UVA 11044 - Searching for Nessy（数学）_searching for nessy uva

本文链接：https://blog.youkuaiyun.com/coco430/article/details/9531613

本文探讨了通过声纳探测技术寻找传说中的尼斯湖水怪的方法。文章提出了一种算法，该算法能够计算出在忽略边缘区域的情况下，用最少数量的声纳设备实现对整个尼斯湖网格的有效覆盖。

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Searching for Nessy

The Loch Ness Monsteris a mysterious and unidentified animal said to inhabit Loch Ness,
a large deep freshwater loch near the city of Inverness in northern Scotland. Nessie is usually categorized as a type of lake monster.
http://en.wikipedia.org/wiki/Loch_Ness_Monster

In July 2003, the BBC reported an extensive investigation of Loch Ness by a BBC team, using 600 separate sonar beams, found no trace of any ¨sea monster¨ (i.e., any large animal, known or unknown) in the loch. The BBC team concluded that Nessie does not exist. Now we want to repeat the experiment.

Given a grid of n rows and m columns representing the loch, 6 $\le$ n, m $\le$ 10000, find the minimum number s of sonar beams you must put in the square such that we can control every position in the grid, with the following conditions:

one sonar occupies one position in the grid; the sonar beam controls its own cell and the contiguous cells;
the border cells do not need to be controlled, because Nessy cannot hide there (she is too big).

For example,

$\textstyle \parbox{.5\textwidth}{\begin{center}\mbox{}\epsfbox{p11044.eps}\end{center}}$ $\textstyle \parbox{.49\textwidth}{\begin{center}\mbox{}\epsfbox{p11044a.eps}\end{center}}$

$\epsfbox{p11044b.eps}$

where X represents a sonar, and the shaded cells are controlled by their sonar beams; the last figure gives us a solution.

Input

The first line of the input contains an integer, t, indicating the number of test cases. For each test case, there is a line with two numbers separated by blanks, 6 $\le$ n, m $\le$ 10000, that is, the size of the grid (n rows and m columns).

Output

For each test case, the output should consist of one line showing the minimum number of sonars that verifies the conditions above.

Sample Input

Sample Output

=====================================

最外面一圈可以不管，中间要全部覆盖，一个声纳可以覆盖周围9个格子，问最少需要多少声纳

横（竖）除以3即为横（竖）方向上最少的数目，因为最外一圈不用管，所以余1余2都可以，即使余数为0声纳数目也不能再少了。最后直接 (a/3) * (b/3) 就是答案。

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int t,a,b;
    cin>>t;
    while(t--)
    {
        cin>>a>>b;
        printf("%d\n",a/3*(b/3));
    }
    return 0;
}