Codeforces 50A(往矩形里填多米诺骨牌)

最新推荐文章于 2021-09-21 08:36:00 发布

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探讨如何在给定尺寸的矩形板上放置尽可能多的标准2x1多米诺骨牌，确保每块骨牌完全覆盖两个方格且不重叠。

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You are given a rectangular board of M × N squares. Also you are given an unlimited number of standard domino pieces of 2 × 1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:

1. Each domino completely covers two squares.

2. No two dominoes overlap.

3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.

Find the maximum number of dominoes, which can be placed under these restrictions.

Input

In a single line you are given two integers M and N — board sizes in squares (1 ≤ M ≤ N ≤ 16).

Output

Output one number — the maximal number of dominoes, which can be placed.

Examples

input
2 4

output
4

input
3 3

output
4

题目大意：

输入MxN大小的矩形，已知多米诺骨牌大小为2x1（可以看做M行N列，多米诺牌大小占2行1列）。求能往矩形里放最多多米诺牌的数量是多少？

解题思路：

令a=M/2，a表示一列里可以放的多米诺牌数量;令sum=a*N，sum表示N列中放的多米诺牌的总数；如果是奇数行，则还需判断最后一行中可放入

多米诺牌的数量，将其加入sum即可求得最多米诺牌总数。

代码实现：

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

int main() {
int N, M; //表示N行M列
int sum=0,a=0;

    while (~scanf("%d%d", &N, &M)) {
       a = N / 2;                                                       //一列里可以放的塔罗牌数
       sum = a * M;                                                 // 乘上总列数
       if (N % 2 == 1) //如果奇数行
           sum += M / 2;                                           //加上这一行中牌的数量
       printf("%d\n", sum);
   }

return 0;
}

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