Strange fuction 2899 （二分+数学求导）

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4717    Accepted Submission(s): 3383

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

2100200

 

Sample Output

-74.4291-178.8534
//一个简单的二分题，但得要先判断一下导数的正负来确定函数的输出
//比较简单（一遍AC）
#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;double fun(double x){return 42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x;}double f(double x,double y){return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;}int main(){int t;double y;scanf("%d",&t);while(t--){scanf("%lf",&y);if(fun(0)>y)printf("0.0000\n");else if(fun(100)<y)printf("%.4lf\n",f(100.0,y));else if(fun(0)<y&&fun(100)>y){double l=0,r=100,mid;while(r-l>1e-10){mid=(l+r)/2;if(fun(mid)-y>1e-6)r=mid;elsel=mid;}printf("%.4lf\n",f(mid,y));}}return 0;}