Hdoj 2899.Strange fuction 题解】的更多相关文章

Problem Description Now, here is a fuction: F(x) = 6 * x^7+8x^6+7x^3+5x^2-yx (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 o…
Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4599    Accepted Submission(s): 3304 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 <=…
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…
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…
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 <=…
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<…
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…
求  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)的最小值 模拟退火,每次根据温度随机下个状态,再根据温度转移 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #define inf (double)0x3f3f3f3f3f3f3f3fll usi…