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…

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 <=…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3809    Accepted Submission(s): 2760 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 <…

Strange fuction Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 661 Accepted Submission(s): 544   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 <=…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2982    Accepted Submission(s): 2202 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 <…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4803    Accepted Submission(s): 3428 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 <=…

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 nu…

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…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=2899 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4802    Accepted Submission(s): 3427 Problem Description Now, here is a fuction:  F(x) = 6 * x^…

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…

本题是一道二分题,但是要利用导数来求最小值.对原函数进行求导,得到的导函数为f(x)=42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;因为0<=x<=100,所以当 y>46802200时,f(x)恒小于0,故F(x)单调递减.当y<46802200时,可知原函数先递减后递增,因此可利用二分找到导函数为零的x的值,再将此x带入原函数就可以得到最小值. #include"iostream" #include"stdi…

在区间(0,100).在恒大二阶导数0.f(x)有极小值.用的最低要求的一阶导数值点: #include<math.h> #include<stdio.h> #include<stdlib.h> #include<string.h> const double eps=1e-6; double f(double x,double y) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x; } dou…

现在,这里有一个功能:  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) 当x在0到100之间时,你能找到最小值吗? 输入 第一行包含一个整数T(1 < = T < = 100),这意味着测试用例的数量.然后T行,每一行只有一个实数Y.(0 < Y < 1e10)当x在0到100之间时,输出值为最小值(精确到小数点后4位). Sample Input 2 100 200 Sample Output -74.4291…

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…

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…

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<…

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;…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4538    Accepted Submission(s): 3261 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 <=…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2098    Accepted Submission(s): 1577 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 <=…

Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10462    Accepted Submission(s): 6996 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 <…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…

二分是一种很有效的减少时间开销的策略, 我觉得单列出二分专题有些不太合理, 二分应该作为一中优化方法来考虑 这几道题都是简单的使用了二分方法优化, 二分虽然看似很简单, 但一不注意就会犯错. 在写二分时, 会遇到很多选择题, 很多"分叉路口", 要根据实际情况选择合适的"路" HDU2199 Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/…

网址:CSUST 7月30日(二分和三分) 这次的比赛是二分&三分专题,说实话以前都没有接触过二分,就在比赛前听渊神略讲了下.......不过做着做着就对二分熟悉了,果然做题是学习的好方法啊~~~~\(≧▽≦)/~啦啦啦,不过我们这组每次都是刚开始垫底,然后才慢慢追上去.......真担心爆零,但是结果还不错,和第一做的题目一样多,差的只是错误率和时间. A:大意是:给你一个直线和一个弧线,(弧线是圆的一部分),求弧线高出直线的高度.POJ 1905   Expanding Rods 解法:先确…

1001 Flooded Island http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=4521 把陆地四面其中三面被海洋环绕的‘@’标记. #include<bits/stdc++.h> using namespace std; typedef long long ll; ][]; ][]={-,,,,,-,,}; int main() { memset(ma,'.',sizeof(ma…

(一)贪心 1.A - 今年暑假不AC “今年暑假不AC?” “是的.” “那你干什么呢?” “看世界杯呀,笨蛋!” “@#$%^&*%...” 确实如此,世界杯来了,球迷的节日也来了,估计很多ACMer也会抛开电脑,奔向电视了. 作为球迷,一定想看尽量多的完整的比赛,当然,作为新时代的好青年,你一定还会看一些其它的节目,比如新闻联播(永远不要忘记关心国家大事).非常6+7.超级女生,以及王小丫的<开心辞典>等等,假设你已经知道了所有你喜欢看的电视节目的转播时间表,你会合理安排吗?(目…

1.题目描述: 1175. Strange Sequence Time limit: 1.0 secondMemory limit: 2 MB You have been asked to discover some important properties of one strange sequences set. Each sequence of the parameterized set is given by a recurrent formula: Xn+1 = F(Xn-1, Xn)…