Can you solve this equation? 详细解答

你的支持是我最大的动力，你的意见是我前进的导航。

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

Recommend

lcy

题目大意就是x∈[0, 100], Y∈[-10^10, 10^10]，求8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = Y 的解。

题目很简单，首先先需要些前期工作，通过一导，二导就会发现，其实对于 f(x) = 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6, x∈[0, 100]时，这个函数是递增的。所以用二分法即可解题。

代码如下

 1 #include <stdio.h>
 2 #include <math.h>
 3 double f (double x)  //for convenience
 4 {
 5     return 8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * x + 6;
 6 }
 7 int main()
 8 {
 9     int T;
10     double x1, x2, x3, y, y1, y2, y3;
11     scanf("%d", &T);
12     while(T--)
13     {
14         x1 = 0;
15         x2 = 100;        
16         scanf("%lf", &y);   //heed
17         y1 = f(x1) - y;
18         y2 = f(x2) - y;
19         if (y1 > 0 || y2 < 0)
20             printf("No solution!\n");
21         else
22         {
23             while (fabs(y1 - y2) >= 0.0001)
24             {
25                 x3 = (x1 + x2) / 2;
26                 y3 = f(x3) - y;
27                 if (y3 >= 0)
28                     x2 = x3;
29                 else
30                     x1 = x3;
31                 y1 = f(x1) - y;
32                 y2 = f(x2) - y;
33             }
34             printf("%0.4f\n", x3);
35         }
36     }
37     return 0;
38 }

 1 #include <stdio.h>
 2 #include <math.h>
 3 double f (double x)  //for convenience
 4 {
 5     return 8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * x + 6;
 6 }
 7 int main()
 8 {
 9     int T;
10     double x1, x2, x3, y, y1, y2, y3;
11     scanf("%d", &T);
12     while(T--)
13     {
14         x1 = 0;
15         x2 = 100;        
16         scanf("%lf", &y);   //heed
17         y1 = f(x1) - y;
18         y2 = f(x2) - y;
19         if (y1 > 0 || y2 < 0)
20             printf("No solution!\n");
21         else
22         {
23             while (fabs(x1 - x2) >= 0.000001)
24             {
25                 x3 = (x1 + x2) / 2;
26                 y3 = f(x3) - y;
27                 if (y3 >= 0)
28                     x2 = x3;
29                 else
30                     x1 = x3;
31             }
32             printf("%0.4f\n", x3);
33         }
34     }
35     return 0;
36 }

注意点：

1、应该花大部分时间在思路上，编码应该快速完成

2、尽量用double，用double时，要注意用"%lf"输入。float 有38位，double有308位

3、这个题目要求x精确到4位，刚开始时，我以为我以x取到4位小数为结束条件，一直错！后来来发现应该以y1，y2很接近为结束结束条件。原因很简单以为x取到4位时，有可能这个y还很不精确。当然，也可以让x1和x2很接近，如上面的second program也是对的。

4、C中有fabs，pow，在math.h头文件中。