[hdu4355]Party All the Time(三分)

解题关键：三分法裸题，一开始一定要保证函数为凸函数或者凹函数。

#include<bits/stdc++.h>
#define eps 1e-6
using namespace std;
int n;
double x[50010],w[50010];
double fuc(double a){//用pow会超时 
    double d,ans=0;
    for(int i=1;i<=n;i++){
        d=fabs(x[i]-a);
        ans+=d*d*d*w[i];
    }
    return ans;
}
double sanfen(){
    double m1,m2,l=-1e6+0.0,r=1e6+0.0;
    while(r-l>eps){
        m1=(l+r)/2;
        m2=(m1+r)/2;
        if(fuc(m1)<fuc(m2))r=m2;
        else l=m1;
    }
    return r;
} 

int main(){
    int t,tt=1;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%lf%lf",x+i,w+i);
        double ans=sanfen();
        if(n==1) printf("Case #%d: %.0f\n",tt++,0.0);
        else printf("Case #%d: %.0f\n",tt++,fuc(ans));
    }
    return 0;
}