HDU 2899 Strange fuction(牛顿迭代)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9333    Accepted Submission(s):
6352

Problem Description

Now, here is a fuction:
  F(x) = 6 *
x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value
when x is between 0 and 100.

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 only one real numbers Y.(0 < Y <1e10)

Output

Just the minimum value (accurate up to 4 decimal
places),when x is between 0 and 100.

Sample Input

2
100
200

Sample Output

-74.4291
-178.8534

Author

Redow

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被eps卡好蛋疼啊。。

首先求函数的最大最小值问题可以转化为导数的零点的问题

然后用牛顿迭代求导数零点就行了

eps不能设的太小

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps = 1e-9;
double y;
double fdd(double x) {
    return 252 * pow(x, 5) + 240 * pow(x, 4) + 42 * x + 10;
}
double fd(double x) {
    return 42 * pow(x, 6) + 48 * pow(x, 5) + 21 * pow(x, 2) + 10 * x - y;
}
double f(double x) {
    return 6 * pow(x, 7) + 8 * pow(x, 6) + 7 * pow(x, 3) + 5 * pow(x, 2) - y  * x;
}
double Newton(double x) {
    while(fabs(fd(x)) > eps)
        x = x - fd(x) / fdd(x);
    return x;
}
int main() {
    int QwQ;
    scanf("%d", &QwQ);
    while(QwQ--) {
        scanf("%lf", &y);
        double ans = 1e15, pos;
        for(int i = 0; i <= 100; i++) {
            double anspos = Newton(i);
            if(anspos >= 0 && anspos <= 100)
                ans = min(ans, f(anspos));
        }
        printf("%.4lf\n", ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zwfymqz/p/9149347.html