hdu 2199 Can you solve this equation?(二分搜索)

最新推荐文章于 2025-08-22 13:45:39 发布

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最新推荐文章于 2025-08-22 13:45:39 发布

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文章标签： java

原文链接：http://www.cnblogs.com/zyx1314/p/3608957.html

本文介绍了如何通过二分搜索法解决给定区间内的四次多项式方程，并讨论了解的存在性。对于每组输入值y，首先检查是否存在解的可能范围，然后使用迭代方法逼近解的精确值。

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Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7156 Accepted Submission(s): 3318

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!

：：很容易知道y=8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 单调递增，那么我们就可以对[0,100]进行二分搜索，找出满足条件的数

代码：

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    double f(double x){
        return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
    }

    int main()
    {
        int T,y;
        scanf("%d",&T);
        while (T--)
        {
            scanf("%d",&y);
            if( f(0) > y || f(100) < y) //f(0)>y或f(100)<y则无解
            {
                printf("No solution!\n");
                continue;
            }
            double l = 1.0, r = 100.0;
            while( r - l > 1e-6)
            {
                double m=(l+r)/2;
                if(f(m)>y) r=m-1e-7;
                else l=m+1e-7;
            }
            printf("%.4lf\n",(l+r)/2);
        }
        return 0;
    }

转载于:https://www.cnblogs.com/zyx1314/p/3608957.html