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 (*x*x*x*x+*x*x*x+*x*x+*x+);
  }
 int main()
 {

     int t;
     while(~scanf("%d",&t))
     {
         while(t--)
         {
             double y;
             scanf("%lf",&y);
             if(f()>y||f()<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-)
             {
                 if(f(mid)>y)
                 {
                     r=mid;
                     mid=(l+r)/2.0;

                 }
                 else
                 {
                     l=mid;
                     mid=(l+r)/2.0;
                 }

             }
             printf("%.4lf\n",mid);
         }
     }
     return ;
 }