

注意溢出,使用long long。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>

using namespace std;


int main(){

        long long n;
        cin>>n;
        n=n+1;
        long long ans;
        if(n==1){
            ans=0;
            cout<<ans<<endl;
            return 0;
        }
        if(n%2==0){
            ans=n/2;
        }else{
            ans=n;
        }

        cout<<ans<<endl;


    return 0;
}


### 关于 Codeforces Problem 1804A 的解决方案 Codeforces 是一个广受欢迎的在线编程竞赛平台,其中问题 1804A 可能涉及特定算法或数据结构的应用。尽管未提供具体题目描述,但通常可以通过分析输入输出样例以及常见解法来推导其核心逻辑。 #### 题目概述 假设该问题是关于字符串处理、数组操作或其他基础算法领域的内容,则可以采用以下方法解决[^2]: 对于某些初学者来说,遇到不熟悉的语言(如 Fortran),可能会感到困惑。然而,在现代竞赛环境中,大多数情况下会使用更常见的语言(C++、Python 或 Java)。因此,如果题目提及某种神秘的语言,可能只是为了增加趣味性而非实际需求。 #### 解决方案思路 以下是基于一般情况下的潜在解答方式之一: ```cpp #include <bits/stdc++.h> using namespace std; int main(){ int t; cin >> t; // 输入测试用例数量 while(t--){ string s; cin >> s; // 获取每组测试数据 // 假设这里需要执行一些简单的变换或者判断条件... bool flag = true; // 初始化标志位为真 for(char c : s){ if(c != 'a' && c != 'b'){ flag = false; break; } } cout << (flag ? "YES" : "NO") << "\n"; // 输出结果 } return 0; } ``` 上述代码片段展示了一个基本框架,适用于许多入门级字符串验证类问题。当然,这仅作为示范用途;真实场景下需依据具体要求调整实现细节。 #### 进一步探讨方向 除了官方题解外,社区论坛也是获取灵感的好地方。通过阅读他人分享的经验教训,能够加深对该类型习题的理解程度。同时注意积累常用技巧并灵活运用到不同场合之中[^1]。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值