楼梯最大高度

Question

给定一个表示正方形块的整数 A。每个方块的高度为 1。任务是使用这些方块创建一个最大高度的楼梯。第一个楼梯只需要一个街区，第二个楼梯需要两个街区，依此类推。查找并返回楼梯的最大高度。

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您的提交因以下输入而失败：A : 92761

您的函数返回以下内容：65536
预期返回值：430

Approach:
We are interested in the number of steps and we know that each step Si uses exactly Bi number of bricks. We can represent this problem as an equation:
n * (n + 1) / 2 = T (For Natural number series starting from 1, 2, 3, 4, 5 …)
n * (n + 1) = 2 * T
n-1 will represent our final solution because our series in problem starts from 2, 3, 4, 5…

Now, we just have to solve this equation and for that we can exploit binary search to find the solution to this equation. Lower and Higher bounds of binary search are 1 and T.

代码

public int solve(int A) {
        int l=1,h=A,T=2*A;
        while(l<=h)
        {
            int mid=l+(h-l)/2;
            if((mid*(mid+1))==T)  
                return mid;
            if((mid*(mid+1))>T && (mid!=0 && (mid*(mid-1))<=T) )
                return mid-1;
            if((mid*(mid+1))>T)
                h=mid-1;
            else
                l=mid+1;
        }
        return 0;
    }

Answer 1

扩展 Matt Timmermans 的评论：

您知道对于n 步骤，您需要(n * (n + 1))/2 块。你想知道，如果给定B 块，你可以创建多少个步骤。

所以你有：

(n * (n + 1))/2 = B
(n^2 + n)/2 = B
n^2 + n = 2B
n^2 + n - 2B = 0

这看起来有点像你会使用 quadratic formula 的东西。

在这种情况下，a=1、b=1 和 c=(-2B)。将数字代入公式：

n = ((-b) + sqrt(b^2 - 4*a*c))/(2*a)
  = (-1 + sqrt(1 - 4*1*(-2B)))/(2*a)
  = (-1 + sqrt(1 + 8B))/2
  = (sqrt(1 + 8B) - 1)/2

所以如果你有 5050 个区块，你会得到：

n = (sqrt(1 + 40400) - 1)/2
  = (sqrt(40401) - 1)/2
  = (201 - 1)/2
  = 100

试试quadratic formula calculator。使用 1 作为 a 和 b 的值，并将 c 替换为给定块数的两倍的负数。所以在上面的例子中，c 是 -10100。

在您的程序中，由于您不能有部分步骤，因此您需要截断结果。

Answer 2

您为什么要使用所有这些公式？一个简单的while() 循环应该可以解决问题，最终，它只是一个简单的高斯和..

public static int calculateStairs(int blocks) {
    int lastHeight = 0;
    int sum = 0;
    int currentHeight = 0; //number of bricks / level
    while (sum <= blocks) {
        lastHeight = currentHeight;
        currentHeight++;
        sum += currentHeight;
    }
    return lastHeight;
}

Answer 3

所以这应该可以完成工作，因为它还返回预期值。如果我错了，请纠正我。

public int solve(int blocks) {
    int current; //Create Variables
    for (int x = 0; x < Integer.MAX_VALUE; x++) { //Increment until return
      current = 0; //Set current to 0

        //Implementation of the Gauss sum
        for (int i = 1; i <= x; i++) { //Sum up [1,*current height*]
            current += i;
        } //Now we have the amount of blocks required for the current height

        //Now we check if the amount of blocks is bigger than
        // the wanted amount, and if so we return the last one
        if (current > blocks) {
            return x - 1;
        }
    }
    return current;
}