[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;
}