HDU-2199-Can you solve this equation?【二分】

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 16553    Accepted Submission(s): 7343

[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!

代码一：

#include<cstdio>
#include<algorithm>
#include<cmath>
#define eps 1e-12
using namespace std;
double f(double x)
{
return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2)+3*x+6;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
double y,mid,maxn;
double l=0.0,r=100.0;
scanf("%lf",&y);
maxn=8*pow(100.0,4.0)+7*pow(100.0,3.0)+2*pow(100.0,2)+3*100+6;
if(y<6||y>maxn)
{
printf("No solution!\n");
continue;
}
int size=50;
while(size--)  //循环次数一定比 50 少
{
mid=(l+r)/2.0;
if(y<f(mid))	r=mid-eps;  //不加 eps 这个精度也能 AC
else	l=mid+eps;
}
printf("%.4lf\n",l);
}
return 0;
}

代码二：

#include<cstdio>
#include<cmath>
double f(double x)
{
return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
}
int main()
{
int t,a;
scanf("%d",&t);
while(t--)
{
double y,l,r,mid;
scanf("%lf",&y);
if(y < 6 || y > 8*100*100*100*100+7*100*100*100+2*100*100+3*100+6)
printf("No solution!\n");
else
{
l=0;
r=100;
a=50;
mid=(l+r)/2;
while(fabs(f(mid)-y)>1e-5)
{
if(f(mid) < y)
{
l=mid+1;
}
else
{
r=mid-1;
}
mid=(l+r)/2;
}
printf("%.4f\n",mid);
}

}
}