NYOJ 503(二分解方程)

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.

输入

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);

输出

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.

样例输入

2
100
-4

样例输出

2.0422
No solution!

exp（-10）=e^-10;
1e-10=10^-10;因此用exp（-10）会wa

#include<stdio.h>
#include<math.h>
double fun(double x)
{

	double y=8*pow(x,4)-7*pow(x,3)+2*pow(x,2)+3*x+6;
	return y;
}

int main()
{
	int a,b,n,T;
	double x1,x2,mid,m;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lf",&m);
		x1=0;
		x2=100;
		if(m<fun(0)||m>fun(100))/*值域符号相同则无解*/
		{
			printf("No solution!\n");
			continue;
		}
		while((x2-x1)>1e-10)/*用exp(-10)就wa*/
		{
			mid=(x2+x1)/2.0;
			if(fun(mid)<m)
				x1=mid;
			else 
				x2=mid;
		}
		printf("%.4lf\n",mid);
	}
}