hdu5432 Pyramid Split

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

这题用到了锥体的计算公式以及二分，二分的判断条件要注意，另外，向下取整可以用printf("%d",(int)a);或者printf("%.0f",floor(a));

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
#define maxn 10050
#define eps 1e-8
double s[maxn],w[maxn],h[maxn],v[maxn];
int n,m;
double cal(double h1)
{
    int i,j;
    double num=0,ans;
    for(i=1;i<=n;i++){
        if(h1>h[i]){
            num+=v[i];
        }
        else{
            ans=1-(h[i]-h1)*(h[i]-h1)*(h[i]-h1)/(h[i]*h[i]*h[i]);
            num+=v[i]*ans;
        }

    }
    return num;

}

int main()
{
    int i,j,T;
    double sum,maxx;
    double l,r,mid;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum=0;maxx=0;
        for(i=1;i<=n;i++){
            scanf("%lf",&h[i]);
            maxx=max(maxx,h[i]);
        }
        for(i=1;i<=n;i++){
            scanf("%lf",&w[i]);
            s[i]=w[i]*w[i];
            v[i]=s[i]*h[i]/3;
            sum+=v[i];
        }
        l=0;r=maxx;
        while(r-l>eps){
            mid=(l+r)/2;
            if(cal(mid)*2>sum){
                r=mid;
            }
            else l=mid;
        }
        printf("%d\n",(int)mid);
    }
    return 0;
}