hdu2199-Can you solve this equation?（浮点型二分查找）

最新推荐文章于 2019-08-13 19:42:15 发布

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#hdu #二分查找

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本文介绍了一个特定多项式方程的数值解法，采用二分法寻找在给定区间内满足条件的根，提供了完整的C++代码实现。

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原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=2199

题目原文：

Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28808 Accepted Submission(s): 12089

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!

题目大意：

给出一个y(x)函数，x属于[0, 100]，求x值，没有值输出“No solution! ”

解题思路：

使用浮点型二分法，EPS注意要高出答案精度两位防止四舍五入后答案错误。

AC代码：

#include <cstdio>
#include <cmath>
using namespace std;

const double EPS = 1e-6; // 题目要求精度4位，EPS最好多两位，从而不影响四舍五入结果

double funY(double x)
{
    return 8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * x + 6;
}

double bsearch(double y)
{
    double left = 0.0, right = 100.0, middle;
    while (left <= right)
    {
        middle = (left + right) / 2;
        if (funY(middle) < y) left = middle + EPS;
        else right = middle - EPS;
    }
    return left;
}

const double MIN = funY(0.0);
const double MAX = funY(100.0);

int main()
{
    int T;
    scanf("%d", &T);
    double y;
    while (T--)
    {
        scanf("%lf", &y);
        if (y < MIN || y > MAX) printf("No solution!\n");
        else printf("%.4lf\n", bsearch(y));
    }
    return 0;
}