一道二分法求解的问题，为什么总是超时呢？求高手解答！
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
2
100
-4
Sample Output
1.6152
No solution!
我的代码如下：
#include<stdio.h>
#include<math.h>
int main()
{
int t,i,m;
float y,x0,x1,x2,f0,f1,f2;
scanf("%d",&t);
for(i=0;i<t;i++)
{
m=0;
x1=0.0;
x2=100.0;
scanf("%f",&y);
f1=8.0*x1*x1*x1*x1+7.0*x1*x1*x1+2.0*x1*x1+3.0*x1+6.0-y;
f2=8.0*x2*x2*x2*x2+7.0*x2*x2*x2+2.0*x2*x2+3.0*x2+6.0-y;
do
{
if(y<6||y>807020288){m=1;break;}
x0=(x1+x2)/2.0;
f0=8.0*x0*x0*x0*x0+7.0*x0*x0*x0+2.0*x0*x0+3.0*x0+6.0-y;
if((f1*f0)<0)
{
x2=x0;
f2=f0;
}
else
{
x1=x0;
f1=f0;
}
}
while(fabs(f0)>=1e-4);
if((x0<0&&x0>100)||m==1)printf("No solution!\n");
else printf("%.4f\n",x0);
}
return 0;
}
交到OJ上去老是超时，为什么啊？要怎么修改？
[解决办法]
pow和double可能是精度的问题，这个可能会影响二分的最后一次位置吧。
[解决办法]
我看关键是double吧
[解决办法]
精度问题。
