Codeforces Round #411 (Div. 2)(A,B,C,D 四水题)
A. Fake NP
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given l and r. For all integers from l to r, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem.
The first line contains two integers l and r (2 ≤ l ≤ r ≤ 109).
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
19 29
2
3 6
3
Definition of a divisor: https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html
The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
题目链接:http://codeforces.com/contest/805/problem/A
分析:直接判断l==r就输出l,否则输出2就可以了!
下面给出AC代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int l,r; 6 scanf("%d%d",&l,&r); 7 if(l==r) 8 cout<<l<<endl; 9 else cout<<2<<endl; 10 return 0; 11 }
作 者:Angel_Kitty
出 处:https://www.cnblogs.com/ECJTUACM-873284962/
关于作者:阿里云ACE,目前主要研究方向是Web安全漏洞以及反序列化。如有问题或建议,请多多赐教!
版权声明:本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文链接。
特此声明:所有评论和私信都会在第一时间回复。也欢迎园子的大大们指正错误,共同进步。或者直接私信我
声援博主:如果您觉得文章对您有帮助,可以点击文章右下角【推荐】一下。您的鼓励是作者坚持原创和持续写作的最大动力!
欢迎大家关注我的微信公众号IT老实人(IThonest),如果您觉得文章对您有很大的帮助,您可以考虑赏博主一杯咖啡以资鼓励,您的肯定将是我最大的动力。thx.
我的公众号是IT老实人(IThonest),一个有故事的公众号,欢迎大家来这里讨论,共同进步,不断学习才能不断进步。扫下面的二维码或者收藏下面的二维码关注吧(长按下面的二维码图片、并选择识别图中的二维码),个人QQ和微信的二维码也已给出,扫描下面👇的二维码一起来讨论吧!!!
欢迎大家关注我的Github,一些文章的备份和平常做的一些项目会存放在这里。

and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
.
浙公网安备 33010602011771号