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.
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
Recommend
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; }
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分类:
题库—HDU
, 数学—微积分—牛顿迭代
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