forehead necklace

People who have seen "Titanic" will never forget the "Ocean Heart (Heart of the Ocean)" necklace that slipped from the bottom of the heroine's finger to the end of the film, and she accompanied Rose for a lifetime, condensing the unforgettable memories. This example through the Photoshop Path tool and the layer style synthesis use, has created a romantic "The Ocean Heart" necklace, the operation is simple a

Few women can resist the charm of jewelry, and pearls with its crystal and elegant won the popular love. There's no such thing as pearls, from a nine-year-old girl to a 90-year-old woman, she can see the light in her dress. In the east, the pearl symbolizes auspicious consummation, New year just passed, give everybody a string of pearl necklace, wish everybody in the New Year peace Ruyi! 1. Create a new document, select the Ellipse tool in the toolbar

Like a stone merge, in order to process the ring, we can copy the input data one after the other. In this way, the final result of the lifting point to find the largest.Pay attention to the calculation of the price here, because our Data[i] only records the head of the bead and the tail of the beads is the head of the next bead.Because it is necessary to calculate dp[i][j]Used to dp[i][k] K is smaller than J so J orderDp[k][j] K ratio i large so I reverse orderK Insert can be for (int2*n-1; I >=

I have been working on noip recently. I have not been able to solve this problem because I am too busy. Ah, ah ...... Clearly, the water is terrible. I made a lesson !! I don't want to engage in oi anymore! Especially funny !!!! Code: # Include Noip2006t1 energy necklace

// The neckcycle (Necklace) // PC/ultraviolet A IDs: 111002/10054, popularity: B, success rate: Low Level: 3 // verdict: accepted // submission date: 2011-10-04 // UV Run Time: 0.556 S // copyright (c) 2011, Qiu. Metaphysis # Yeah dot net // [Problem description] // my little sister had a beautiful neck1_made of colorful beads. each two // successive beads in the neckdeskshared a common color at their meeting point, // as shown below: /// ---- green -

Source: http://poj.org/problem? Id = 1286 Q: I used three colors to color a n-length necklace. How many methods are there... Idea: continue to use the bare ploya to make water healthier... Re is pitted by n = 0, WA is pitted by long ,,, Code: # Include

Rokua Topic LinksTest instructionsGiven a long 5w static sequence, ask for 20w times, each time ask to find the number of elements in a range Dyeing problem God, I've just been able to do it, and I feel like we can use a unified approach.The first is to figure out the position of the nearest previous element, which is the same as an element, so that only the last position within the interval is less than the lower bound is the number of elementsIt's a very reasonable look, but I can't see

() { while(~SCANF ("%d%d",n,m) {memset (f),0,sizeoff); for(intI=1; i) { intU,v; scanf"%d%d",u,v); F[U][V]=1; } if(n==0) {printf ("0\n");Continue;} intans=0x7fffffff; for(intI=1; iN; Do{memset (g,0,sizeofg); for(intI=1; i) { for(intj=0; j) { if(j==0f[i][p[n]]==0f[i][p[1]]==0) g[j][i-1]=1; Else if(j!=0f[i][p[j]]==0f[i][p[j+1]]==0) g[j][i-1]=1; }} ans=min (ans,n-Maxmatch ()); } while(Next_permutation (p+1, P+

MO team algorithm.#include #include#include#include#includeusing namespacestd;Const intMAXN =50000+Ten;intcnt[1000000+Ten];intA[MAXN], POS[MAXN];intN, M, Ans, L, R;structx{intL, R, id; intans;} s[200000+Ten];intans[200000+Ten];BOOLcmpConstXa,Constxb) { if(POS[A.L] = = POS[B.L])returna.rB.R; returna.lB.L;}BOOLCMP1 (ConstXa,Constxb) { returna.idb.id;}intMain () { while(~SCANF ("%d", N)) {intSZ = sqrt (n); memset (CNT,0,sizeofCNT); for(inti =1; I ) {scanf ("%d", A[i]); Pos[i]= I/sz; } scanf

2. R, then obviously the first point is removed.Then we can record the next position of each color, then the first point is removed, it is equal to the next position of the point becomes the first point, the next position of the next position becomes the second point,Since we only turn the second point that appears into 1, it becomes the point of the first point (that is, the next position), we want to change it back to 0, the next position in the next location, we have to change it back to 1.T

){ +Memset (G,0,sizeof(g)); -memset (D,0,sizeof(d)); the *scanf"%d",n); $ for(i=1; i){Panax Notoginsengscanf"%d%d",u,v); -g[u][v]++;//here is the input has a heavy edge, so the point of the heavy edge corresponding to the degree of the corresponding plus 1, remember that Hangzhou Electric has a problem to be judged heavy theg[v][u]++; +d[v]++; Ad[u]++; the } + for(i=1; i -; i++){ - if(d[i]%2) Break; $ } $ - if(kase>1) print

; + } - $ voidADD (intLintR) $ { -l=l/2+1, r=r/2-1; - if(L>r)return ; theq[++tot]=(Node) {l,r}; - }Wuyi voidManacher () the { -m=2*n+1; Wu for(intI=1; i) - { Abouta[i1]=S[i]; $a[i1|1]='#'; - } -a[0]='+', a[m+1]='-', a[1]='#'; - intmx=0, id; A for(intI=1; i) + { the if(mx>i) P[i]=min (mx-i,p[id*2-i]); - Elsep[i]=1; $ while(A[i-p[i]]==a[i+p[i]]) p[i]++; theADD (i-p[i],i+p[i]); the if(P[I]+IGT;MX) mx=i+p[i],id=i; the } the } - in

Test instructionsA sequence with a length of N and a M query;The number of different numbers in the interval to ask;nExercisesThe online algorithm is too advanced and will not, so this problem will be done offline;Analysis data range, m query can be stored completely, 1000000 of the number can also be hash barely discrete;Then consider the number of a range of numbers, probably 1-r minus 1-(L-1) ;But because there are duplicate numbers, for the number of repetitions, we should only calculate bet

The main topic: give a string, find out how many palindrome is at least the number of the composition. Palindrome strings can overlap.Idea: All palindrome strings in the original string are counted, and then into some intervals, the problem is transformed into a range of problems.CODE:#include Bzoj 3790 Magic Necklace hash+ two points

n is only 350, so we can enumerate each point directly as the break point, and then count the maximum value of its left and right color beads, and then take the maximum of those values.The code is as follows:/*id:15674811lang:c++task:beads*/#include #include #include #include using namespace Std;intMain () {Ofstream fout ("Beads.out"); Ifstream Fin ("Beads.in");Char Str[ +];intLen while(Fin>>len) {fin>>Str+1;intans=0; for(intI=1; iinttmp=0;intj=i,t=i;CharCh=Str[j]; while(true) {if(St

) { intret=0; while(p) ret+=s[p],p-=lowbit (P); returnRet;}intMain () {scanf ("%d",N); for(intI=1; i) {scanf ("%d", A +i); if(!m[a[i]]) m[a[i]]=++CNT; A[i]=M[a[i]]; } scanf ("%d", m), cnt=0; for(intI=1, l,r;i) {scanf ("%d%d",l,R); MEM[CNT].U=L,MEM[CNT].T=I,MEM[CNT]. next=tab[r],tab[r]=mem[cnt],cnt++; } for(intI=1; i){ if(Pos[a[i]]) Update (pos[a[i]],-1); Pos[a[i]]=i,update (I,1); for(Hash *p=tab[i];p; p=p->next) Ans[p->t]=sum (N)-sum (p->u-1); } for(intI=1; i"%d\n", Ans[i])

"Topic link" http://www.lydsy.com/JudgeOnline/problem.php?id=3790"The main topic"Ask at least a few palindrome string can constitute a string, overlapping parts can be combinedExercisesWe first use Manacher to process the longest palindrome string in each position,Then the problem is converted to the minimum line to cover the whole range, that is the classic DP problem,Dp[i]=min (dp[j]+1) (the left end of the segment I-1 and the right end of the J segment has intersection)Use a tree-like array t

Matrix Fast power.#include using namespacestd;Const Long Longmod=1e9+7;Const Long Longinf=1e18;intT;Long LongN;structmatrix{Long Longa[5][5]; intR, C; Matrixoperator*(Matrix b);}; Matrix X, Y, Z; Matrix Matrix::operator*(Matrix B) {matrix C; memset (C.A,0,sizeof(C.A)); intI, j, K; for(i =1; I ) for(j =1; J ) for(k =1; K ) C.a[i][j]= (C.a[i][j] + (a[i][k] * b.a[k][j])%mod)%MoD; C.R=r; C.c=B.C; returnC;}voidinit () {n= N-2; memset (X.A,0,sizeofx.a); memset (Y.A,0,sizeofy.a); m

Title Link hh NecklaceThis problem can be directly on the chairman of the tree template#include Of course, you can do it with the MO team algorithm.#include Bzoj 1878 [Sdoi2009]hh's Necklace (Chairman tree or MO team algorithm)

++) - { to if(I j) Swap (Y[i], y[j]); +K = len/2; - while(J >=k) the { *J-=K; $K/=2;Panax Notoginseng } - if(J K; the } + } A voidFFT (Complex y[],intLenintOn ) the { + Change (y, len); - for(inth =2; H 1) $ { $Complex wn (cos (-on*2*pi/h), sin (-on*2*pi/h)); - for(intj =0; J h) - { theComplex W (1,0); - for(intK = J; K 2; k++)Wuyi { theComplex U =Y[k]; -Complex t = w * y[k + h/2]; WuY[k] = U +T;