Error Curves Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu SubmitStatusPracticeUVALive 5009 Appoint description: Description Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a met…

Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1198    Accepted Submission(s): 460 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. She pa…

Error Curves 思路:这个题的思路和上一个题的思路一样,但是这个题目卡精度,要在计算时,卡到1e-9. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 10100 #define eps 1e-9 using namespace std; int T,n; double ans; double l,r,mid1,mid2;…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3714 懂了三分思想和F(x)函数的单调性质,这题也就是水题了 #include "stdio.h" //最后得到的F(x)函数要么是单调的,要么是先减后增的,有了这个性质,就可以三分了~ #include "string.h" #define N 10005 #define eps 1e-9 #define INF 0x3fffffff struct node { in…

http://acm.hdu.edu.cn/showproblem.php?pid=3714 [题意]: 题目意思看了很久很久,简单地说就是给你n个二次函数,定义域为[0,1000], 求x在定义域中每个x所在的n个函数的最大值的最小值.很拗口吧,显然这题不是组队或者耐心的做是不知道性质的,至少我没看出来.网上说是三分,我画了几个图,确实是.根据二次函数的性质,增长的快慢已经确定了,那的确是单峰的.那就OK了.另外eps的问题1e-8还是wa,1e-9AC.想了下,因为有系数a,b,c的缘故,一…

题目传送门 /* 三分:凹(凸)函数求极值 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; const double EPS = 0.0000000001; struct F { double a, b, c; }f[MAXN]; int n; dou…

Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which has many interesting properties. In order to test the algorithm's efficiency, she collects many datas…

Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4137    Accepted Submission(s): 1549 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. Shepay…

F - Error Curves Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which h…

Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which has many interesting properties. In order to test the algorithm's efficiency, she collects many datas…

Problem Description Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which has many interesting properties. In order to test the algorithm's efficiency, she…

UVA 1476 1476 - Error Curves 题目链接 题意:给几条下凹二次函数曲线.然后问[0,1000]全部位置中,每一个位置的值为曲线中最大值的值,问全部位置的最小值是多少 思路:三分法,因为都是下凹函数,所以全部曲线合并起来.仍然是一个下凹函数.满足单峰.用三分求极值 代码: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using n…

                                       Error Curves Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which has many interesting properties.In order to test…

Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 6241    Accepted Submission(s): 2341 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. Shepay…

题目: Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1151    Accepted Submission(s): 440 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. She…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3714 题意:求n个二次函数在[0,1000]的最小值. 三分枚举. #include <bits/stdc++.h> using namespace std; ; ; double a[maxn], b[maxn], c[maxn]; int n; double f(double x) { ; ; i <= n; i++) { double y = a[i]*x*x+b[i]*x+c[i];…

这是一道典型的DFS题目.幻想有n个箱子,每次都向箱子里扔一个数,(当然第一个是必定是1,因为题目要求按字典序输出).判断输出的条件就是,当我移动到第n+1个箱子的时候,就要return了,当然还要判断是否能输出..怎么判断能否输出呢?题目说:组成一个圈子,a[j]+a[j+1]都必须是质数.但是,如果我们在n+1的时候再判断这个,已经很迟了,递归次数太多了.所以,应该是当满足这个条件,我才递归. 当然,a[1]+a[n]是需要在输出的时候判定的 #include <stdio.h> #inc…

//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits…

给出的曲线要么是开口向上的抛物线要么是直线,但所定义的F(x)的图形一定是下凸的. 注意一点就是求得是极小值,而不是横坐标,样例也很容易误导人. #include <cstdio> #include <algorithm> using namespace std; + ; int n; double a[maxn], b[maxn], c[maxn]; double F(double x) { ]*x*x+b[]*x+c[]; ; i < n; i++) ans = max(…

对x的坐标三分: #include<cstdio> #include<algorithm> #define maxn 10009 using namespace std; double a[maxn],b[maxn],c[maxn]; int n; double f(double x) { double ans=-999999999.0,t; ;i<n;i++) { t=a[i]*x*x+b[i]*x+c[i]; ans=max(ans,t); } return ans; }…

大意: 给你n个二次函数Si(x),F(x) = max{Si(x)} 求F(x)在[0,1000]上的最小值. S(x)=ax^2+bx+c       (0<=a<=100, |b|,|c|<=5000) 简单分析一下可知函数F(x)的图形是下凸函数,可以采用三分法求最值. CODE: #include <cstdio> #include <algorithm> using namespace std; const int maxn = 10000 + 10;…

题意 PDF 分析 因为这些函数都可以看成下凸的,所以总函数也是下凸的(可用反证法证明). 三分答案即可,时间复杂度\(O(100)\) 代码 #include<bits/stdc++.h> #define rg register #define il inline #define co const template<class T>il T read(){ rg T data=0,w=1;rg char ch=getchar(); while(!isdigit(ch)) {if(…

画出函数图像后,发现是一个类似V字型的图. 可以用三分法找图像最低点 WA了一串,之后发现是读入优化迷之被卡. /*by SilverN*/ #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<cmath> using namespace std; ; int a[mxn],b[mxn],c[mxn]; int n; double f…

已知n条二次曲线si(x) = ai*x^2 + bi*x + ci(ai ≥ 0),定义F(x) = max{si(x)},求出F(x)在[0,1000]上的最小值. 链接:传送门 分析:最大值最小,我们可以利用二分来解,但是有一个更牛的方法叫:“三分法”,这个方法的应用范围是凸函数,可以看一个图像: L和R是边界,m1,m2是三等分点,如果f(m1) < f(m2),那么最小值肯定在[l,m2]内,注意,不是[l,m1]因为如果m1在最低点右边,那么就会矛盾,同理,如果f(m2) < f(…

#include<stdio.h> #define mi 1e-9 #define N 11000 struct node{ double x,y,z; }a[N]; int n; double Max(double a,double b) { return a>b?a:b; } double ff(double h) { double ma; int i; for;i<=n;i++) ma=Max(ma,a[i].x*h*h+a[i].y*h+a[i].z); return ma…

最大值最小问题,三分....竟然排第六当时..... #include<stdio.h> #include<string.h> #define max 10000+10 #define Max(x,y) (x>y?x:y) #define Min(x,y) (x<y?x:y) #define inf 1e-8 typedef long long LL; int a[max],b[max],c[max]; int t,n; double func(int a,int b,…

Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which has many interesting properties. In order to test the algorithm's efficiency, she collects many datas…

Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4928    Accepted Submission(s): 1867 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. Shepay…

原文地址:https://cn.mathworks.com/help/fuzzy/examples/chaotic-time-series-prediction.html?requestedDomain=www.mathworks.com This example shows how to do chaotic time series prediction using ANFIS. Time Series Data The data is generated from the Mackey-Gl…

1.LA 5694 Adding New Machine 关键词:数据结构,线段树,扫描线(FIFO) #include <algorithm> #include <cstdio> #include <cstring> #include <string> #include <queue> #include <map> #include <set> #include <ctime> #include <cm…