RT , 所有点全部秒RE , 本地下载的测试点全过

可以排除是变量名的原因

就是这样一份丑陋的代码

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#define enter putchar('\n')
#define space putchar(' ')
#define int long long
using namespace std;
const int inf = 0x3f3f3f3f;
void error()
{
    printf("error!!!\n");
    enter;
}
int n, m, t;
int Sentences, Complexity, Now_complexity, Num_for, Now_for, Max_complexity;
int Vis_variable[105], Now_variable[105], Complexity_variable[105];
bool Flag , Flag2;
string S1, S2;
signed main()
{
    cin >> t;
    while (t--)
    {
        Sentences = 0, Complexity = 0, Num_for = 0, Now_for = 0, Max_complexity = 0, Flag = false , Now_complexity = 0 , Flag2 = false;
        memset(Vis_variable, 0, sizeof(Vis_variable));
        memset(Now_variable, 0, sizeof(Now_variable));
        memset(Complexity_variable, 0, sizeof(Complexity_variable));
        cin >> Sentences >> S1;
        for (int i = 4; i < S1.size() - 1; i++)
            Complexity = Complexity * 10 + S1[i] - '0';
        // cout <<S1<<endl <<Complexity << endl;
        while (Sentences--)
        {
            cin >> S1;
            if (S1[0] == 'F')
            {
                Num_for++;
                cin >> S2;
                if (Vis_variable[S2[0] - 'a'])
                    Flag2 = true; //ERR
                else
                    Vis_variable[S2[0] - 'a'] = true, Now_variable[Num_for] = S1[0] - 'a';
                cin >> S1 >> S2;
                if (S1[0] != 'n' && S2[0] == 'n' && !Flag)
                    Now_complexity++, Complexity_variable[Num_for]++;
                else if ((S1.size() == S2.size() && S1 > S2) || S1.size() > S2.size() || (S1[0] == n && S2[0] != n))
                    Flag = true, Now_for = Num_for;
            }
            if (S1[0] == 'E')   
            {
                Max_complexity = max(Max_complexity, Now_complexity);
                Vis_variable[Now_variable[Num_for]] = false;
                Now_complexity -= Complexity_variable[Num_for];
                Complexity_variable[Num_for] = 0;
                Num_for--;
                if (Now_for && Num_for < Now_for)
                    Flag = false, Now_for = 0 , Flag2 = false;
            }
            
        }
        if (Num_for > 0 || Flag2) //too much fors
        {
            // if(Num_for)
            //     cout << "Num_for : " << Num_for << endl;
            // else
            //     cout << "Flag2 : " << Flag2 << endl;
            printf("ERR\n");
            continue;
        }
        if (!Num_for && Max_complexity == Complexity)
            printf("Yes\n");
        if (!Num_for && Max_complexity != Complexity)
            printf("No\n");
    }
    return 0;
}
/*
1
2 O(n^2)
F a 60 n
E
*/

QwQ