HDU 2199

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23060 Accepted Submission(s): 10003

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!

Author

Redow

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lcy | We have carefully selected several similar problems for you: 2899 2289 2298 3400 1551

这是一道比较基础的纯粹二分算法的的题目。这一题要求的就是8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y，给定Y在0-100内找出一个数X是这个式子成立，当然也就是找出一个double数字使左边和右边最相近。我是按照二分，然后判断mid带入上面式子之后比10^-8小，但是经过实践是不行的，可以过案例但是会ＷＡ，我不是很明白可能是精度的问题，所以不要这样做。正确的做法是：从 low＝0和high=100往中间挤，也就是当low和high的差很小时，mid的差别也很小，那么mid代入上式和Y的差距也就小了，实践证明开到e-8，e-7都是可以的。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <string>
 4 #include <algorithm>
 5 #include <cmath>
 6 #define eps 1e-6 
 7 using namespace std;
 8 double A(double x)
 9 {
10     return (8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * pow(x, 1) + 6);
11 }
12 int main()
13 {
14     int n;
15     long long y, t;
16     cin >> n;
17     while (n--)
18     {
19         cin >> y;
20         if (A(0)>y || A(100)<y)
21         {
22             cout << "No solution!" << endl;
23             continue;
24         }
25         if (y == 0) cout << 0 << endl;
26         else
27         {
28             double l = 0, r = 100;
29             double mid;
30             while (r - l >=eps)
31             {
32                 mid = (l + r) / 2;
33                  if (A(mid) > y)
34                 {
35                     r = mid;
36                 }
37                 else l = mid;
38             }
39             printf("%.4lf
", mid);
40         }
41     }
42     return 0;
43 }