求n位数字构成的第k个回文数

int main() {
    int T;
    cin>>T;
    while(T--){
        int n, k;
        cin>>n>>k;
        int temp;
        if(n%2==0){
            temp = pow(10, n/2-1);
            temp += k-1;
            string out = to_string(temp), out2(out.rbegin(), out.rend());
            cout<