Can you solve this equation? (二分法）

Can you solve this equation?

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 3 Accepted Submission(s) : 2

Problem Description

Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

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 a real number Y (fabs(Y) <= 1e10);

Output

For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input

2100-4

Sample Output

1.6152No solution!

Author

Redow

 

函数fabs的功能是求指定参数的绝对值
1e-8 表示的数据是 1*10的负8次方
fabs(x)>1e-8
表达的意思就是x的绝对值是否大于一个很小数值0.00000001
常常用来表示迭代计算的停止条件

 

#include<iostream>#include<algorithm>#include<cmath>using namespace std;#define f(x) (8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6)int main(int ac,char *av[]){   double n,Y,res,min,max,mid;   cin>>n;   while(n--)   {      cin>>Y;      if(f(0.0)>Y||f(100.0)<Y)      {          cout<<"No solution!"<<endl;          continue;      }       min=0.0,max=100.0;       while(max-min>1e-6)       {           mid=(min+max)/2.0;           res=f(mid);           if(res>Y)             max=mid-1e-6;            else             min=mid+1e-6;       }       printf("%.4lf\n",(min+max)/2.0);   }    return 0;}