杭电 2199 Can you solve this equation?

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Can you solve this equation?

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : Accepted Submission(s) :

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!

类似这种题，都用二分法，书上有详细的解释，我在这简单的说说，二分法数学早学了，并不是所有求根都能用二分法求，本题有它的特殊性，只能求一个范围内有且仅有的唯一根，由两端的正负差异用循环把根逼出来，这就是原理，就是公式复杂了点，代码也就难看了点，慢慢看吧！

代码如下：

#include<stdio.h>#include<math.h>int main (){    int T;    double x0,x1,x2,y0,y1,y2,Y;    scanf("%d",&T);    while(T--)    {        x1=0;x2=100;        scanf("%lf",&Y);        y1=8.0*pow(x1,4)+7.0*pow(x1,3)+2.0*pow(x1,2)+3.0*x1+6.0-Y;        y2=8.0*pow(x2,4)+7.0*pow(x2,3)+2.0*pow(x2,2)+3.0*x2+6.0-Y;        if(y1*y2>=0)            printf("No solution!\n");        else        {            x0=(x1+x2)/2.0;            y0=8.0*pow(x0,4)+7.0*pow(x0,3)+2.0*pow(x0,2)+3.0*x0+6.0-Y;            while(fabs(y0)>=1e-4)            {                if(y0==0.0)                    break;                if(y1*y0<0)                {                    x2=x0;                    x0=(x1+x0)/2.0;                    y0=8.0*pow(x0,4)+7.0*pow(x0,3)+2.0*pow(x0,2)+3.0*x0+6.0-Y;                }                if(y2*y0<0)                {                    x1=x0;                    x0=(x2+x0)/2.0;                    y0=8.0*pow(x0,4)+7.0*pow(x0,3)+2.0*pow(x0,2)+3.0*x0+6.0-Y;                }            }            printf("%.4lf\n",x0);        }    }    return 0;}

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