Codeforces Round #511 (Div. 1) T2 Little C Loves 3 II

最新推荐文章于 2020-11-01 17:23:50 发布

转载最新推荐文章于 2020-11-01 17:23:50 发布 · 136 阅读

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原文链接：http://www.cnblogs.com/forever-/p/9736068.html

题目

Little C loves number «3» very much. He loves all things about it.
Now he is playing a game on a chessboard of size $\times m$ . The cell in the $x$ -th row and in the $y$ -th column is called $(x, y)$ . Initially, The chessboard is empty. Each time, he places two chessmen on two different empty cells, the Manhattan distance between which is exactly $3$ . The Manhattan distance between two cells $x_i,y_i)$ and $x_j,y_j)$ is defined as $x_i-x_j|+|y_i-y_j|$ .
He want to place as many chessmen as possible on the chessboard. Please help him find the maximum number of chessmen he can place.
Input
A single line contains two integers $n$ and $m$ ( $\leq n,m \leq 10^9$ ) — the number of rows and the number of columns of the chessboard.
Output
Print one integer — the maximum number of chessmen Little C can place.
Examples
inputCopy
2 2
outputCopy
0
inputCopy
3 3
outputCopy
8
Note
In the first example, the Manhattan distance between any two cells is smaller than $3$ , so the answer is $0$ .
In the second example, a possible solution is $(1, 1) (3, 2)$ , $(1, 2) (3, 3)$ , $(2, 1) (1, 3)$ , $(3, 1) (2, 3)$ .

和史前巨佬ldx还有dzy神仙一起做的

这道题…

可以说是打表吧

首先我们可以观察到，对于任意一个 $1 * 6$ 或者 $2 * 4$ 的格子，都是可以填满的，

那也就说如果有一边长能被6或者4给整除，那就是可以填满的

然后对于5*5以下的了，我们可以直接打表预处理出来（因为有一些特殊情况吧）

然后。。

就被2*7给hack了

然后特判了一个2*7就过了

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int tr[5][5]={ {0,0,0,2,4},
			   {0,0,4,8,10},
			   {0,4,8,12,14},
			   {2,8,12,16,18},
			   {4,10,14,18,24}};
int main() {
	int n,m;
	cin>>n>>m;
	if(n<m) swap(n,m);
	ll ans=0;
	if(m==1)ans+=n/6*6,n%=6,ans+=tr[0][n-1];
	else if(n%4==0 || m%4==0)ans=(ll)n*m;
	else if(n%6==0 || m%6==0)ans=(ll)n*m;
	else if(n<=5 && m<=5)ans=tr[n-1][m-1];
	else if(n==7 && m==2)ans=12;else ans=(ll)n*m/2*2;
	cout<<ans<<'\n';
}

反正dzy神仙带我飞,什么都不怕

转载于:https://www.cnblogs.com/forever-/p/9736068.html