Uva 11044 Searching for Nessy

最新推荐文章于 2025-05-03 20:49:46 发布

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本文探讨了一种使用声纳束探测尼斯湖水怪的方法。针对一个n行m列的网格地图，目标是最小化声纳数量，确保每个位置都能被至少一个3x3范围内的声纳覆盖。

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

题目大意：

问n*m的地图最多放多少个3*3的矩形。

题目解析：

水题，简单输出(m/3)*(n/3)就好了

#include<iostream>
#include<math.h>

using namespace std;

int main() {
	int t;
	int m,n;
	while(cin >> t) {
		while(t--) {
			cin>>m>>n;
			cout<<(m/3)*(n/3)<<endl;
		}	
	}
	return 0;
}