tips

vector<int> v;
int (int x){
    return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
int main(){
 for(int i = 1;i <= n;i++){
      v.push_back(a[i]);
   }
    sort(v.begin, v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
    a[i] = getid(a[i]);
}

for(int i = 1;i <= n;i++){
            tmp[i] = s[i];
}
sort(tmp + 1,tmp + n + 1);
int len = unique(tmp + 1,tmp + 1 + n) - tmp - 1;
for(int i = 1;i <= n;i++){
  s[i] = lower_bound(tmp + 1,tmp + len + 1,s[i]) - tmp;
}

__builtin_popcount(n) 返回1的二进制中有多少个1
next_permutation(a + 1, a + n + 1);