HDOJ-2199-Can you solve this equation?(二分查找)

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11044    Accepted Submission(s): 5083

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

#include<stdio.h>#include<math.h>double f(double x){return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}double binary_search(double x,double y,double n){double mid=(x+y)/2;if(y-x<=1e-7)return x;   //刚开始写成 if(fabs(n-f(mid))<=1e-7) return mid; 结果超时,原来还是没理解透二分查找的具体退出边界条件!!!else if(f(mid)>=n) return binary_search(x,mid+1e-20,n);   //错了n遍,写成了mid+1e-7!!!else return binary_search(mid+1e-20,y,n);    }int main(){int T;scanf("%d",&T);while(T--){double y,t;scanf("%lf",&y);        if(y<f(0)||y>f(100)){        printf("No solution!\n");        continue;        }t=binary_search(0,100,y);printf("%.4lf\n",t);}return 0;}