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Ahmed_Negm

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Feb 26th, 2023
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  1. #include <bits/stdc++.h>
  2. #include <ext/pb_ds/assoc_container.hpp>
  3. #include <ext/pb_ds/tree_policy.hpp>
  4. using namespace std;
  5. using namespace __gnu_pbds;
  6. #define ll long long
  7. #define OO 2'000'000'000
  8. #define ull unsigned long long
  9. #define nl '\n'
  10. #define sz(x) (ll)(x.size())
  11. #define all(x) x.begin(),x.end()
  12. #define rall(s)  s.rbegin(), s.rend()
  13. #define getline(s) getline(cin>>ws,s)
  14. #define ceill(n, m) (((n) / (m)) + ((n) % (m) ? 1 : 0))
  15. #define pi  3.141592653589793
  16. #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
  17. #define multi_ordered_set tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
  18.  
  19.  
  20. void Fast_IO(){
  21. ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  22. // freopen("filename.in", "r", stdin);
  23. // freopen("filename.txt", "w", stdout);
  24. #ifndef ONLINE_JUDGE
  25. freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
  26. #endif
  27. }
  28.  
  29.  
  30.  
  31.  
  32. int dx[] = { 2, 1, -1, -2, -2, -1, 1, 2 };
  33. int dy[] = { 1, 2, 2, 1, -1, -2, -2, -1 };
  34.  
  35.  
  36.  
  37.  
  38. void solve(){
  39.   ll n; cin>>n;
  40.   vector<ll>v(n);
  41.   for(int i=0;i<n;i++)cin>>v[i];
  42.   ll maxi = *max_element(all(v));
  43.   ll mini = *min_element(all(v));
  44.   ll ans = 0;
  45.   if(maxi == mini){
  46.         cout<<n*(n+1)/2<<nl;
  47.         return;
  48.   }
  49.     ll idx = -1;
  50.     ll cnt = 0;
  51.     for(int i=0;i<n;i++){
  52.         if(v[i] == mini) cnt++;
  53.         if(v[i] == maxi){
  54.             idx = i;
  55.             break;
  56.         }
  57.     }
  58.     ans+= cnt *(n-idx);
  59.     cnt = 0;
  60.     idx = -1;
  61.     ll min_idx = -1;
  62.     reverse(all(v));
  63.     for(int i=0;i<n;i++){
  64.         if(v[i] == mini and idx == -1) cnt++;
  65.         if(v[i] == maxi and idx == -1){
  66.             idx = i;
  67.         }
  68.         if(v[i] == mini and idx != -1){
  69.             min_idx = i;
  70.             break;
  71.         }
  72.     }
  73.     if(min_idx == -1){
  74.         ans += cnt * (n-idx);
  75.     }else ans += cnt * (min_idx-idx);
  76.  
  77.     cout<<ans<<nl;
  78. }
  79.  
  80. int main(){
  81.     Fast_IO();
  82. int t =1;
  83. cin>>t;
  84. while(t--){
  85. solve();
  86. }
  87. return 0;
  88. }
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