HDOJ 2199 Can you solve this equation?（二分）

最新推荐文章于 2018-05-11 14:41:14 发布

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最新推荐文章于 2018-05-11 14:41:14 发布

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本文介绍了一种使用二分法在指定区间内精确求解特定多项式方程的方法，并提供了实现该算法的C++代码示例。

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Can you solve this equation?

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：给出方程y = 8 * x ^ 4 + 7 * x ^ 3 + 2 * x ^ 2 + 3 * x + 6，其中fabs(y) <= 1e10，求x在区间[0,100]内的解，输出4位小数。

解题思路：在区间[0,100]内二分x，将x带入方程，判断结果与y的大小关系，若ans > y,则ub = mid;若ans < y,则lb = mid。由于fabs(y) <= 1e10，所以经过40次二分即可带到精度要求。

代码如下：

#include <cstdio>
#include <cmath>

double y;

double fun(double x)
{
    return 8 * pow(x,4.0) + 7 * pow(x,3.0) + 2 * pow(x,2.0) + 3 * x + 6;
}

int main()
{
    double lb,ub;
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%lf",&y);
        if(fun(0) <= y && y <= fun(100)){
            lb = 0,ub = 100;
            for(int i = 0;i < 40;i++){
                double mid = (ub + lb) / 2.0;
                if(fun(mid) > y)
                    ub = mid ;
                else
                    lb = mid ;
            }
            printf("%.4lf\n",lb);
        }else{
            printf("No solution!\n");
        }

    }
    return 0;
}