There is rectangle of size 𝑁×𝑀N×M with opposite corners at (0,0)(0,0) and (𝑁,𝑀)(N,M); and a special point (𝑥+0.5,𝑦+0.5) (0≤𝑥<𝑁,0≤𝑦<𝑀)(x+0.5,y+0.5) (0≤x<N,0≤y<M).Two players play a game on the rectangle where each player takes alternate turns. In his/her turn, the player will choose a line, either 𝑥=𝑘x=k or 𝑦=𝑘y=k, such that:𝑘k is an integer;The chosen line divides the current rectangle into two non-empty parts.The part of rectangle that does not consist of the special point, is discarded for further moves.The player who cannot make a move loses. If both players play optimally, determine the number of special points such that the first player wins.Input FormatThe first line of input will contain a single integer 𝑇T, denoting the number of test cases.Each test case contains two space-separated integers 𝑁N and 𝑀M — the lengths of the sides of the rectangle.Output FormatFor each test case, print the number of special points (𝑥+0.5,𝑦+0.5)(x+0.5,y+0.5) for which the first player wins the game.Note that 0≤𝑥<𝑁0≤x<N and 0≤𝑦<𝑀0≤y<M.Constraints1≤𝑇≤1041≤T≤10 4 1≤𝑁,𝑀≤1061≤N,M≤10 6 The sum of 𝑁N as well as the sum of 𝑀M over all test cases does not exceed 10610 6 .Sample 1:InputOutput42 12 23 51 120100Explanation:Test case 11 : There are 22 possible special points (0.5,0.5)(0.5,0.5) and (1.5,0.5)(1.5,0.5). In both cases, the first player can select the line 𝑥=1x=1 on his first move, and then the second player is left with no moves. Thus, the first player wins for both the special points.Test case 22 : There are 44 possible special points, all of them end up losing for the first player.
Question
There is rectangle of size 𝑁×𝑀N×M with opposite corners at (0,0)(0,0) and (𝑁,𝑀)(N,M); and a special point (𝑥+0.5,𝑦+0.5) (0≤𝑥<𝑁,0≤𝑦<𝑀)(x+0.5,y+0.5) (0≤x<N,0≤y<M).Two players play a game on the rectangle where each player takes alternate turns. In his/her turn, the player will choose a line, either 𝑥=𝑘x=k or 𝑦=𝑘y=k, such that:𝑘k is an integer;The chosen line divides the current rectangle into two non-empty parts.The part of rectangle that does not consist of the special point, is discarded for further moves.The player who cannot make a move loses. If both players play optimally, determine the number of special points such that the first player wins.Input FormatThe first line of input will contain a single integer 𝑇T, denoting the number of test cases.Each test case contains two space-separated integers 𝑁N and 𝑀M — the lengths of the sides of the rectangle.Output FormatFor each test case, print the number of special points (𝑥+0.5,𝑦+0.5)(x+0.5,y+0.5) for which the first player wins the game.Note that 0≤𝑥<𝑁0≤x<N and 0≤𝑦<𝑀0≤y<M.Constraints1≤𝑇≤1041≤T≤10 4 1≤𝑁,𝑀≤1061≤N,M≤10 6 The sum of 𝑁N as well as the sum of 𝑀M over all test cases does not exceed 10610 6 .Sample 1:InputOutput42 12 23 51 120100Explanation:Test case 11 : There are 22 possible special points (0.5,0.5)(0.5,0.5) and (1.5,0.5)(1.5,0.5). In both cases, the first player can select the line 𝑥=1x=1 on his first move, and then the second player is left with no moves. Thus, the first player wins for both the special points.Test case 22 : There are 44 possible special points, all of them end up losing for the first player.
Solution
Para resolver este problema, debemos analizar cómo se puede jugar el juego de manera óptima y determinar en qué casos el primer jugador puede ganar. Aquí está el procedimiento paso a paso:
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Entender el juego:
- Los jugadores eligen líneas verticales u horizontales que dividen el rectángulo en dos partes no vacías.
- La parte que no contiene el punto especial se descarta.
- El jugador que no puede hacer un movimiento pierde.
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Identificar el punto especial:
- El punto especial está en la forma (x+0.5, y+0.5) donde 0 ≤ x < N y 0 ≤ y < M.
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Condiciones de victoria:
- Si el rectángulo tiene dimensiones N x M, el primer jugador gana si puede forzar al segundo jugador a no tener movimientos válidos.
- Esto ocurre si el primer jugador puede dividir el rectángulo de tal manera que el segundo jugador se quede sin opciones.
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Análisis de los movimientos:
- Si N y M son ambos impares, el primer jugador siempre puede ganar. Esto se debe a que siempre puede dividir el rectángulo en dos partes de tal manera que el segundo jugador no tenga un movimiento válido.
- Si N o M es par, el segundo jugador puede siempre replicar el movimiento del primer jugador en el otro eje, forzando al primer jugador a quedarse sin movimientos.
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Implementación:
- Para cada caso de prueba, contar los puntos especiales (x+0.5, y+0.5) donde tanto x como y son impares.
Aquí está el código que implementa esta lógica:
def count
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