Alice and Bob are playing a game with marbles; you may have played this game in childhood. The game is playing by alternating turns. In each turn a player can take exactly one or two marbles.
Both Alice and Bob know the number of marbles initially. Now the game can be started by any one. But the winning condition depends on the player who starts it. If Alice starts first, then the player who takes the last marble looses the game. If Bob starts first, then the player who takes the last marble wins the game.
Now you are given the initial number of marbles and the name of the player who starts first. Then you have to find the winner of the game if both of them play optimally.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case contains an integer n (1 ≤ n < 231) and the name of the player who starts first.
Output
For each case, print the case number and the name of the winning player.
Sample Input
Output for Sample Input
3
1 Alice
2 Alice
3 Bob
Case 1: Bob
Case 2: Alice
Case 3: Alice
直接找循环节
/*************************************************************************
> File Name: LightOJ1020.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年06月09日 星期二 18时46分32秒
************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
int main() {
int t, icase = 1;
scanf("%d", &t);
while (t--) {
int n;
string s;
cin >> n >> s;
n %= 3;
cout << "Case " << icase++ << ": ";
if (s[0] == ‘A‘) {
if (n == 1) {
printf("Bob\n");
}
else {
printf("Alice\n");
}
}
else {
if (n == 0) {
printf("Alice\n");
}
else {
printf("Bob\n");
}
}
}
return 0;
}