Light OJ 1020

简单推理题；

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int T, n;
    string Name;
    cin >> T;
    for(int kase = 1; kase <= T; ++kase)
    {
        cin >> n >> Name;
        if(Name == "Bob")
        {
            printf("Case %d: %s\n",kase, n % 3 ? "Bob" : "Alice");
        }
        else
        {
            printf("Case %d: %s\n",kase, n % 3 == 1 ? "Bob" : "Alice");
        }
    }
    return 0;
}