萌新求助关于 F 题

#include <bits/stdc++.h>
using namespace std;
int number[100];
int ss[314514];
bool cmp(int a, int b){
  return a > b;
}
int main() {
  string s; cin >> s;
  int cnt = 0;
  for (int i = 0; i < s.size(); ++i) {
    cnt += (s[i] - '0');
   // cout << cnt << ' ';
    //cout << s[i] << ' ';
    number[s[i] - '0']++;
    ss[i + 1] = (s[i] - '0');
  //  cout << ss[i];
  }
  // cout << cnt;
//   for (int i = 1; i <= s.size(); ++i) {
//     cout << ss[i];
//   }
 // cout << "nmsl";
  if (cnt % 3 != 0) {
    cout << -1;
    return 0;
  }
  int A = 0, B = 0, C = 0;
  bool flg = false;
  for (int i = 0; i <= 9; ++i){
    for (int j = 0; j <= 9; ++j){
      for (int k = 0; k <= 9; ++k){
        if ((i*100 + j*10 + k) % 125 == 0  && number[i] && number[j] && number[k]){
          if (i == j) if (number[i] < 2) break;
          if (i == k) if (number[i] < 2) break;
          if (j == k) if (number[j] < 2) break;
          if (i == j && j == k) if (number[i] < 3) break;
          A = i, B = j, C = k;
          flg = true; goto exit;
        }
      }
    }
  }
  exit:;
  if (!flg) {
    cout << -1;return 0;
  }
  int n = s.size();
  bool AA = 0, BB = 0, CC = 0;
  for (int i = 1; i <= n; ++i) {
    if (ss[i] == A && !AA) ss[i] = 10, AA = 1;
    else if (ss[i] == B && !BB) ss[i] = 10, BB = 1;
    else if (ss[i] == C && !CC) ss[i] = 10, CC = 1;
  }
  // for (int i = 1; i <= n; ++i) cout << ss[i];
  sort(ss + 1, ss + n + 1, cmp);
//   for (int i = 1; i <= n; ++i) {
//     cout << ss[i] << ' ';
//   }
  for (int i = 4; i <= n; ++i) {
    cout << ss[i] ;
  }
  cout << A <<  B << C << endl;
}

思路大概就是先判断能不能被 整除，然后如果可以就判断能不能取到 个数组成一个数，使其能被 整除。

结果 /kel