CodeForces - 1610A Anti Light‘s Cell Guessing

Posted 2022-04-16 海岛Blog

A. Anti Light’s Cell Guessing
time limit per test1 second
memory limit per test256 megabytes

You are playing a game on a n×m grid, in which the computer has selected some cell (x,y) of the grid, and you have to determine which one.

To do so, you will choose some k and some k cells (x1,y1),(x2,y2),…,(xk,yk), and give them to the computer. In response, you will get k numbers b1,b2,…bk, where bi is the manhattan distance from (xi,yi) to the hidden cell (x,y) (so you know which distance corresponds to which of k input cells).

After receiving these b1,b2,…,bk, you have to be able to determine the hidden cell. What is the smallest k for which is it possible to always guess the hidden cell correctly, no matter what cell computer chooses?

As a reminder, the manhattan distance between cells (a1,b1) and (a2,b2) is equal to |a1−a2|+|b1−b2|.

Input
The first line of the input contains a single integer t (1≤t≤10⁴) — the number of test cases. The description of test cases follows.

The single line of each test case contains two integers n and m (1≤n,m≤10⁹) — the number of rows and the number of columns in the grid.

Output
For each test case print a single integer — the minimum k for that test case.

Example
input
2
2 3
3 1
output
2
1
Note
In the first test case, the smallest such k is 2, for which you can choose, for example, cells (1,1) and (2,1).

Note that you can’t choose cells (1,1) and (2,3) for k=2, as both cells (1,2) and (2,1) would give b1=1,b2=2, so we wouldn’t be able to determine which cell is hidden if computer selects one of those.

In the second test case, you should choose k=1, for it you can choose cell (3,1) or (1,1).

问题链接：CodeForces - 1610A Anti Light’s Cell Guessing
问题简述：（略）
问题分析：（略）
AC的C++语言程序如下：

/* CodeForces - 1610A Anti Light's Cell Guessing */

#include <bits/stdc++.h>

using namespace std;

int main()

    int t;
    scanf("%d", &t);
    while (t--) 
        int n, m;
        scanf("%d%d", &n, &m);
        if (n == 1 && m == 1)
            printf("0\\n");
        else
            printf("%c\\n", min(n, m) == 1 ? '1' : '2');
    

    return 0;