hdu2199Can you solve this equation?(解方程+二分)

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25633    Accepted Submission(s): 11018

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!

题意：知道y值，要求算出x。并且x只能在0~100之间，不然就输出No solution!

题解：因为x只能在0~100之间。所以在知道y值得情况下判断有没有解的方法是：如果给出的Y值比f(0)还小，那他肯定没有0~100之间的解。因为解在0~100之间都是正数。同理也不能大于f(100);

其实上面的可以用这个函数在0~100之间单调递增来解释，比较清楚

剩下的就是二分来找解，看一下代码还是挺容易理解的

#include<bits/stdc++.h>
using namespace std;
double f(double x)
{
    return (8*x*x*x*x+7*x*x*x+2*x*x+3*x+6);
 } 
int main()
{
    
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            double y;
            scanf("%lf",&y);
            if(f(0)>y||f(100)<y)
            {
                printf("No solution!\n");
                continue;
            }
            double l,r;
            l=0.0;r=100.0;
            double mid=50.0;
            while(fabs(f(mid)-y)>1e-5)
            {
                if(f(mid)>y)
                {
                    r=mid;
                    mid=(l+r)/2.0;
                    
                }
                else
                {
                    l=mid;
                    mid=(l+r)/2.0;
                }
                
            }
            printf("%.4lf\n",mid);
        }
    }
    return 0;
}