hdu 2199 2899 (二分，三分)

HDU2199   Can you solve this equation?

[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!

二分法：

#include<stdio.h>
#include<math.h>
double f(double x)
{
return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2.0)+3*x+6;
}
int main()
{
int T;
double y,l,r,m;
scanf("%d",&T);
while(T--)
{
scanf("%lf",&y);
if(f(0.0)<=y&&f(100.0)>=y)
{
l=0;
r=100;
while(r-l>=1e-6)
{
m=(r+l)/2;
if(f(m)<y)
l=m+1e-7;
else
r=m-1e-7;
}
printf("%.4lf\n",(l+r)/2);
}
else
printf("No solution!\n");
}
return 0;
}

 

 

HDU2899  Strange fuction

[align=left]Problem Description[/align]
Now, here is a fuction:

  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)

Can you find the minimum value when x is between 0 and 100.
 

 
[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 only one real numbers Y.(0 < Y <1e10)
 

 
[align=left]Output[/align]
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
 

 
[align=left]Sample Input[/align]

2
100
200

 

 
[align=left]Sample Output[/align]

-74.4291
-178.8534

 

 

// 三分法
#include<stdio.h>
#include<math.h>
double f(double x,double y)
{
return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0)-y*x;
}
int main()
{
int T;
double y,l,r,lt,rt,y1,y2;
scanf("%d",&T);
while(T--)
{
scanf("%lf",&y);
l=0.0;
r=100.0;
while(r-l>1e-6)
{
lt=(2*l+r)/3;
rt=(l+2*r)/3;
y1=f(lt,y);
y2=f(rt,y);
if(y1<y2)
r=rt;
else
l=lt;
}
printf("%.4lf\n",f(r,y));
}
return 0;
}

//二分法
#include<stdio.h>
#include<math.h>
double f(double x,double y)
{
return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0)-y*x;
}
double fd(double x)
{
return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*x;
}
int main()
{
int T;
double l,r,m,y;
scanf("%d",&T);
while(T--)
{
scanf("%lf",&y);
l=0.0;
r=100.0;
while(r-l>1e-6)
{
m=(l+r)/2;
if(fd(m)-y<=0)
l=m;
else
r=m;
}
printf("%.4lf\n",f((l+r)/2,y));
}
return 0;
}