HDU—2199—Can you solve this equation?—【二分】【精度控制】

最新推荐文章于 2021-02-10 11:42:51 发布

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最新推荐文章于 2021-02-10 11:42:51 发布

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分类专栏： HDU 文章标签： ACM HDU

本文链接：https://blog.youkuaiyun.com/u013795055/article/details/38661285

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本文介绍了一个特定多项式方程的数值解法，通过二分查找算法在指定区间内精确求解给定方程的实数根，并提供了两种实现方式：递归版与非递归版。

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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): 8567 Accepted Submission(s): 3948

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!

思路：

设f(x)=8*x^4+7*x^3+2*x^2+3*x+6 ，定义域为[0,100]

求导可证明f(x)在区间上单调增，所以在 f(0)<=f(x)<=f(100) ，

所以二分查找Y，当然先判断下Y是否满足：f(0)<=Y<=f(100)

上代码：

递归版：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;

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

double BiSearch(double Y,double left,double right)
{
    double mid=(left+right)/2;
    if(right-left>=10e-7)   //如果是-6就错了。。。卡精度的呀。。。为什么！！！
    {
        if(Equation(mid)==Y)
            return mid;
        if(Equation(mid)>Y)
            return BiSearch(Y,left,mid);
        if(Equation(mid)<Y)
            return BiSearch(Y,mid,right);
    }
    return mid;
}

int main()
{
    double x,Y;
    int cases;
    scanf("%d",&cases);
    while(cases--)
    {
        scanf("%lf",&Y);
        if(Y<Equation(0) || Y>Equation(100))
            printf("No solution!\n");
        else
            printf("%.4lf\n",BiSearch(Y,0,100));
    }
    return 0;
}

非递归版：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;

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

int main()
{
    double x,Y;
    int cases;
    scanf("%d",&cases);
    while(cases--)
    {
        scanf("%lf",&Y);
        if(Y<Equation(0) || Y>Equation(100))
            printf("No solution!\n");
        else
        {
            double left=0,right=100,mid;
            while(right-left>10e-12)    //还是卡精度，干脆就严格点。。。
            {
                mid=(left+right)/2;
                if(Equation(mid)>Y)
                    right=mid;
                else
                    left=mid;
            }
            printf("%.4lf\n",mid);
        }
    }
    return 0;
}

怎么能保证不在精度上出问题？？？SOS！！！