题目描述

Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 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!

大意

求解方程的简单水题

思路

简单的水题，使用二分法即可。但需要注意浮点数比较时的精度，我写的是1e-5，如果精度太高的话会超时的

1. #include<iostream>
2. #include<stdio.h>
3. #include<cmath>
4. #include<iomanip>
5. using namespace std;
6. double m, res = 0;
7. int ji = 0;
8. double f(double res)
9. {
10. return res*res*res*res * 8 + res*res*res * 7 + res*res * 2 + res * 3 + 6;
11. }
12. int main()
13. {
14. //cin.sync_with_stdio(false);
15. //freopen("date.in", "r", stdin);
16. //freopen("date.out", "w", stdout);
17. int N, temp;
18. double b, e, tem = 50;
19. scanf("%d", &N);
20. for (int i = 0; i<N; i++){
21. b = 0, e = 100, tem = 50;
22. cin >> m;
23. if (m<6 || m>807020306)
24. //cout << "No solution!" << endl;
25. printf("No solution!\n");
26. else{
27. while (fabs(f(tem) - m) >= 1e-5)
28. if (f(tem)>m){
29. e = tem;
30. tem = (b + e) / 2;
31. }
32. else{
33. b = tem;
34. tem = (b + e) / 2;
35. }
36. //cout << fixed << setprecision(4) << tem << endl;
37. printf("%0.4f\n", tem);
38. }
39. }
41. }