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MariusPure

skaiciavimas laiko

Apr 29th, 2019
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C++ 0.15 KB | None | 0 0
  1.     for (int i = 0; i < n; i++)
  2.     {
  3.         val[i] = (m[i].pabV * 60 + m[i].pabM) - (m[i].pradV * 60 + m[i].pradM);
  4.         min[i] = val[i] % 60;
  5.         val[i] /= 60;
  6.     }
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