HDU 4355
思路：三分法求f(x)极值。
f(x)是指位置为x时的愤怒值之和，，是一个三次函数，且存在极值点使f(x)最小。
code：
/** @author Novicer* language : C++/C*/#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>#define INF 2147483647#define cls(x) memset(x,0,sizeof(x))#define rise(i,a,b) for(int i = a ; i <= b ; i++)using namespace std;const double eps(1e-8);typedef long long lint;const int maxn = 500000 + 5;double x[maxn];double w[maxn];int n;double f(double pos){double sum = 0.0;for(int i = 1 ; i <= n ; i++){sum += (double)pow(abs(pos – x[i]) , 3) * w[i];}return sum;}double ts(double L , double R){double ans = -1;for(int i = 1 ; i < 30 ; i++){double mid1 = (L+R)/2.0;double mid2 = (mid1 + R)/2.0;if(f(mid1) >= f(mid2))L = mid1;elseR = mid2;}ans = L;return ans;}int main(){//freopen("input.txt","r",stdin);int t ; cin >> t; int kase = 1;while(t–){cin >> n;double left = 1e7 , right = -1e7;for(int i = 1 ; i <= n ; i++){scanf("%lf%lf",&x[i],&w[i]);left = min(left , x[i]);right = max(right , x[i]);}printf("Case #%d: %.0f\n",kase++ , f(ts(left,right)));}return 0;}
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