CF - 794B. Cutting Carrot - 数学_follow carrot算法-优快云博客

解决如何将一个等腰三角形（象征胡萝卜）通过n-1条垂直于高的直线等分为n个相同面积的部分的问题。

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1.题目描述：

Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area.

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

$\text{[math]}$ Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold.

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $\text{[math]}$ .

   Examples
 
     input
   
3 2

     output
   
1.154700538379 1.632993161855

     input
   
2 100000

     output
   
70710.678118654752

Note

Definition of isosceles triangle: https://en.wikipedia.org/wiki/Isosceles_triangle.

2.题意概述：

给你一个等腰三角形，要你用n-1条垂直于高的直线将它们分成n等份，问这些直线的坐标。

3.解题思路：

既然是n等份，即每份的面积为总面积的1/n，因为三角形更好算，考虑从上往下分，前i份组成的三角形所占面积为i/n，即Si/Sn=i/n。

而由相似三角形容易得Si/Sn=Xi^2/n^2，那么再利用三角形面积公式就出来了。

PS:开始没想到那个相似三角形公式，而是推了一遍带tan的数，可能是爆精度了，样例都没过QAQ

4.AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define lson root << 1
#define rson root << 1 | 1
#define lent (t[root].r - t[root].l + 1)
#define lenl (t[lson].r - t[lson].l + 1)
#define lenr (t[rson].r - t[rson].l + 1)
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
double n, h;
double ans[1001];
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int a,b;
	while (~scanf("%d%d", &a, &b))
	{
		for (int i = 1; i < a; i++)
		{
			double temp = i;
			if (i == 1)
				printf("%.10f", b*sqrt(temp / a));
			else
				printf(" %.10f", b*sqrt(temp / a));
		}
		puts("");
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}