Solution: 题解 CF1029F Multicolored Markers

令(c=a+b)

一个整数边矩形的边长一定是矩形面积的因数

于是这个问题等价于求最大(ileqslantsqrt c,i|c)，使得有(fleqslant i,f|a)且(frac{a}{f}leqslant frac{c}{i})，或有(gleqslant i,g|b)且(frac{b}{g}leqslant frac{c}{i})，于是输出(2(i+frac{c}{i}))

既然是求最大(i)，就让(i)从(lfloorsqrt c floor)开始往下枚举

不失一般性，可以假设(fleqslantsqrt a,gleqslantsqrt b)

那么如果矩形(a)可以嵌入，就必有(fleqslant i)且(frac{a}{f}leqslant frac{c}{i})

鉴于(f)与(frac{a}{f})在(fgeqslant 1)时负相关，可以把(f)设为(i)，然后向下寻找(a)的最大因数，判断(frac{a}{f})是否可行

另外因为外循环中(i)是下降的，所以(f)在每次循环开始时应设为(min(f,i))

(g)同理

Time complexity: (O(sqrt a))

Memory complexity: (O(1))

细节见代码（(1.46)s / (12.00)KB）

//This program is written by Brian Peng.
#pragma GCC optimize("Ofast","inline","no-stack-protector")
#include<bits/stdc++.h>
#define int long long
using namespace std;
#define Rd(a) (a=read())
#define Gc(a) (a=getchar())
#define Pc(a) putchar(a)
int read(){
	int x;char c(getchar());bool k;
	while(!isdigit(c)&&c^'-')if(Gc(c)==EOF)exit(0);
	if(c^'-')k=1,x=c&15;else k=x=0;
	while(isdigit(Gc(c)))x=(x<<1)+(x<<3)+(c&15);
	return k?x:-x;
}
void wr(int a){
	if(a<0)Pc('-'),a=-a;
	if(a<=9)Pc(a|'0');
	else wr(a/10),Pc((a%10)|'0');
}
signed const INF(0x3f3f3f3f),NINF(0xc3c3c3c3);
long long const LINF(0x3f3f3f3f3f3f3f3fLL),LNINF(0xc3c3c3c3c3c3c3c3LL);
#define Ps Pc(' ')
#define Pe Pc('
')
#define Frn0(i,a,b) for(int i(a);i<(b);++i)
#define Frn1(i,a,b) for(int i(a);i<=(b);++i)
#define Frn_(i,a,b) for(int i(a);i>=(b);--i)
#define Mst(a,b) memset(a,b,sizeof(a))
#define File(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
int a,b,c,f,g;
signed main(){
	c=Rd(a)+Rd(b),f=sqrt(a),g=sqrt(b);
	Frn_(i,sqrt(c),1)if(!(c%i)){
		f=min(f,i),g=min(g,i);
		while(a%f)--f;
		if(a/f<=c/i)wr((i+c/i)<<1),exit(0);
		while(b%g)--g;
		if(b/g<=c/i)wr((i+c/i)<<1),exit(0);
	}
	exit(0);
}