hdu2199 二分搜索

这是很简单的二分搜索；需要注意的是精度。一开始我用的是right-left<1e-4,测试样例中100我输出的最后一位为1，不合要求。

后来自己写了个四舍五入的一段，但交上去WA，于是改成1e-6，AC了。

代码题目如下：

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!

代码：

#include<stdio.h>
#include<math.h>
double equo(double a);
int y,num;
int main()
{
    double left=0.0,right=100.0,mid=0;
    int ok=0,s;
    double an1,an2,an3,t;
    scanf("%d",&num);
    while(num--)
    {
        left=0.0;
        right=100.0;
        scanf("%d",&y);
        an1=equo(left);
        an2=equo(right);
        while((an2-an1)>1e-6)
        {
            if(y<6||y>807020288)
            {ok=1;break;}
            mid=(left+right)/2;
            an3=equo(mid);
            if(an3*an1<0)
            {
                an2=an3;
                right=mid;
            }
            else
            {
                left=mid;
                an1=an3;
            }
        }
        if((mid<0&&mid>100)||ok==1)printf("No solution!\n");
         else 
                 printf("%.4f\n",mid);
    }
    return 0;
}
double equo(double a)
{
    double b;
    b=8*pow(a,4)+7*pow(a,3)+2*pow(a,2)+3*a+6-y;
    return b;
}