三分（1）

Party All the Time

Time Limit: 2000ms

Memory Limit: 32768KB

64-bit integer IO format: %I64d      Java class name: Main

Submit Status

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

#include<cstdio>
#include<algorithm>
using namespace std;
//const int eps = 1e-9;
const int maxn = 5e4+5;
struct node{
    double x;
    double y;//!!!
}a[maxn];
int n;
double work(double s){
    double cnt = 0;
    for(int i = 0; i < n; i++){
        double res = s - a[i].x;
        cnt += fabs(res*res*res)*a[i].y;//pow(res,3);TLE!
    }
    return cnt;
}
int main(){
    int T,cas = 0;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
            scanf("%lf%lf",&a[i].x,&a[i].y);
        double l = a[0].x,r = a[n-1].x;
        for(int i = 0; i < 200; i++){//or (while(l + eps < r)){}
            double lmid = (l+r)/2;
            double rmid = (lmid + r)/2;
            if(work(lmid) <= work(rmid))
                r = rmid;
            else l = lmid;
        }
　　　　//or ans = min(work(l),work(r));
        printf("Case #%d: %0.f\n",++cas,work(l));//浮点小数精度15~16位，而题中位数可能高达18位，直接输出的方法不可取，但能AC。
    }
    return 0;
}