mountain beads

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BZOJ2081 [Poi2010]beads

) { - returnH2[L]-h2[r +1] * b[r-l +1]; $ } $ - SetS; - intCalcintSZ) { the s.clear (); - for(inti =1; i + SZ-1sz)WuyiS.insert (min (Calc1 (i, i + sz-1), CALC2 (i, i + sz-1))); the return(int) s.size (); - } Wu - intMain () { About intI, t; $n =read (); - for(i =1; I read (); - for(i = b[0] =1; I 1] *Base; - for(i =1; I i) AH1[i] = h1[i-1] *Base+A[i]; + for(i = n; i;--i) theH2[i] = h2[i +1] *Base+A[i]; -Ans =0; $ for(i =1; I i) { thet =Calc (i); the

BZOJ2081: [Poi2010]beads

Violent enumeration $k$, for a substring, calculates its positive hash value and the inverse hash value, taking the minimum value to its final hash.For $k$, there are $\lfloor\frac{n}{k}\rfloor$ substrings, and the number of different substrings can be counted by calculating their final hash value.Time Complexity $o (n\log N) $.#include   BZOJ2081: [Poi2010]beads

Super Bouncing Beads

one line: Outputs an integer that represents the number of squares that the Bessie can select.Sample inputSample Input4 32 12 34 1Sample outputSample Output5Data range and TipsData Size Hint11Sample DescriptionBessie can choose to stand in any one of the lattice: (2,1), (2,3), (3,2), (bis), (4,3). In the image below, Bessie is labeled as "*" in the same grid as the other cows.. . . . . . . .B. B. =======> *. * .. B. . =======>. B. .B. B. * . B.Ideas:You c

POJ 1509 Glass beads suffix automaton

To give a circular string, ask where to start is the dictionary order of the string is minimal.Idea: The minimum representation and suffix automata, but I was to learn the suffix automata to write this problem, I did not learn the minimum notation.The procedure is simple, first establish a suffix automaton, and then from the root start along the Tranc pointer from the A->Z, the point that arrives at the end of the string is the smallest dictionary order, the starting point as long as the length

POJ topic 1286 Necklace of Beads (Polya theorem)

POJ topic 1286 Necklace of Beads (Polya theorem)Necklace of Beads Time Limit:1000 MS Memory Limit:10000 K Total Submissions:7061 Accepted:2942 DescriptionBeads of red, blue or green colors are connected together into a circular neckcycle of n beads (n InputThe input has several lines, and each line contains the input dat

"POJ1509" Glass beads

"POJ1509" Glass beads"Title description" Given the string s, and the provision of the end-to-end ring, to find the most dictionary order.The input input has more than one data, and the first line contains only positive integer n, which indicates that there are n sets of data. Each data includes one row, and the string is entered.Output for each data, the output line of data indicates which letter begins with the dictionary order."The Solution" violenc

[ACM] poj 1286 necklace of beads (polyA counting, directly set the formula)

Necklace of beads Time limit:1000 ms Memory limit:10000 K Total submissions:6547 Accepted:2734 DescriptionBeads of red, blue or green colors are connected together into a circular neckcycle of N beads (n InputThe input has several lines, and each line contains the input data N. -1 denotes the end of the input file. OutputThe output shoshould contain the output data

POJ 1509 Glass Beads suffix Automatic Machine

POJ 1509 Glass Beads suffix Automatic Machine Give a circular string and ask where the Lexicographic Order of the string is the minimum. Idea: the least representation method and the bare Question of the suffix automatic mechanism. However, I did not learn the least representation method when I wrote this question to learn the suffix automatic mechanism. The procedure is very simple. First, create a suffix automatic machine, and then start from the r

ALG: beads at the beginning and end

# Include /* Three difficulties: 1) Design a stack structure to meet the following conditions: the time complexity of Min, push, and pop operations is O (1) 3) design a system to deal with word collocation. For example, if Chinese and people can be used together, the Chinese people and China are both effective. Requirements:1) the number of queries per second may exceed;2) the word order is 10 W;3) each word can be combined with up to wordsWhen users enter the Chinese people, they need to re

Poj 2409 let it bead/1286 neckquota of beads polyA count

Q: use k different colors to dye a n-length necklace. Question: 1. Rotation replacement: There are n rotation replacements, which are rotated in sequence: 0, 1, 2, ''' n-1. For each rotation replacement, the number of cyclic factors obtained after cyclic decomposition is gcd (n, I ). 2. Flip replacement: parity. When an odd number is used, the rotation axis is equal to (vertex + link to the Edge End Point). There are n vertices in total, so there are n replications, the number of factors after e

M beads in N colors to find the shortest continuous fragment containing N colors

Original question: There is a string of beads (the beginning and the end are not connected), a total of m, each bead has a color, and the total number of colors does not exceed n (n Question Analysis: I. Brute Force Search 1. The simplest method is brute force search. Scanning the nth position one by one starts to contain the shortest color range of n. the time complexity is O (M ^ 2 ). 2. Start scanning from I. Each time a new color appears, t

[Poj1509] glass beads [suffix automatic mechanism]

Question Returns the minimum representation of a string. Analysis Of course, this question can be done in the smallest notation! But I want to learn the suffix! We multiply the length of this string by two, then create Sam, and then run the smallest letter on Sam every time to find the length of N and then stop. If u is stopped, ANS = sT [u]. Len-n + 1    1 #include View code [Poj1509] glass beads [suffix automatic mechanism]

Bzoj 1501 noi2005 smart beads game dancing-links (DLX)

Given a 10x10 triangle board and 12 kinds of parts, each part can only be placed once, and can be rotated and flipped. Some parts have been placed on the top, and a solution is proposed, place 12 parts on the board without Overlap First of all, this question is about DLX, but it is really disgusting to create a graph. We need to enumerate up to eight orientation positions for each part. At first, I wanted to process all the code, but later I found that I could not do it. So I chose to input all

Xuzhou Copper Mountain near the full set of "reputation first" (have called Sister health) also service one night real how much

"Composition Ren said: ssk493" door Xuzhou Copper Mountain sister find real service "No ☆ NEED ☆ SET ☆ Gold" "Composition ren words: ssk493" [Xuzhou Copper Mountain Hotel real sister find real service "Composition Ren said: ssk493" [Xuzhou Copper Mountain beauty Beautiful service "Composition Ren said: ssk493" [ Xuzhou Copper

Three realms of design: see the Mountain Trilogy

In retrospect of his meditation experience, Castle Peak said: "When not meditation 30 years ago, see Mountain is mountain, see water is water." And then, having witnessed the knowledge, there was a place in it. See the mountain is not a mountain, see water is not water. Now we have a rest place, according to the

Steps for setting confidence in mountain migration

There are many great figures who have summed up the steps for setting confidence in the mountains, which are generally divided into six stages. First, people must have a dream: This is the first step of confidence and prudence. Confidence in mountain migration often begins with a dream. There is no doubt that the biggest energy in the world is a novel idea. There is a saying that a great person can only have a great idea. Confidence begins with an ide

Introduction to Mountain climbing algorithm and simulated annealing algorithm

Transferred from: http://www.cnblogs.com/chaosimple/archive/2013/06/10/3130664.htmlI. Mountain climbing algorithm (hill climbing)This paper introduces the algorithm of mountain climbing before simulated annealing. The mountain climbing algorithm is a simple greedy search algorithm, which chooses an optimal solution from the adjacent solution space of the current

Yuelu Mountain at night

Came to Changsha so many days, except for business trips, most of the time in the company, sister home and Hotel 3.1 line of life, today cooling, longing for a rain did not come, it is a day of wind blowing. After dinner in the evening, it was intended to hike back to the hotel in cool weather, just to relieve the tension of the work. Sister call, Night together to climb Yuelu Mountain, with my idea of walking home coincides with, compared to climb t

Weekend mountain climbing

This is the fifth time during the weekend's mountain training. This time, only two people were involved because of the temporary tasks of other associates. We decided to go through one of the green fields and capture the mountains from the qianjuntai pit to the qingshuijian, this route is 1.5 in the green field, and should not be difficult (for those people ).Because we heard that there was bus No. 929 heading for qianjuntai pit at a.m., we all arrive

Yuelu Mountain. Fog

The beauty of Yuelu mountain, words can not be described, ink can not be portrayed.Since September 2014 to South Central South University to study, today nearly two time, climb Yuelu mountain dozens of times. However, I have never met the prosperity of today. Forest, fog diffuse, far from the mountains of white clouds, in which, such as Mu Fairyland. The fog is blowing, the whole body is refreshing, the moo

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