Codeforces Round #506 (Div. 3) 1029 F. Multicolored Markers

CF-1029F

题意：

　　a，b个小正方形构造一个矩形，大小为（a+b），并且要求其中要么a个小正方形是矩形，要么b个小正方形是矩形。

思路：　

　　之前在想要分a，b是否为奇数讨论，后来发现根本不需要。只用枚举（a+b）大小的矩形的边长，并暴力判断（注意暴力判断的顺序）能否成立，更新答案。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <iomanip>
#include   <cstdlib>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include    <cctype>
#include     <queue>
#include     <cmath>
#include      <list>
#include       <map>
#include       <set>
//#include <unordered_map>
//#include <unordered_set>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
//__gnu_pbds::cc_hash_table<int,int>ret[11];    //这是很快的hash_map
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFFLL;  //2147483647
const ll nmos = 0x80000000LL;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18

const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/
            ll a,b, sum;
            bool check(ll x,ll y){
                for(ll i=x; i>=1; i--){
                    if(a%i == 0 && a/i <= y)return true;
                    if(b%i == 0 && b/i <= y)return true; 
                }  
                return false;
            }

int main(){
            
            cin>>a>>b;
            sum = a + b;
            ll ans = inff;

            for(ll i=1ll; i * i <= sum; i++){
                 if(sum % i == 0){
                    ll x = i, y = sum/i;
                    if(check(x,y))
                    {
                        ans = min(ans, 2ll*(x + y));
                    }
                }
            }
            cout<<ans<<endl;
            return 0;       
}