C程序，用于在给定长度下从2D平面的原点到形式(d，0)的点所需的跳数

给定三个正整数a，b和d 。您当前位于无限2D坐标平面上的原点(0，0)。您可以从当前位置以等于或等于a或b的欧几里得距离跳到2D平面上的任何点。任务是找到从(0，0)达到(d，0)所需的最小跳数。

例子：

Input : a = 2, b = 3, d = 1 
Output : 2
First jump of length a = 2, (0, 0) -> (1/2, √15/2)
Second jump of length a = 2, (1/2, √15/2) -> (1, 0)
Thus, only two jump are required to reach 
(1, 0) from (0, 0).

Input : a = 3, b = 4, d = 11 
Output : 3
(0, 0) -> (4, 0) using length b = 4
(4, 0) -> (8, 0) using length b = 4
(8, 0) -> (11, 0) using length a = 3

#include 
using namespace std;
  
// Return the minimum jump of length either a or b
// required to reach (d, 0) from (0, 0).
int minJumps(int a, int b, int d)
{
    // Assigning maximum of a and b to b
    // and assigning minimum of a and b to a.
    int temp = a;
    a = min(a, b);
    b = max(temp, b);
  
    // if d is greater than or equal to b.
    if (d >= b)
        return (d + b - 1) / b;
  
    // if d is 0
    if (d == 0)
        return 0;
  
    // if d is equal to a.
    if (d == a)
        return 1;
  
    // else make triangle, and only 2
    // steps required.
    return 2;
}
  
int main()
{
    int a = 3, b = 4, d = 11;
    cout << minJumps(a, b, d) << endl;
    return 0;
}

输出：

3

有关更多详细信息，请参阅完整文章：给定长度以达到从2D平面原点到形式(d，0)的点所需的跳数。

想要从精选的最佳视频中学习和练习问题，请查看《基础知识到高级C的C基础课程》。