题意:给一个字符串t ,求与这个序列刚好有m个位置字符不同的由两个相同的串拼接起来的字符串 s,要求字典序最小的答案. 分析:按照贪心的想法,肯定在前面让字母尽量小,尽可能的填a,但问题是不知道前面填了那么多a之后后面能否填完(因为对于那些s[i]!=s[i+n/2]的位置,必定会有一次花费) 于是就想到用dp[i][j]表示i~n这段位置,花费j是否合法 贴上转移: dp[n/][]=; -;i>=;--i) ]) { ;j<=m;++j) dp[i][j]|=dp[i+][j]; ;j&l…

Square Distance  Accepts: 73  Submissions: 598  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 一个字符串被称为square当且仅当它可以由两个相同的串连接而成. 例如, "abab", "aa"是square, 而"aaa", "abba"不是. 两个长…

Problem Description A string is called a square string if it can be obtained by concatenating two copies of the same string. For example, "abab", "aa" are square strings, while "aaa", "abba" are not. Hamming distanc…

题意: 给一个字符串t ,求与这个序列刚好有m个位置字符不同的由两个相同的串拼接起来的字符串 s, 要求字典序最小的答案    分析: 把字符串折半,分成0 - n/2-1 和 n/2 - n-1 dp[i][j] 表示 第i位及之后的总代价为j可不可行 从第 n/2-1 位推回第 0 位, 若dp[0][m] = 1,则存在 然后贪心对每一位从'a'试到'z',选取接下来存在解的字符 #include <cstdio> #include <algorithm> #include…

题意:一个字符串被称为square当且仅当它可以由两个相同的串连接而成. 例如, "abab", "aa"是square, 而"aaa", "abba"不是. 两个长度相同字符串之间的 hamming distance是对应位置上字符不同的位数. 给定一行字符串和 m,输出字典序最小的字符串. 析:首先先用dp判断能不能形成这样的字符串,然后再打印出来,dp[i][j] 表示 i - 中间的数能不能改 j 个字符得到,最后打印…

The chi squared distance d(x,y) is, as you already know, a distance between two histograms x=[x_1,..,x_n] and y=[y_1,...,y_n] having n bins both. Moreover, both histograms are normalized, i.e. their entries sum up to one.The distance measure d is usu…

题目链接:hdu_5903_Square Distance 题意: 给你一个长度为n的a串,一个数m,现在让你构造一个长度也为n的b串,使这个串是由两个相同的串拼起来的,并且和a串对应的位不同的数量为m 题解: 1.可以知道构造的串前面和后面都是相同的,所以只需要构造前半段就行了,当然你可以分类讨论,然后构造 2.设dp[i][j]表示考虑到第i个字符已经有j个与a串对应位不同,然后状态转移方程看代码,注意的是这里要倒着转移回去 因为要满足最小的字典序,并且对应不同的位数为m,倒着转移构造m,最…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8800    Accepted Submission(s): 5991 Problem Description People in Silverland use square coins. Not only they have square shapes but…

$dp$预处理,贪心. 因为$t$串前半部分和后半部分是一样的,所以只要构造前一半就可以了. 因为要求字典序最小,所以肯定是从第一位开始贪心选择,$a,b,c,d,...z$,一个一个尝试过去,如果发现某字符可行,那么该位就选择该字符. 第$i$位选择字符$X$可行的条件: 记这一位选择字符$X$的情况下,对$dis$的贡献为$Q$,$1$至$i-1$位对$dis$贡献和为$F$: 如果第$i+1$位至第$\frac{n}{2}$位,对$dis$的贡献可以凑出$m-Q-F$,那么该位选择$X$可…

小喵的唠叨话:这次的博客,真心累伤了小喵的心.但考虑到知识需要巩固和分享,小喵决定这次把剩下的内容都写完. 小喵的博客:http://www.miaoerduo.com 博客原文: http://www.miaoerduo.com/deep-learning/基于caffe的deepid2实现(下).html ‎ 四.数据的重整,简单的划分 前面的Data层用于生成成对的输入数据,Normalization层,用于将feature归一化,那么之后是不是就可以使用ContrastiveLoss层进…

Week1: Machine Learning: A computer program is said to learn from experience E with respect to some class of tasks T and performance measure P, if its performance at tasks in T, as measured by P, improves with experience E. Supervised Learning:We alr…

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. Example 1: Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4. Example 2: Input: n = 13 Output: 2 Explanation: 13 = 4 + 9. 解题…

A Beginner’s Guide to Eigenvectors, PCA, Covariance and Entropy Content: Linear Transformations Principal Component Analysis (PCA) Covariance Matrix Change of Basis Entropy & Information Gain Resources This post introduces eigenvectors and their rela…

## Nearest Neighbor Search ## Input file: standard input Output file: standard output Time limit: 1 second Memory limit: 1024 megabytes Bobo has a point p and a cube C in 3-dimension space. The point locates at coordinate (x0, y0, z0), while C = {(x,…

Week 1: Machine Learning: A computer program is said to learn from experience E with respect to some class of tasks T and performance measure P, if its performance at tasks in T, as measured by P, improves with experience E. Supervised Learning:We al…

Candela to lux calculation with distance in meters The illuminance Ev in lux (lx) is equal to the luminous intensity Iv in candela (cd), divided by the square distance from the light source d2 in square meters (m2): Ev(lx) = Iv(cd) / (d(m))2 So lux =…

关于triplet loss的原理.目标函数和梯度推导在上一篇博客中已经讲过了.详细见:triplet loss原理以及梯度推导.这篇博文主要是讲caffe下实现triplet loss.编程菜鸟.假设有写的不优化的地方,欢迎指出. 1.怎样在caffe中添加新的layer 新版的caffe中添加新的layer.变得轻松多了.概括说来.分四步: 1)在./src/caffe/proto/caffe.proto 中添加 相应layer的paramter message. 2)在./include/…

这节课主要讲述了RBF这类的神经网络+Kmeans聚类算法,以及二者的结合使用. 首先回归的了Gaussian SVM这个模型: 其中的Gaussian kernel又叫做Radial Basis Function kernel 1)radial:表示输入点与center点的距离 2)basis function:表示‘combined’ 从这个角度来看,Gaussian Kernel SVM可以看成许多小的radial hypotheses的线性组合(前面的系数就是SV的alphan和yn)…

About this Course This course will teach you how to build models for natural language, audio, and other sequence data. Thanks to deep learning, sequence algorithms are working far better than just two years ago, and this is enabling numerous exciting…

##Advice for Applying Machine Learning Applying machine learning in practice is not always straightforward. In this module, we share best practices for applying machine learning in practice, and discuss the best ways to evaluate performance of the le…

from heapq import *; from collections import *; import random as rd; import operator as op; import re; data = [2,2,6,7,9,12,34,0,76,-12,45,79,102]; s = set(); for num in data: s.add(data.pop(0)); if s.__len__() == 4: break; heap = []; for n in s: hea…

论文:Towards End-to-End Lane Detection: an Instance Segmentation Approach 代码:https://github.com/MaybeShewill-CV/lanenet-lane-detection 参考:车道线检测算法LaneNet + H-Net(论文解读) 数据集:Tusimple Overview 本文提出一种端到端的车道线检测算法,包含 LanNet + H-Net 两个网络模型.其中 LanNet 是一种将语义分割和对…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values o…

题目链接: http://codeforces.com/contest/161/problem/D D. Distance in Tree time limit per test 3 secondsmemory limit per test 512 megabytes 问题描述 A tree is a connected graph that doesn't contain any cycles. The distance between two vertices of a tree is th…

Problem Distance in tree 题目大意 给出一棵树,求这棵树上有多少个最短距离为k的点对. Solution 这个题目可以用点分治来做,然而我到现在还是没有学会点分治,所以只好用树形dp了. 这个题目,我们可以将其转化为一个个子树中搞事情,再慢慢合并. 设f[i][j]为以i为根的子树中距离根距离为j的点有多少个. 对于一个点u,我们枚举它的子节点v,则我们可以计算出经过u-v这条边的答案 我们枚举j=1->k,则ans+=f[u][j]*f[v][k-j-1]; 枚举完以后…

题目如下: #include <iostream> #include <string> #include <vector> using namespace std; // use this struct to store square subsequence, 4 positions and 1 length struct SqSb { // take square subsequence as two subsquence s0 and s1 int s00; //…

[CF161.D] Distance in Tree time limit per test 3 seconds memory limit per test 512 megabytes A tree is a connected graph that doesn't contain any cycles. The distance between two vertices of a tree is the length (in edges) of the shortest path betwee…

题目链接  Square Subsets 这是白书原题啊 先考虑状压DP的做法 $2$到$70$总共$19$个质数,所以考虑状态压缩. 因为数据范围是$70$,那么我们统计出$2$到$70$的每个数的个数然后从$2$考虑到$70$. 设$dp[x][mask]$为考虑到$x$这个数的时候,$x$这个数和之前的所有数中,选出某些数,他们的乘积分解质因数,所有的指数对$2$取模之后, 状态为$mask$的方案数. 然后就可以转移了……这个状压DP花了我好几个小时……真是弱啊 哦对最后还要特判$1$的…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9903    Accepted Submission(s): 6789 Problem Description People in Silverland use square coins. Not only they have square shapes but…

Free from square Problem Description There is a set including all positive integers that are not more then n. HazelFan wants to choose some integers from this set, satisfying: 1. The number of integers chosen is at least 1 and at most k. 2. The produ…