AtCoder Beginner Contest 112 D - Partition

被C题坑了，就没有再看D题，还是比较有意思的。

Problem Statement

You are given integers N and M.

Consider a sequence aa of length N consisting of positive integers such that a1+a2+...+aN = M. Find the maximum possible value of the greatest common divisor of a1,a2,...,aN.

Constraints

• All values in input are integers.
• 1≤N≤10^5
• N≤M≤10^9

思路：假设最大公约数为g,则M=tg(存在t为正整数)，a1=t1g,a2=t2g...an=tng（存在t1,t2...tn为正整数），于是只需要找出来能被M整除结果大于N的最大数即可。 

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=m/n;i>=1;i--)
	{
		if(m%i==0)
		{
			int t=m/i;
			if(t>=n)
			{
				cout<<i<<endl;
				break;
			}
		}
	}
}