1. 青蛙跳跳FrogJmp Count minimal number of jumps from position X to Y.

package com.code;

public class Test03_1 {
public int solution(int X, int Y, int D) {
int res = (Y-X)/D+((Y-X)%D==0?0:1);
return res;
}

public static void main(String[] args) {
Test03_1 t03 = new Test03_1();
System.out.println(t03.solution(10, 85, 30));

}
}

/**

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position
greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

class Solution { public int solution(int X, int Y, int D); }

that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

X = 10
Y = 85
D = 30
the function should return 3, because the frog will be positioned as follows:

after the first jump, at position 10 + 30 = 40
after the second jump, at position 10 + 30 + 30 = 70
after the third jump, at position 10 + 30 + 30 + 30 = 100
Assume that:

X, Y and D are integers within the range [1..1,000,000,000];
X ≤ Y.
Complexity:

expected worst-case time complexity is O(1);
expected worst-case space complexity is O(1).
*
*
*
*/