hdu 5432 Pyramid Split【二分查找】

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本文深入探讨了一种独特的数学挑战——通过精确的几何操作将多个金字塔分割成两部分，使得每一部分的体积相等。利用剑‘TuLong’进行平面切割，作者揭示了解决该问题的关键步骤和算法，包括输入解析、体积计算以及找到平均切割平面的高度。通过实例分析和代码实现，展示了如何在有限的时间内高效地解决这一问题。

Pyramid Split

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 880 Accepted Submission(s): 352

Problem Description

Xiao Ming is a citizen who's good at playing,he has lot's of gold cones which have square undersides,let's call them pyramids.

Anyone of them can be defined by the square's length and the height,called them width and height.

To easily understand,all the units are mile.Now Ming has

n pyramids,there height and width are known,Xiao Ming wants to make them again to get two objects with the same volume.

Of course he won't simply melt his pyramids and distribute to two parts.He has a sword named "Tu Long" which can cut anything easily.

Now he put all pyramids on the ground (the usdersides close the ground)and cut a plane which is parallel with the water level by his sword ,call this plane cutting plane.

Our mission is to find a cutting plane that makes the sum of volume above the plane same as the below,and this plane is average cutting plane.Figure out the height of average cutting plane.

Input

First line:

T, the number of testcases.

(1≤T≤100)

Then

T testcases follow.In each testcase print three lines :

The first line contains one integers

n(1≤n≤10000), the number of operations.

The second line contains

n integers

A1,…,An(1≤i≤n,1≤Ai≤1000) represent the height of the

ith pyramid.

The third line contains

n integers

B1,…,Bn(1≤i≤n,1≤Bi≤100) represent the width of the

ith pyramid.

Output

For each testcase print a integer - **the height of average cutting plane**.

(the results take the integer part,like 15.8 you should output 15)

Sample Input

2
2
6 5
10 7
8
702 983 144 268 732 166 247 569
20 37 51 61 39 5 79 99

Sample Output

1
98

这个题AC的我好伤心、我的思路是每一个正四棱锥都找到那个高度，然后加在一起除n得到结果，但是最终还是因为精度问题。无限的wa、

最后找到各路大牛的思路，从整体找到这个高度，就能AC了：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#define EPS 1e-5
using namespace std;
double x[11000];
double h[11000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        double r=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%lf",h+i);
            r=max(r,h[i]);
        }
        double v=0;//这里在写的时候，忘记=0，奇葩的是编译器竟然没有爆数据、、然后一直在wa啊wa
        for(int j=0;j<n;j++)
        {
            scanf("%lf",&x[j]);
            v+=x[j]*x[j]*h[j]/3;
        }
        double l=0;
        double mid;
        while(r-l>EPS)
        {
            mid=(l+r)/2;
            double vv=0;
            for(int j=0;j<n;j++)
            {
                if(h[j]>mid)
                vv+=(h[j]-mid)*(h[j]-mid)*(h[j]-mid)*x[j]*x[j]/h[j]/h[j]/3;
            }
            if(vv*2>v)
            l=mid;
            else
            r=mid;
        }
        printf("%d\n",(int)(l+EPS));
    }
    return 0;
}