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
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:
1 #include <stdio.h> 2 #include <string.h> 3 #include <math.h> 4 #include <stdlib.h> 5 #include <algorithm> 6 using namespace std; 7 8 int main () 9 { 10 int l,r; 11 while (~scanf("%d %d",&l,&r)) 12 { 13 if (l == r) 14 printf("%d ",l); 15 else 16 printf("2 "); 17 } 18 return 0; 19 }