题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3714
题意：求n个二次函数在[0,1000]的最小值。
三分枚举。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 10100; 5 const double eps = 1e-9; 6 double a[maxn], b[maxn], c[maxn]; 7 int n; 8 9 double f(double x) { 10 double ret = -1000000000; 11 for(int i = 1; i <= n; i++) { 12 double y = a[i]*x*x+b[i]*x+c[i]; 13 ret = max(ret, y); 14 } 15 return ret; 16 } 17 18 double ts() { 19 double lo = 0, hi = 1000.0; 20 double ret1, ret2; 21 while(hi - lo >= eps) { 22 double mid = (lo + hi) / 2.0; 23 double mimid = (mid + hi) / 2.0; 24 ret1 = f(mid); 25 ret2 = f(mimid); 26 if(ret1 > ret2) lo = mid; 27 else hi = mimid; 28 } 29 return ret1; 30 } 31 32 int main() { 33 //freopen("in", "r", stdin); 34 int T; 35 scanf("%d", &T); 36 while(T--) { 37 scanf("%d", &n); 38 for(int i = 1; i <= n; i++) { 39 scanf("%lf %lf %lf", &a[i], &b[i], &c[i]); 40 } 41 printf("%.4lf ", ts()); 42 } 43 return 0; 44 }