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  1. //SOLUTIE 2_0_3_3 OPTIMIZARE DUPA SFATURI MARIAN CALL 26-11-2020
  2. //2021-11-04
  3.  
  4.  
  5.  
  6. #include <iostream>
  7.  
  8. using namespace std;
  9.  
  10. int main() {
  11.     int n, m, mt[501][501], k;
  12.  
  13.     cin >> n >> m;
  14.     for(int i = 1; i <= n; i++){
  15.         for(int j = 1; j <= m; j++){
  16.             cin >> mt[i][j];
  17.         }
  18.     }
  19.     cin >> k;
  20.     int x, i;
  21.     for(int p = 1; p <= k; p++){
  22.         cin >> x;
  23.  
  24.         if (x < mt[1][1] || x > mt[n][m]) {
  25.             cout << "0NU\n";
  26.             //break;
  27.         } else { // mt[1][1] <= x && x <= mt[n][m] ...val x se incadreaza in interv de valori ale elem matricei
  28.             //int i, j;
  29.             if (x - mt[1][1] <= mt[n][m] - x) { // x e mai aproape de coltul din stg sus
  30.                 i = 1;
  31.                 //int left = 1, right = m;
  32.                 while (mt[i][1] <= x && x <= mt[i][m]) {
  33.                     int left = 1, right = m;
  34.                     while (left < right) {
  35.                         int mid = (left + right) / 2;
  36.                         if (mt[i][mid] < x) {
  37.                             left = mid + 1;
  38.                         } else { // x <= mt[i][mid]
  39.                             right = mid;
  40.                         }
  41.                     }
  42.                     cout << "x= " << x << " ";
  43.                     if (mt[i][left] == x) {
  44.                         cout << "1DA\n";
  45.                         //break;
  46.                     } else {
  47.                         cout << "1NU\n";
  48.                         //break;
  49.                     }
  50.                     if (i < n) {
  51.                         i++;
  52.                     }
  53.                 }
  54.             } else { // x - mt[1][1] > mt[n][m] - x ...adica x e mai aproape de coltul din dr jos
  55.                 i = n;
  56.                 while (mt[i][1] <= x && x <= mt[i][m]) {
  57.                     int left = 1, right = m;
  58.                     while (left < right) {
  59.                         int mid = (left + right) / 2;
  60.                         if (mt[i][mid] < x) {
  61.                             left = mid + 1;
  62.                         } else { // x <= mt[i][mid]
  63.                             right = mid;
  64.                         }
  65.                     }
  66.                     cout << "x= " << x << " ";
  67.                     if (mt[i][left] == x) {
  68.                         cout << "2DA\n";
  69.                         //break;
  70.                     } else {
  71.                         cout << "2NU\n";
  72.                         //break;
  73.                     }
  74.                     if (i > 1) {
  75.                         i--;
  76.                     }
  77.                 }
  78.             }
  79.         }
  80.  
  81.     }
  82.     return 0;
  83. }
Success #stdin #stdout 0.01s 5348KB
comments (?)
stdin
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3   3
1 	2 	3
2 	5 	6
5 	6 	7
5
2   4   10   7   5
stdout
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x= 2 1DA
x= 2 1DA
0NU
x= 7 2DA
x= 5 2DA
x= 5 2DA
https://ideone.com/h7SNbZ
language:
C++ (gcc 8.3)
created:
2 years ago
visibility:
public

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