也是超简单

#include<bits/stdc++.h>
using namespace std;
#define int long long
int T , C;
int f(int t , int x) {
    //我们以f(t,x) = t/2作为开端
    if (t + t / 2 + 1 <= T) {
        return f(t + t / 2 + 1, x);
    }
    else {
        return t / 2 - (T - t);
    }
}

//第t轮结束，下标为x
int g(int t , int x) {
    if (t == 0 && x == 0) {
        return 0;
    }
    if (x > t / 2) {
        return x;
    }
    else if (x == t/2){
        return g(t - 1,0);
    }
    else {
        //t--,x++
        //t + x 是一个定值
        //已知递归到 x == t / 2
        int tot = (t + x);
        int nxt_x = tot / 3;
        int nxt_t = tot - nxt_x;
        if (nxt_x == nxt_t / 2) {
            return g(nxt_t, nxt_x);
        }
        else {
            return g(nxt_t - 1 , nxt_x + 1);
        }
    }
}
signed main() {
    int q;
    cin >> q;
    for (int i = 1 ; i <= q ; i ++) {
        int op;
        cin >> op;
        if (op == 1) {
            cin >> T >> C;
            if (C > T / 2) {
                cout << C << endl;
            }
            else if (3 * C > T) {
                //从 t / 2 == c 开始，每一轮都会减去，一共减去c轮
                cout << C - (T - 2 * C + 1) << endl;
            }
            else {
                cout << f(3 * C  ,C) << endl;
            }
        }
        else {
            cin >> T >> C;
            cout << g(T , C) << endl;
        }
    }
}//就是按照题目要求一步一步算就能出结果