공부하자/Codility
[Codility] Lesson3. FrogJmp (C#)
ceste
2020. 11. 17. 16:16
문제: app.codility.com/programmers/lessons/3-time_complexity/frog_jmp/
FrogJmp coding task - Learn to Code - Codility
Count minimal number of jumps from position X to Y.
app.codility.com
if (X >= 1 && X <= 1000000000 && Y >= 1 && Y <= 1000000000 && D >= 1 && D <= 1000000000 && X <= Y)
{
double calc = (Y - X) / D;
if ((Y - X) % D == 0)
return (int)calc;
else
return (int)calc + 1;
}
return 0;