Disport with Jelly

数字x左边有x - L个数字，右边有R - x个数字。
可以看成有两堆石子，分别有x - L个、R - x个。
玩家每次可以取多个或者取完，但至少取一个。
至此就是一个nim博弈。
考虑异或和等于0即可。
或者考虑对称博弈，每次先手取多少，对手跟着在另一边取多少，因为两边的数目一样，所以
当x - L = R - x时一定是后手Bob赢。

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

#ifdef LOCAL
#define debug(x) cout << "[" __FUNCTION__ ": " #x " = " << (x) << "]\n"
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
#define hash_ 1000000009
#define Continue(x) { x; continue; }
#define Break(x) { x; break; }
const int mod = 998244353;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define gc p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;
inline int read(){ static char buf[1000000], *p1 = buf, *p2 = buf; register int x = false; register char ch = gc; register bool sgn = false; while (ch != '-' && (ch < '0' || ch > '9')) ch = gc; if (ch == '-') sgn = true, ch = gc; while (ch >= '0'&& ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc; return sgn ? -x : x; }
ll fpow(ll a, int b, int mod) { ll res = 1; for (; b > 0; b >>= 1) { if (b & 1) res = res * a % mod; a = a * a % mod; } return res; }

int main()
{
#ifdef LOCAL
    freopen("E:/input.txt", "r", stdin);
#endif
    int kcase = 0;
    int t;
    cin >> t;
    while (t--)
    {
        int x, L, R;
        cin >> L >> R >> x;
        L = x - L;
        R = R - x;
        int flag = 0;
        if (L == R)
            ;
        else
            if (L == 0 || R == 0)
                flag = 1;
            else
                if (L != R)
                    flag = 1;
        printf("Case #%d: %s\n", ++kcase, flag ? "Alice" : "Bob");
    }
    return TIME;
}