hdu Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 108 Accepted Submission(s): 96

Problem Description Now, here is a fuction:   F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) Can you find the minimum value when x is between 0 and 100.

Input The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

Output             Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input 2 100 200

Sample Output -74.4291 -178.8534 分析： 求函数的最小值，首先求导的导函数为：G(x) = 42 * x^6+48*x^5+21*x^2+10*x-y (0 <= x <=100) 分析导函数的，导函数为一个单调递增的函数。如果导函数的最大值小于0，那么原函数在区间内单调递减。 即F(100)最小；如果但函数的最小值大于0，那么原函数在区间内单调递增，即F(0)最小。如果导函数既有正又有负 又由于导函数是单增函数，所以必有先负后正，即原函数必有先减后增的性质。求出导函数的零点就是原函数的最小值点。 求导函数最小值方法是2分法. #include <iostream> #include <stdio.h> #include <math.h> using namespace std; double f(double x,double y) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x; } double func(double x, double y) { return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y; } int main() { int t; double y,left,right,mid,ans1,ans2,ans3; scanf("%d",&t); while(t--) { scanf("%lf",&y); if(func(0.0,y)>=0) printf("%.4f\n",f(0.0,y)); else if(func(100.0,y)<=0) printf("%.4f\n",f(100.0,y)); else { left = 0.0; right = 100.0; mid = 50.0; ans1 = func(left,y); ans2 = func(mid,y); ans3 = func(right,y); while(fabs(ans1-ans2)>0.0001) { if(ans2>0) { ans3 = ans2; right = mid; mid = (left+right)/2; }else { ans1 = ans2; left = mid; mid = (left+right)/2; } ans2 = func(mid,y); } printf("%.4f\n",f(mid,y)); } } return 0; }