hdu 4355 Party All the Time 三分

Problem Description
In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebrate the harvest every year. But there is one thing you may not know that they hate walking so much that they would prefer to stay at home if they need to walk a long way.According to our observation,a spirit weighing W will increase its unhappyness for S3*W units if it walks a distance of S kilometers.
Now give you every spirit’s weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.

Input
The first line of the input is the number T(T<=20), which is the number of cases followed. The first line of each case consists of one integer N(1<=N<=50000), indicating the number of spirits. Then comes N lines in the order that x[i]<=x[i+1] for all i ( 1 < = i < N). The i-th line contains two real number : Xi,Wi, representing the location and the weight of the i-th spirit. ( |xi|<=106, 0< wi<15 )

Output
For each test case, please output a line which is “Case #X: Y”, X means the number of the test case and Y means the minimum sum of unhappyness which is rounded to the nearest integer.

Sample Input
1
4
0.6 5
3.9 10
5.1 7
8.4 10

Sample Output
Case #1: 832

Author
Enterpaise@UESTC_Goldfinger

Source
2012 Multi-University Training Contest 6

设该点坐标为x,,sum=（fab(x-x1))^3*w+（fab(x-x2))^3*w+（fab(x-x3))^3*w+…+（fab(x-xn))^3*w
对这个式子进行2次求导有..6(fab(x-xi))+1..恒大于0,,因此不满意度呈凹性。所以用3分。。
注意0.f已经四舍五入了。。不需要所谓的 rounded to the nearest integer.了。。（被坑一次）
fab(x)要耗时。。提出来。。。(少用）

#include<cstdio>#include<cstring>#include<algorithm>#include<cstdlib>#include<cmath>#define EPS 1e-5using namespace std;int n;double a[50005];double s[50005];double man(double x){    double sum=0.0;    for(int i=1;i<=n;i++)    {        double t=fabs(x-a[i]);        sum+=s[i]*t*t*t;    }    return sum;}int main(){    int T;    scanf("%d",&T);    for(int t=1;t<=T;t++)    {        scanf("%d",&n);        for(int i=1;i<=n;i++) scanf("%lf %lf",&a[i],&s[i]);        double l=a[0];        double r=a[n];        while(r-l>=EPS)        {            double d1=l+(r-l)/3.0;            double d2=r-(r-l)/3.0;            if(man(d1)<man(d2)) r=d2;            else l=d1;        }       printf("Case #%d: %.0f\n",t,man(l+(r-l)/3.0));    }    return 0;}