题目链接:https://vjudge.net/problem/HDU-3374 String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3646    Accepted Submission(s): 1507 Problem Description Give you a string with length N,…

String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4828    Accepted Submission(s): 1949 Problem Description Give you a string with length N, you can generate N strings by left shifts…

String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1512    Accepted Submission(s): 668 Problem Description Give you a string with length N, you can generate N strings by left shifts…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=3374 题目: String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3029    Accepted Submission(s): 1230 Problem Description Give you a string wi…

/** 题目:hdu3667 Transportation 拆边法+最小费用最大流 链接:http://acm.hdu.edu.cn/showproblem.php?pid=3667 题意:n个城市由m条有向边连接.要从城市1运输k流量到城市n.每条边有可以运输的流量容量,以及费用系数ai. 费用系数指该条边的费用为该条边的运输流量x的平方乘以ai.即ai*x^2. 如果无法运输k流量,输出-1,否则输出从城市1运输k流量到城市n的最小花费. 思路:拆边法+最小费用最大流 假设从u->v 容量为…

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings:String Rank SKYLONG 1KYLONGS 2YLONGSK 3LONGSKY 4ONGSKYL 5NGSKYLO 6GSKYLON 7and lexicographically…

#include <iostream> #include <cstring> #include <cstdio> using namespace std; int nxt[1000005], len; char a[1000005]; void mknxt(){ int i=0, j; nxt[0] = j = -1; while(i<len){ if(j==-1 || a[i]==a[j]) nxt[++i] = ++j; else j = nxt[j]; }…

题意: 给出一个字符串,问这个字符串经过移动后的字典序最小的字符串的首字符位置和字典序最大的字符串的首字符的位置,和能出现多少次最小字典序的字符串和最大字典序的字符串 解析: 能出现多少次就是求整个字符串能出现几次循环 然后就是最大最小表示法..有点厉害... #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #in…

目录 最小角回归法 一.举例 二.最小角回归法优缺点 2.1 优点 2.2 缺点 三.小结 更新.更全的<机器学习>的更新网站,更有python.go.数据结构与算法.爬虫.人工智能教学等着你:https://www.cnblogs.com/nickchen121/ 最小角回归法 最小角回归相当于前向选择法和前向梯度法的一个折中算法,简化了前项梯度法因\(\epsilon\)的迭代过程,并在一定程度的保证了前向梯度法的精准度. 通常用最小角回归法解决线性模型的回归系数.对于一个有\(m\)个样…

String Problem Description Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings: String Rank SKYLONG 1 KYLONGS 2 YLONGSK 3 LONGSKY 4 ONGSKYL 5 NGSKYLO…

hdu 3374 String Problem 最小表示法 view code#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <string> using namespace std; const int N = 10010; int n; char s[105]; map<…

Problem Description Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings:String Rank SKYLONG 1KYLONGS 2YLONGSK 3LONGSKY 4ONGSKYL 5NGSKYLO 6GSKYLON 7an…

http://acm.hdu.edu.cn/showproblem.php?pid=3374 String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2045    Accepted Submission(s): 891 Problem Description Give you a string with lengt…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3374 题意很简单,输出的是最小字典序的编号,最小字典序个数,最大字典序编号,最大字典序个数. 可以想一下因为是循环移动的, 所以最大字典序个数和最小字典序个数相等=循环节: 本题有一个最大最小表示法: 这里简单介绍对字符串最小表示的方法: (1)  利用两个指针p1, p2.初始化时p1指向s[0], p2指向s[1]. (2)  k = 0开始,检验s[p1+k] 与 s[p2+k] 对应的字符是…

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings: String Rank SKYLONG 1 KYLONGS 2 YLONGSK 3 LONGSKY 4 ONGSKYL 5 NGSKYLO 6 GSKYLON 7 and lexicograp…

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings: String Rank SKYLONG 1 KYLONGS 2 YLONGSK 3 LONGSKY 4 ONGSKYL 5 NGSKYLO 6 GSKYLON 7 and lexicograp…

做一个高产的菜鸡 传送门:HDU - 3374 题意:多组输入,给你一个字符串,求它最小和最大表示法出现的位置和次数. 题解:刚刚学会最大最小表示法,amazing.. 次数就是最小循环节循环的次数. #include<bits/stdc++.h> using namespace std; int nt[1000100],b[1000100]; char a[1000100]; void kmp_nt(int m) { int i,j; i = 0; nt[0] = j =-1; while(…

String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1492    Accepted Submission(s): 662 Problem Description Give you a string with length N, you can generate N strings by left shifts.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3374 题目大意:输出最大和最小的是从哪一位开始的,同时输出最小循环节的个数. 这里简单介绍对字符串最小表示的方法: (1)  利用两个指针p1, p2.初始化时p1指向s[0], p2指向s[1]. (2)  k = 0开始,检验s[p1+k] 与 s[p2+k] 对应的字符是否相等,如果相等则k++,一直下去,直到找到第一个不同,(若k试了一个字符串的长度也没找到不同,则那个位置就是最小表示位置,…

HDU 3374 String Problem (KMP+最大最小表示) String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1602    Accepted Submission(s): 714 Problem Description Give you a string with length N, you c…

String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2706    Accepted Submission(s): 1140 Problem Description Give you a string with length N, you can generate N strings by left shift…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=3374 [题目大意] 给出一个字符串,求出最小和最大表示是从哪一位开始的,并且输出数量. [题解] 最小最大表示可以用最小最大表示法解决,数量则可以发现就是该字符串的循环节,可以用nxt数组求解. [代码] #include <cstring> #include <cstdio> #include <algorithm> const int N=1000010; usin…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3374 题目大意:给出一个字符串,依次左移一个单位形成一堆字符串,求其字典序最小和最大的字符串需要左移多少位,以及一共有几个这样的字符串(例如0101->1010->0101). 解题思路:首先可以确定两个字符串出现的次数应该相同,即循环节数目,这个比较容易得到.然后就是最大最小字符串如何得到的问题了,直接暴力肯定超时.... 所以这里找到了比较好的方法,转载:http://blog.csdn.ne…

题意:给你一个字符串,这个字符串可以这样操作:把第一个字符放到最后一个形成一个新的字符串,记原式Rank为1,每操作一步Rank+1,问你这样操作得出的最小字典序的字符串的Rank和这样的字符串有几个,最大字典序的字符串的Rank和这样的字符串有几个. 思路:手动模拟操作复杂度O(n^2)果断超时,引入一种专门计算此情况的方法,复杂度O(n). 这里只说最小表示: 我们先拿两个指针i,j,分别指向s[0],s[1],将k初始化为0.然后我们循环计算s[i + k]是否等于s[j + k],直到找…

Problem - 3374   KMP求循环节. http://www.cnblogs.com/wuyiqi/archive/2012/01/06/2314078.html   循环节推导的证明相当的好,这题是很裸的套算法的题. 代码如下: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; ; char buf…

前面的文章对线性回归做了一个小结,文章在这: 线性回归原理小结.里面对线程回归的正则化也做了一个初步的介绍.提到了线程回归的L2正则化-Ridge回归,以及线程回归的L1正则化-Lasso回归.但是对于Lasso回归的解法没有提及,本文是对该文的补充和扩展.以下都用矩阵法表示,如果对于矩阵分析不熟悉,推荐学习张贤达的<矩阵分析与应用>. 1. 回顾线性回归 首先我们简要回归下线性回归的一般形式: \(h_\mathbf{\theta}(\mathbf{X}) = \mathbf{X\theta…

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some). For example 0110 express a necklace, you can…

String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2000    Accepted Submission(s): 875 Problem Description Give you a string with length N, you can generate N strings by left shifts.…

On Octorber 21st, HDU 50-year-celebration, 50-color balloons floating around the campus, it's so nice, isn't it? To celebrate this meaningful day, the ACM team of HDU hold some fuuny games. Especially, there will be a game named "crashing color ballo…

后缀数组+RMQ+二分 后缀数组二分确定第K不同子串的位置 , 二分LCP确定可选的区间范围 , RMQ求范围内最小的sa Boring String Problem Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 661    Accepted Submission(s): 183 Problem Description In thi…