HDU2199:Can you solve this equation?（二分查找）

最新推荐文章于 2019-08-13 19:42:15 发布

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最新推荐文章于 2019-08-13 19:42:15 发布

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分类专栏：搜索文章标签： ACM 搜索二分查找

本文链接：https://blog.youkuaiyun.com/qq_43555854/article/details/86600774

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本文介绍了一种使用二分查找法解决特定多项式方程的方法，该方程为8x^4+7x^3+2x^2+3x+6=Y，在0到100的范围内寻找方程的解。文章提供了完整的AC代码实现，展示了如何精确地找到方程的根，或判断在指定范围内方程是否有解。

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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): 27695 Accepted Submission(s): 11702

Problem Description
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!

注意条件精度不要设的太小，否则会超时。

下面附上ac代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <ctype.h>
#include <queue>
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long int ll;
double f(double x) {
    return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6;
}
int main() {
    int t;
    cin >> t;
    while(t--) {
        double x;
        cin >> x;
        if(x >= f(100) || x <= f(0)) {
            cout << "No solution!" << endl;
        } else {
            double l = 0;
            double r = 100;
            double m = (l + r) / 2;
            while(r - l > 1e-6) { //精度略高一点就行
                m = (l + r) / 2;
                if(f(m) > x) {
                    r = m;
                } else {
                    l = m;
                }
            }
            printf("%.4f\n",(l + r) / 2);
        }
    }
    return 0;
}