题解：首先完成数据的读入，然后高斯消元求秩按题意解即可

#pragma GCC optimize(2, 3, "Ofast", "inline")
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100;using matrix = double[maxn][maxn];
using vect = array<double, maxn>;const double eps = 1e-8;
int row[maxn], var[maxn];
int one_possible(matrix A, int n, int m, vect &ans)
{memset(row, -1, sizeof(row));int r = 0;for (int c = 0; c < m && r < n; ++c){int x = r;for (int i = x + 1; i < n; ++i)if (fabs(A[i][c]) > fabs(A[x][c]))x = i;if (x != r)for (int j = 0; j <= m; ++j)swap(A[x][j], A[r][j]);if (fabs(A[r][c]) < eps)continue;for (int k = r + 1; k < n; ++k)for (int j = m; j >= c; --j)A[k][j] -= A[k][c] / A[r][c] * A[r][j];row[c] = r++;}for (int i = r; i < n; ++i)if (fabs(A[i][m]) > eps)return -1;for (int c = m - 1; c >= 0; --c){int x = row[c];if (x < 0)ans[c] = 0;else{for (int i = x - 1; i >= 0; --i)A[i][m] -= A[i][c] / A[x][c] * A[x][m];ans[c] = A[x][m] / A[x][c];}}return r;
}
unordered_map<string,int> Yuansu;
int cnt=0;
void work_in(matrix &M,string s,int index)
{string yuansu="";double num=0;for(int i=0;i<s.size();){while(isalpha(s[i])){// cout<<s[i]<<endl;yuansu+=s[i];i++;}while(isdigit(s[i])){num=num*10+(s[i]-'0');i++;}if(Yuansu.find(yuansu)==Yuansu.end()){Yuansu[yuansu]=cnt++;}// M[Yuansu[yuansu]][0]=1;M[Yuansu[yuansu]][index]=num;// cout<<yuansu<<": "<<Yuansu[yuansu]<<" :"<<num<<endl;num=0;yuansu="";}// print_M(M,cnt,m);
}
void print_M(matrix M,int n,int m)
{for(int i=0;i<n;i++){for(int j=0;j<=m;j++){cout<<M[i][j]<<" ";}cout<<endl;}
}
void work()
{int q;cin>>q;while(q--){matrix M={0};cnt=0;Yuansu.clear();int m;cin>>m;string s1;for(int j=0;j<m;j++){cin>>s1;work_in(M,s1,j);}for(int j=0;j<cnt;j++){M[j][m]=0;}// print_M(M,cnt,m);vect ans;int r=one_possible(M,cnt,m,ans);if(r==-1){cout<<"N"<<endl;continue;}if(m-r>0){cout<<"Y"<<endl;}else{cout<<"N"<<endl;}// cout<<"r,m:"<<r<<" "<<m<<endl;}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);work();return 0;
}