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  • acm课程练习2--1001

    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! ");
    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 ", tem);
    38. }
    39. }
    40. }




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  • 原文地址:https://www.cnblogs.com/liuzhanshan/p/5405713.html
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