hdu Can you solve this equation

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 164 Accepted Submission(s): 94

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!

这道题用的是二分搜索法

二分法原理

给定方程f(x)=0,设f(x)在区间[a,b]连续,且f(a)f(b)<0,则方程f(x)在(a,b)内

至少有一根,为便于讨论,不妨设方程f(x)=0在(a,b)内只有一实根

采取使有根区间逐步缩小,从而得到满足精度要求的实根

的近似值。取[a,b]区间二等分的中点x0 =(a+b)/2,

若f(x0)=0,则x0是f(x)=0的实根若f(a)f(b)<0 成立,

则必在区间(a, x0)内,取a1=a,b1= x0;

否则必在区间(x0,b)内,取a1= x0,b1=b,

这样,得到新区间(a1,b1),其长度为[a,b]的一半,

如此继续下去,进行k次等分后,得到一组不断缩小的区间,

[a,b],[a1,b1],......[ak,bk].

分析本题得：函数是一个单调递增的函数；所以此函数的0点只有一个。本题的精度为0.0001，保证小数点后前四位相同。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 using namespace std;
 5 double y;
 6 double func(double x)
 7 {
 8     return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
 9 }
10 int main()
11 {
12     int t;
13     double left,right,mid,ans1,ans2,ans3;
14     cin>>t;
15     while(t--)
16     {
17         cin>>y;
18         if(func(0)>y||func(100)<y)
19         cout<<"No solution!"<<endl;
20         else
21         {
22             left = 0.0;
23             right = 100.0;
24             mid = 50.0;
25             ans1 = func(left)-y;
26             ans2 = func(mid)-y;
27             ans3 = func(right)-y;
28             while(fabs(ans1-ans2)>0.0001)
29             {
30                 if(ans2>0)
31                 {
32                     right = mid;
33                     ans3 = ans2;
34                     mid = (left+right)/2;
35                 }else
36                 {
37                     left = mid;
38                     ans1 = ans2;
39                     mid = (left+right);
40                 }
41                 ans2 = func(mid)-y;
42             }
43             printf("%.4f\n",mid);
44         }
45     }
46     return 0;
47 }