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



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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);

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
100
-4

[align=left]Sample Output[/align]

1.6152
No solution!

[align=left]Author[/align]
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;
}