相当于y是个常数求 F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)这个函数的最小值,令F‘ = 0,得出x,y的方程,用二分法解方程得x0(易证得x0>=0 && x0<=100),则F‘(x0) = 0,由F‘‘ 在[0-100]上恒大于0,所以F‘在[0-100]上单增,所以F‘(x)<0(x<x0),F‘(x)>0(x>x0),所以F(x)在x=x0处取得最小值,所以本题主要就是二分求解方程的x0,然后直接带入x0,y计算即可。
#include<cstdio> #include<cmath> #include<cstdlib> const double eps = 1e-6; double cal(double x){ return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*x; } double gao(double x,double y){ return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y; } int main() { int t; double low,high,y,x,res; scanf("%d",&t); while(t--){ scanf("%lf",&y); low=0.0; high=100.0; while(high-low>eps){ x=(low+high)/2; res=cal(x); if(res<y) { low = x + 1e-8; }else { high = x - 1e-8; } } printf("%0.4lf\n",gao(x,y)); } return 0; }
时间: 2024-10-24 23:19:49