A. Fake NP
time limit per test  

1 second

memory limit per test  

256 megabytes

 

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.

Input

The first line contains two integers l and r (2 ≤ l ≤ r ≤ 109).

Output

Print single integer, the integer that appears maximum number of times in the divisors.

If there are multiple answers, print any of them.

 
input
19 29
output
2
input
3 6
output
3
Note:

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}.

题目大意:

     给定一个范围,输出一个整数x,(使得这个范围内能整除x的整数个数最多)

解题思路:

     刚开始写的时候没想辣么多,处理超级复杂,之后仔细想想其实只有两种情况

     1、当l和r相同时任意输出一个。  2、当l和r不同时输出2(所有的偶数都能被2整除)

AC代码:

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
using namespace std; int main ()
{
int l,r;
while (~scanf("%d %d",&l,&r))
{
if (l == r)
printf("%d\n",l);
else
printf("2\n");
}
return ;
}

Codeforces Round #411 A. Fake NP的更多相关文章

  1. Codeforces Round #411 (Div. 2)(A,B,C,D 四水题)

    A. Fake NP time limit per test:1 second memory limit per test:256 megabytes input:standard input out ...

  2. Codeforces Round #411 (Div. 2)

    来自FallDream的博客,未经允许,请勿转载,谢谢. 由于人傻又菜 所以这次又滚去div2了  一堆结论题真的可怕 看见E题不是很有思路   然后就去大力搞F题  T了最后一个点 真的绝望   但 ...

  3. Codeforces Round #411 (Div. 2) 【ABCDE】

    A. Fake NP 题意:给你l,r,让你输出[l,r]里面除1以外的,出现因子数量最多的那个数. 题解:如果l==r输出l,否则都输出2 #include<bits/stdc++.h> ...

  4. Codeforces Round #411 (Div. 2) A-F

    比赛时候切了A-E,fst了A Standings第一页只有三个人挂了A题,而我就是其中之一,真™开心啊蛤蛤蛤 A. Fake NP time limit per test 1 second memo ...

  5. Codeforces Round #411 div2

    A. Fake NP 题意:询问一个区间[L,R]出现次数最多的正整数因子(>1). 一个区间内一个因子P出现次数大概为[R/P]-[(L-1)/P],约等于(R-L+1)/P,P取2时最优.注 ...

  6. Codeforces Round 411 Div.2 题解

    A Fake NP standard input/output s, MB Submit Add to favourites x3673 B -palindrome standard input/ou ...

  7. Codeforces Round #411 (Div. 1) D. Expected diameter of a tree

    题目大意:给出一个森林,每次询问给出u,v,问从u所在连通块中随机选出一个点与v所在连通块中随机选出一个点相连,连出的树的直径期望(不是树输出-1).(n,q<=10^5) 解法:预处理出各连通 ...

  8. Codeforces Round #411 div 2 D. Minimum number of steps

    D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input st ...

  9. Codeforces Round #411 (Div. 2) C. Find Amir

    C. Find Amir time limit per test   1 second memory limit per test   256 megabytes   A few years ago ...

随机推荐

  1. c# Config配置文件读写

    using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.C ...

  2. Oracle批量插入数据SQL语句太长出错:无效的主机/绑定变量名

    Oracle数据库,用mybatic批量插入数据: <insert id="saveBatch" parameterType="io.renren.entity.N ...

  3. 一些排序 (python实现)

    ►快速排序 <时间复杂度O(n㏒n)> def partition(li,left,right): tmp = li[left] while left < right: while ...

  4. 矩阵&&高斯消元

    矩阵运算: \(A\times B\)叫做\(A\)左乘\(B\),或者\(B\)右乘\(A\). 行列式性质: \(1.\)交换矩阵的两行(列),行列式取相反数. \(2.\)某一行元素都\(\ti ...

  5. jenkins 重置构建历史

    item = Jenkins.instance.getItemByFullName("98")//THIS WILL REMOVE ALL BUILD HISTORYitem.bu ...

  6. Pytorch学习笔记(二)——Tensor

    一.对Tensor的操作 从接口的角度讲,对Tensor的操作可以分为两类: (1)torch.function (2)tensor.function 比如torch.sum(a, b)实际上和a.s ...

  7. css提取数据2个常用方法

    提取标签里的内容 所谓数据就是HTML里标签的内容,如下面红色字体,就是标签内容 <title>我只是个实验 - SCRAPY</title> 提取标签里的数据,标签可以是ti ...

  8. oracle模糊查询提高效率的方法

    转载:https://blog.csdn.net/weixiaohuai/article/details/83513957 https://blog.csdn.net/chihen/article/d ...

  9. 2.7 Go channel

    channel简单示例 package main import ( "fmt" "time" ) //channel的创建,发送,接收 func channe1 ...

  10. JRebel idea 热部署

    下载链接:https://pan.baidu.com/s/1CUvEarKVARJF46LJ2W90Lg 提取码:s46w 下载完以后是个ZIP(jr-ide-intellij-7.1.5_13-17 ...