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Dec 13th, 2018
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  1. #include <stdio.h>
  2.  
  3. char line[1 << 10];
  4. int x[1 << 10][1 << 10], now[1 << 10];
  5.  
  6. int minChanges(int N)
  7. {
  8.  
  9. }
  10.  
  11. int main()
  12. {
  13. int i, j, test, Q, N, M;
  14.  
  15. freopen("data.in", "r", stdin);
  16. freopen("data.out", "w", stdout);
  17.  
  18. scanf("%d", &Q);
  19. for (test = 1; test <= Q; test ++)
  20. {
  21. scanf("%d%d", &N, &M);
  22. for (i = 1; i <= N; i ++)
  23. {
  24. gets(x + 1);
  25. for (j = 1; j <= M; j ++)
  26. x[i][j] = line[j] - '0';
  27. }
  28.  
  29. memset(now, 0, sizeof(now));
  30. for (i = 1; i <= N; i ++)
  31. for (j = 1; j <= M; j ++)
  32. now[i] += x[i][j];
  33. int sol = minChanges(N);
  34. if (sol == -1)
  35. {
  36. printf("Case %d: impossible\n", test);
  37. continue;
  38. }
  39. memset(now, 0, sizeof(now));
  40. for (j = 1; j <= M; j ++)
  41. for (i = 1; i <= N; i ++)
  42. now[j] += x[i][j];
  43. int sol2 = minChanges(M);
  44. if (sol2 == -1)
  45. {
  46. printf("Case %d: row %d\n", test, sol);
  47. continue;
  48. }
  49. printf("Case %d: both %d\n", test, sol + sol2);
  50. }
  51.  
  52. return 0;
  53. }
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