#include <bits/stdc++.h>//打表存所有必败态
using namespace std;
int a[1000005];
int main() {
    memset(a, 0, sizeof(a));
    a[0] = 0;
    for (int i = 1, k = 1; i <= 1000000; ++i)
        if (a[i] == 0) {
            a[i] = i + k;
            if (i + k <= 1000000) a[i + k] = -1;
            k++;
        }
    int T; scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        int n, m; scanf("%d%d", &n, &m); if (n > m) swap(n, m);
        if (a[n] == m) puts("Farmer John");
        else puts("Bessie");
    }
    return 0;
}

#include <bits/stdc++.h>//打表存所有必败态（unordered_map版）
using namespace std;
unordered_map<int, int> a;
int main() {
    a[0] = 0;
    for (int i = 1, k = 1; i <= 1000000; ++i)
        if (a.find(i) == a.end()) {
            a[i] = i + k;
            a[i + k] = i;
            k++;
        }
    int T; scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        int n, m; scanf("%d%d", &n, &m);
        puts(a[n] == m ? "Farmer John" : "Bessie");
    }
    return 0;
}

用公式，当 $n = \lfloor \frac{ \sqrt(5) + 1 }{2} \times (m - n) \rfloor$ 时，$(n, m)$ 为必败态

#include <bits/stdc++.h>
using namespace std;
int main() {
    int T; scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        int n, m; scanf("%d%d", &n, &m); if (n > m) swap(n, m);
        double ans = (sqrt(5.0) + 1) * 0.5 * (m - n);
        if (n == (int)ans) puts("Farmer John");
        else puts("Bessie");
    }
    return 0;
}