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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. #define watch2(x , y) cerr<<(#x)<<" = "<<(x)<<"  "<<(#y)<<" = "<<(y)<<"\n"
  4. int SD(){int x; scanf("%d" , &x); return x;}
  5. const int AKA = 1e6 + 6 , OO = 1e9;
  6. int a[AKA];
  7. int main()
  8. {
  9.     int t = SD();
  10.     while(t--)
  11.     {
  12.         int n  = SD() , cnt = 0 , mnV = OO , f = 0;
  13.         for(int i = 0 ; i < n ; ++i)
  14.             a[i] = SD() , cnt += (a[i] == -1) , a[i] != -1 ? mnV = min(mnV , a[i]) : 0;
  15.         for(int i = 0 ; i < n - 1 ; ++i)
  16.             if(a[i] != -1 && a[i + 1] != -1)
  17.                 f = 1;
  18.         if(cnt == n)
  19.             puts("0 42");
  20.         else if(cnt == n - 1)
  21.             printf("0 %d\n" , mnV);
  22.         else if(f == 1)
  23.         {
  24.             int m = -1 , k = -1;
  25.             for(int i = 0 ; i < n - 1 ; ++i)
  26.                 if(a[i] != -1 && a[i + 1] != -1)
  27.                     m = max(m , abs(a[i] - a[i + 1]));
  28.             vector<int> v;
  29.             for(int i = 0 ; i < n ; ++i)
  30.                 if(a[i] == -1)
  31.                     v.push_back(m);
  32.                 else
  33.                     v.push_back(a[i]);
  34.             for(int i = 0 ; i < v.size() - 1 ; ++i)
  35.                 k = max(k , abs(v[i] - v[i + 1]));
  36.             printf("%d %d\n" , m , k);
  37.         }
  38.         else
  39.         {
  40.               vector<int> v;
  41.               for(int i = 0 ; i < n ; ++i)
  42.                     if(a[i] != -1)
  43.                         v.push_back(a[i]);
  44.               int m = -1 , k = -1;
  45.               for(int i = 0 ; i < v.size() - 1 ; ++i)
  46.                 m = max(m , (int)( (v[i] + v[i + 1])  / 2.00));
  47.               vector<int> vv;
  48.             for(int i = 0 ; i < n ; ++i)
  49.                 if(a[i] == -1)
  50.                     vv.push_back(m);
  51.                 else
  52.                     vv.push_back(a[i]);
  53.             for(int i = 0 ; i < vv.size() - 1 ; ++i)
  54.                 k = max(k , abs(vv[i] - vv[i + 1]));
  55.             printf("%d %d\n" , k , m);
  56.         }
  57.     }
  58. }
  59.  
Success #stdin #stdout 0s 4344KB
comments (?)
stdin
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1
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5 1 2 3 4 -1
stdout
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4 4
https://ideone.com/Ue0vwn
language:
C++14 (gcc 8.3)
created:
4 years ago
visibility:
public

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