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.
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!
Author
Redow
Recommend
lcy
思路：简单的二分法求解。
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
int b=0;
int f(double x1,double x2,double y)
{
double f,f1,f2,x;
f1=8*x1*x1*x1*x1 + 7*x1*x1*x1 + 2*x1*x1 + 3*x1 + 6 - y ;
f2=8*x2*x2*x2*x2 + 7*x2*x2*x2 + 2*x2*x2 + 3*x2 + 6 - y ;
if(f1*f2>0)
{
b=1;
return 0 ;
}
do
{
x=(x1+x2)/2;
f=8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6 - y ;
if(f*f1>0)
{
x1=x;
f1=f;
}
else if(f*f2>0)
{
x2=x;
f2=f;
}
}while(fabs(f)>1e-6);
printf("%.4lf\n",x);
return x;
}
int main()
{
int n;
double y,l,t,x;
scanf("%d",&n);
while(n--)
{
l=0;
t=100;
scanf("%lf",&y);
x=f(l,t,y);
if(b==1)
printf("No solution!\n");
}
return 0;
}