Strange fuction Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5933 Accepted Submission(s): 4194 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) C…

http://acm.hdu.edu.cn/showproblem.php?pid=2899 Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4865    Accepted Submission(s): 3468 Problem Description Now, here is a fuction:  F…

题目链接:http://acm.hdu.edu.cn/showproblem.pihp?pid=2899 题目大意:找出满足F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)的x值.注意精确度的问题. 求满足条件的x的最小值!!求导,利用单调性来找到最小值. #include <iostream> #include <cstdio> #include <cmath> using namespace std;…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=2899 还可三分.不过只写了模拟退火. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<ctime> #include<cmath> #include<cstdlib> #define db double using…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=2899 模拟退火: 怎么也过不了,竟然是忘了写 lst = tmp ... 还是挺容易A的. 代码如下: #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<cstdlib> #include<…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2278    Accepted Submission(s): 1697 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 <=…

1.题意:给一个函数F(X)的表达式,求其最值,自变量定义域为0到100 2.分析:写出题面函数的导函数的表达式,二分求导函数的零点,对应的就是极值点 3.代码: # include <iostream> # include <cstdio> # include <cmath> using namespace std; ; double Y; int sgn(double x) { ; ) ; ; } double F(double x) { )+)+)+)-Y*x;…

题目链接 \(Description\) 求函数\(F(x)=6\times x^7+8\times x^6+7\times x^3+5\times x^2-y\times x\)在\(x\in \left[0,100\right]\)时的最小值. \(Solution\) \(x\geq 0\)时\(F(x)\)为单峰凹函数,三分即可. 而且由此可知\(F(x)\)的导数应是单增的.函数最值可以转化为求导数零点问题,于是也可以二分求\(F'(x)\)的零点,或者用牛顿迭代求. 峰值函数最值也可…

Working out time limit per test  2 seconds memory limit per test  256 megabytes input  standard input output  standard output Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where th…

三分可以用来求单峰函数的极值. 首先对一个函数要使用三分时,必须确保该函数在范围内是单峰的. 又因为凸函数必定是单峰的. 证明一个函数是凸函数的方法: 所以就变成证明该函数的一阶导数是否单调递增,或者其二阶导数是否大于0. #include<stdio.h> #include<math.h> ; double js(double x,double y){ *pow(x,)+*pow(x,)+*pow(x,)+*pow(x,)-y*x; } int main(){ int n; do…