Codeforces Gym 100500E IBM Chill Zone

Codeforces Gym 100500E IBM Chill Zone

A relaxing, fun way to unwind nightly with old and new friends at the ACM-ICPC World Finals is to stop by the IBM Chill Zone! A great way to participate in interactive games and interesting conversation with innovative IBMers and attendees from all over the world, the IBM Chill Zone is always a favorite for all. Many games can be found in the chill zone, including board game, human-size chess board, real-life angry birds, and stones game.
The stones game was invented by one of ICPC world finalists. It a 2-player game. It consists of a line of n stones and at each move a player should remove k consecutive stones, and if the player can not make any moves he loses, for example if we have n = 6, k = 2 (stones are represented by ’*’, and empty spaces by ’.’):
First player now has no valid moves, so he loses, but note that he did not play optimally here.

Given n, and k assume both players play optimally well, determine if the first player is losing or winning.

Input
The first line will be the number of test cases T. Each of the following T lines will contain 2 numbers n, k.

1≤T ≤100 1 ≤ n ≤ 50 1≤k≤n

Output
For each test case print a single line containing: Case_x:_y
x is the case number starting from 1.

y is either ’Winning’, or ’Losing’ without the quotes (winning if the first is winning, losing otherwise) Replace underscores with spaces.

Examples

Input

2 

5 2 

5 3

Output

Case 1: Losing

Case 2: Winning

题意：给连续n个石子，两人轮流拿石子，每次必须取连续的k个石子，在最优策略下询问先手胜负。

做法：SG函数，根据题意写出对应的SG函数，注意VIS数组开在SG转移里头，保证所有的状态之间不会互相干扰。

/*written by tomriddly*/#include <bits/stdc++.h>using namespace std;int n, m;int f[55];inline int sg(const int &x){    //printf("%d %d\n", x, f[x]);    if (f[x] != -1)        return f[x];    bool vis[50*50 + 5];    memset(vis, false, sizeof(vis));    for (int i = 1; i + m - 1 <= x; i++)    {        // printf("%d: %d\n", x, sg(i - 1) ^ sg(x - i - m + 1));        vis[sg(i - 1) ^ sg(x - i - m + 1)] = true;    }    int i = 0;    while (vis[i])        i++;    return f[x] = i;}int main(){    int fuck;    scanf("%d", &fuck);    for (int cas = 1; cas <= fuck; cas++)    {        memset(f, -1, sizeof(f));        scanf("%d%d", &n, &m);        for (int i = 0; i < m; i++)            f[i] = 0;        if (sg(n))            printf("Case %d: Winning\n", cas);        else            printf("Case %d: Losing\n", cas);    }    return 0;}