牛客网暑期ACM多校训练营（第五场） max

链接：https://www.nowcoder.com/acm/contest/143/G
来源：牛客网
 

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

Give two positive integer c, n. You need to find a pair of integer (a,b) satisfy 1<=a,b<=n and the greatest common division of a and b is c.And you need to maximize the product of a and b

输入描述:

The first line has two positive integer c,n

输出描述:

Output the maximum product of a and b.

If there are no such a and b, just output -1

示例1

输入

复制

2 4

输出

复制

8

说明

a=2,b=4

备注:

1<=c,n<=10^9

#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<set>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#define LL long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define lson o<<1
#define rson o<<1|1
using namespace std;
int main()
{
	LL  c, n;
	cin >> c >> n;
	if(c > n)
		puts("-1");
	else if(n < c * 2)
		printf("%lld\n", c * c);
	else
		printf("%lld\n", (n / c) * c * (n / c - 1) * c);
    return 0;
}