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  1. #include <iostream>
  2. #include <vector>
  3. #include <algorithm>
  4. using namespace std;
  5.  
  6. int main() {
  7.     int N;
  8.     cin >> N;
  9.     vector<int> soldiers(2 * N);
  10.     vector<pair<int, int>> pos(N + 1, {-1, -1});
  11.  
  12.     for (int i = 0; i < 2 * N; i++) {
  13.         cin >> soldiers[i];
  14.         if (pos[soldiers[i]].first == -1) {
  15.             pos[soldiers[i]].first = i;
  16.         } else {
  17.             pos[soldiers[i]].second = i;
  18.         }
  19.     }
  20.  
  21.     long long defenseValue = 0;
  22.  
  23.     for (int num = 1; num <= N; num++) {
  24.         int left = pos[num].first;
  25.         int right = pos[num].second;
  26.  
  27.         vector<int> between;
  28.         for (int i = left + 1; i < right; i++) {
  29.             between.push_back(soldiers[i]);
  30.         }
  31.  
  32.         sort(between.begin(), between.end());
  33.  
  34.         // 計算有多少不同的數字在between中出現至少兩次
  35.         int cnt = 0;
  36.         for (int i = 0; i < between.size(); ) {
  37.             if (between[i] > num) {
  38.                 int j = i;
  39.                 while (j < between.size() && between[j] == between[i]) j++;
  40.                 if (j - i >= 2) cnt++;
  41.                 i = j;
  42.             } else {
  43.                 i++;
  44.             }
  45.         }
  46.         defenseValue += cnt;
  47.     }
  48.  
  49.     cout << defenseValue << endl;
  50.     return 0;
  51. }
Success #stdin #stdout 0.01s 5288KB
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https://ideone.com/MOzruk
language:
C++14 (gcc 8.3)
created:
20 hours ago
visibility:
public

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