forehead necklace

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: number of different forms, in each line cor

Accepted neck.pdf Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 1022 accepted submission (s): 406Problem descriptioni Have n precious stones, and plan to use K of them to make a neck.pdf For my mother, but she won't accept a neck1_which is too heavy. Given The value and the weight of each precious stone, please help me find Out of the most valuable necklace my mother will accept. Inputthe first l

Maintain a ring. Each vertex has a color and provides six operations: 1. Rotate the ring clockwise K 2. Flip along the diameter of point 1 3. Swap two beads 4. Dye a segment 5. query the color segments on the ring. 6. query the color segments in a specific interval. The general processing method for color segments is to record three values in each interval, including the number of color segments, the color of the Left endpoint, and the color of the right endpoint, if the color of the Left endpoi

The implementation code is as follows:#include #includestring>#includeusing namespacestd;intMain () {intN, M; CIN>> N >>m; int*a = (int*)malloc(n *sizeof(n)); for(inti =0; I ) Cin>>A[i]; //Define M operations for(intx =0; x ) { intT1, T2; strings; CIN>>s; if(s = ="DEL") {cin>>T1; for(inti =0; I ) { if(A[i] = =T1) { for(intj = i; J 1; J + +) A[j]= A[j +1]; Break; }} N--; } Else if(s = ="ADD") {n++; CIN>> T1 >>T2; for(inti =0;

Do noip questions -- Training DP Search for confidence first # Include Noip2006 energy necklace

Welcome to the original source-- Blog Park-zhouzhendong go to the blog Park to see the puzzleTopic Portal-BZOJ1878Test Instructions SummaryA sequence of length n is given, using M-times to ask how many different numbers are in the interval li~ri.0 SolvingThere are many ways to do this.Here we introduce the tree-like array and the team, which are offline algorithms.tree-like arrayWe sorted the sequence according to R from small to large.Then walk from left to right.Add a number to the current st

This article ljh2000Author Blog: http://www.cnblogs.com/ljh2000-jump/Reprint please specify the source, infringement must investigate, retain the final interpretation right!Title Link: HDU5730Positive solution: divide and Cure $fft$Problem Solving Report:Divide and cure $+fft$ template problem.Easy to find a $o (n^2) $ $dp$ equation, is a convolution form, in the division process, with the left to update the right, do a $fft$.Be careful not to empty the array every time.#include 　　HDU5730 Shell

"Topic link" http://acm.hdu.edu.cn/showproblem.php?pid=5730"The main topic"Give An array w, representing the weights of the fields of different lengths, such as W[3]=5, if the field length is 3, then its weight is 5, and now there is a field of length n, the sum of the values of the field weights obtained by different splits. Exercises　　Note Dp[i] is the answer to the length of I, dp[i]=sum_{j=0}^{i-1}dp[j]w[i-j], found to be a convolution of the formula, so the operation can be optimized by FFT

#include 　　C + + Template implementation single-necklace list

Take the query offline, read all the queries, and then sort all the queries in ascending orderWhen dealing with the I query, ensure that all and just the same points of the right range ri from the first number to the I queryAnd the position of the same point at each point is the last point within the RI range, due to theAll queries have a right range greater than RI, so deleting the same point does not affect the subsequent query#include #include #include #include using namespace Std;const int M

intMax_n = 1e5+9;7typedef pairint,int>P;8P q[max_n*4];9 intq1[max_n*4];Ten intvis[max_n*Ten]; One Long LongVec[max_n]; A Long LongRes[max_n]; - Long Long out[max_n*4]; - voidAddintXLong Longnum) the { - for(; xx)) -RES[X] + =num; - } + Long LongSumintx) - { + Long LongAns =0; A for(;x>0; x-= (x-x)) atAns + =Res[x]; - returnans; - } - intMain () - { - intn,m,t; inCin>>T; - while(t--) to { +Cin>>N; -memset (Res,0,sizeof(res)); thememset (Vis,0,sizeof(Vis)); *

Question: Given a sequence, how many different numbers are in an intervalA positive solution is a tree array that sorts all the intervals by the left endpoint and then counts the first number of each color at the left end of the line. Maintain with a tree arrayI'm writing about the MO team. Obviously it's a slow m√m, but it's still a complicated one to accept.At first, the discretization array opens up a small number of seconds re ... On my knees#include Bzoj 1878 SDOI2009 hh

);}inlineEoperator*= (e A,Conste b) {return(A = a * b);}inlineEoperator/(e A,Const Double B) {returnE (a.r/b,a.i/b);}inlineEoperator/= (e A,Const Double B) {return(A = A/b);}intN,M,L,R[MAXN];voidFFT (e* A,intf) { for(inti =0; I if(I for(inti =1; I 1) {E wn (cos (pi/i), f * sin (pi/i)); for(intj =0; J 1)) {E W (1,0); for(intK =0; K if(f = =-1) for(inti =0; I intN,A[MAXN],F[MAXN];voidSolveintLintR) {if(L = = r) {F[l] = (F[l] + a[l])% P;return; }intMID = L + R >>1; Solve (L,mid); n = mid-l +1;

Topic Links:http://poj.org/problem?id=1286Main topic:Given 3 colors of beads, the number of each color beads are unlimited, the beads are made of the length of the necklace of N.Ask how many kinds of non-repeating necklaces can be made, and the final result will not exceed the range of int type data. and twoThe necklace is the same, when and only if the two necklaces by rotating or flipping can be coinciden

I made a water question yesterday, and today it is a relatively water application. The beads of n necklaces are given. There are two colors at both ends of the beads. The adjacent beads on the necklace must match the color to determine whether they can be pieced together into a one-day necklace. It was quite watery, but at first I thought of the entire necklace a

NecklaceFrog have \ (n\) gems arranged in a cycle, whose beautifulness is \ (a_1, a_2, \dots, a_n\). She would like to remove some gems to make them into a beautiful necklace without changing their relative order.Note that a beautiful necklace can is divided into \ (3\) consecutive parts \ (X, y, z\), where \ (x\) consists of gems with non-decreasing beautifulness, \ (y\) is the only perfect

First necklace. The implementation principle of the IMG src attribute to the same picture Effect Chart: Copy Code code as follows: A second necklace. Implementation principle of the IMG src attribute loop set the picture Effect Chart: Copy Code code as follows: A third necklace. Implementation o

There is a long color necklace that requires you to quickly maintain a data structure. He can: 1. Put the last K of the sequence in front. 2. Set the interval to 2 ~ The CNT beads are flipped. 3. Swap the beads of position I and position J. 4. dye the interval I to J into K. 5. Number of color blocks in the output sequence 6. output the number of color blocks from I to J. Idea: splay. Some of the hard-to-solve problems are to always think that this i

Necklace of BeadsTest instructions: In three colors to the length of N (n Idea: Pure equivalence class calculation problem;Emphasis: Conversion of rotation and rollover to displacement operations;Rotation: The length of the interval is enumerated, that is 0 Flip: The odd even, and then find the number of symmetric axis and the number of cycles in each case can be;The above-calculated a+b is just the sum of the number of fixed points, the final divided

Title effect: some beads necklace. Beads have different colors. We asked several times how many different colors had been in a certain range.Thinking: This problem has taught me to learn smart offline practices. Sort the problem by the left endpoint. The processed position is the first occurrence of each color. Where the next occurrence of each color is. Then, 1 to the CNT cycle, where there is a problem with the left endpoint is the current node, jus