hack cf

]][i-1] +1;101cnt[3-a[i]][i] = cnt[3-a[i]][i-1];102 }103cnt[1][n+1] = cnt[2][n+1] =Int_max;104 theID =A[n];106 intan = C[id], bn = c[1]+c[2]-an ;107 108 //if (an 109 //puts ("0"); the //return 0;111 // } the 113 intI, J; the vpii ans; the the for(i=1; ii) {117j =Calc (i);118 if(j)119 ANS.PB (MP (J, I)); - }121 122 sort (All (ans));123 124n =SZ (ans); theprintf"%d\n", n);126Rep (I,0, N) {127printf"%d%d\n", Ans[i].fir,

the BOOLFlag =true; the thescanf"%d", n); - theRep (I,0, N) theRep (J,0, N) thescanf"%d", m[i][j]);94 theRep (I,0, N) { the if(M[i][i]! =0) { theFlag =false;98 Goto_output; About } -Rep (J, i+1, N) {101 if(M[i][j]!=m[j][i] | | m[i][j]==0) {102Flag =false;103 Goto_output;104 } the }106 }107 108 Kruskal ();109 theRep (I,0, N) {111Rep (J, i+1, N) { the if(M[i][j]! =T

color, for n pots lined up in a straight line of the flower coloring. The color of the adjacent flowers is different, and the color used is exactly K . Ask a total of several coloring schemes (results except 10e9+7 take remainder). Problem Solving Ideas: First you can m m K C (M, k) and then by trying to discover, the first one has k Choice, second reason cannot be the same as the first one , only (k-1) (k-1) kx (K-1) ^ (n-1) k K Then there is a problem of repu

Test instructions: Give a string, where only F, A, n three letters, ask the least exchange how many times can make all a before all F.Solution: Greedy. First preprocessing how many F on the left side of each bit has a, for each a must be at least the left of the number of times to switch to the F number of its left, and for N to compare the number of F to the left and the number of right a, the small portion of the exchange to the other end of N. The answer to both cases is the final answer.Code

input 58 2 7) 3 11100 Sample output Case 1:3case 2:0 Tips first set of examples: The following 4 books bound together, binding 3 times, composed of 8 13 The second set of examples: only one book, no binding How to solve the problem: we define dp[i] as the minimum number of steps to organize the first stacks of books into non-descending sequences. Defi

) { -T[p].next[id] = + +L; -++T[P].N; the } -p =T[p].next[id];Wuyi++i; the } - } Wu - intDfsintRT) { About inti; $ intSt =0; - - if(T[RT].N = =0) - return 2; A + for(i=0; i -; ++i) { the if(T[rt].next[i]) -St |=DFS (T[rt].next[i]); $ } the Switch(ST) { the Case 0: the return 3; the Case 1: - return 2; in Case 2: the return 1; the Case 3: About return 0; the default: the ret

) { $ if(!Visit[i]) { -p[k++] =i; - for(J=i*i; ji) -VISIT[J] =true; A } + } the - for(i=1; ii) { $ for(j=0; jj) { the if(I%p[j] = =0) { theMask[i] |= (1j); the } the } - } in the #ifndef Online_judge theprintf"p[%d] =%d\n", M, P[m]); About #endif the } the the intMain () { +Ios::sync_with_stdio (false); - #ifndef Online_judge theFreopen ("data.in","R", stdin);BayiFreopen ("Data.out","W", stdout); th

(strcmp (S,S2) >0) strcpy (S,S2); - the for(intI=1; i) * { $ Charc=str[i-1][0];Panax Notoginseng for(intj=1; J) -str[i][j-1]=str[i-1][j]; thestr[i][len-1]=C; + strcpy (S2,str[i]); A intnum=Ten-(s2[0]-'0'); the for(intj=0; J) + { - intk= (s2[j]-'0'+num)%Ten; $s2[j]=k+'0'; $ } - if(strcmp (S,S2) >0) strcpy (S,S2); - } theprintf"%s\n", S); - }Wuyi return 0; the}Vie

141b-hopscotchLet's bust the "level" 0≤ i ≤106, with which assumedly the stone could hit. Let's find the minimal number of square on the. Then we can understand, how many squares there is on this level:one or both. Then we check with one or both ifs (if on this level, squares) if the stone is in corresponding square or not. If the stone is inside then output the answer. If we didn ' t find any square, where the stone is, output "-1".The official meaning is probably 6 levels, one to enumerate pa

[++cns]=Ss; return; } intt1=fnum-Nownum; Rep (I,nowpos,nn-t1+1){ intNss=ss+ (11)); Finds (Nn,fnum,nownum+1, i+1, NSs); }}voidCalcintNnintS) {//The total length is NN, and the state S is calculated in the position of 1, which exists in global pointer P "". cnp=0; Rep (I,1, nn) { intT= (11)); if((st) >0) {p[++cnp]=i; } }}voidinit () {CnS=0; CnP=0; Finds (N,2,0,1,0); MEM (DP,0); Rep (I,1, CnS) { intss=S[i]; Calc (N,SS); Rep (J,2, CnP) { if(g[p[1]][p[j]]=

1Because the first article did not write test instructions led to most of the problems of God has been completely unaware of what is said ... So we have to re-open an article to protect the peace ..."303A" test instructions: for the arrangement of three lengths of $n (nKonjac konjac cultivation of CF Orange name Program 2

; - if(Mark[u]) -nd[u].mx =0; - for(i=0; Ii) { Av =E[u][i]; + if(V! =FA) { theTMP = DFS (V, u) +1; - if(TMP >=nd[u].mx) { $ //find the fathest distance from p[*] to u theND[U].MX2 =nd[u].mx; thend[u].mx =tmp; the}Else if(tmp >nd[u].mx2) { theND[U].MX2 =tmp; - } in } the } the About //-1 means no P in the path the returnnd[u].mx; the } the + voidDFS2 (intUintFaintpmx) { - intI, V; the intTMP =Max (PMX, nd[u

#include#include#include#include#include#includestring>#includeusing namespacestd;#defineN 110000#defineMet (A, b) memset (A, B, sizeof (a))#defineINF 0x3f3f3f3fConst Long LongMax =2000000000; typedefLong LongLL; LL A[n], b[n]; LL N, K; ll Judge (ll mid) {ll I, K1=k, k2=K; for(i=1; i) { if(b[i]mid) {K1-= (A[i]*mid-B[i]); if(k10) return-1; } } for(i=1; i) { if(B[i]1) ) {K2-= (a[i]* (mid+1) -B[i]); if(k20) return 0; } } return 1;}int

, directly exit.1#include 2#include 3#include 4#include 5#include 6#include 7#include Set>8#include 9#include Ten#include One#include A#include -#include -#include the using namespacestd; - #defineINF 100000 -typedef unsignedLong Longul; - Const intmaxn=10010; +Mapint>MP; - intMain () + { A ul N; atscanf"%i64u",n); - intflag=1; - while(n>1){ -mp[n]=1; - if(n%2==0) n=n/2; - Elsen=3*n+3; in if(mp[n]==1) {flag=0; Break;} - } to if(flag==1) printf ("tak\

1, CF #371 (Div. 2) C. Sonya and Queries map application, also available trie2, Summary: Start directly with the array traversal, decisive t a pitchTest instructions: Number of T, odd change 1, even change 0, then match with ask.#include #defineMax (A, b) a>b?a:b#defineF (I,A,B) for (int i=a;i#defineMes (A, b) memset (A,b,sizeof (a))#defineINF 0x3f3f3f3f#defineLL Long Longusing namespacestd;Const intn=110000, max=1000100; LL Cal (ll m) {intk=0; LL sum

)? ≠? (aJ,?bJ,?CJ).The next line contains a integer- q (1?≤? Q≤?100), denoting the number of the queries.Then followsQLines, containing space-separated, integers- ui and vi (1?≤? u i,? v i? ≤? N ). It's Guaranteed that ui? ≠? v I .OutputFor each query, print the answer-a separate line.Sample Test (s) input4 51 2 11 2 22 3 12 3 32 4 331 23 41 4Output210Input5 71 5 12 5 13 5 14 5 11 2 22 3 23 4 251 55 12 51 51 4Output11112NoteLet ' s consider the first sample. The figure above shows the f

Write PPC recently ProgramAfter repeated debugging procedures, I summarized some methods to improve the debugging efficiency. 1. First, you 'd better have a real PDA. The simulator runs much slower than the real PDA. 2. Recommended PDA controller professional. It is used to remotely control the PDA. It is placed on the desktop and set to start when the PDA is connected. You can view the screen on the PDA in real time and provide functions such as soft start, task management, file management,

You can go to CF to find the question. Analysis process: If it is an even number, it is directly equal to dividing by 2, for example, F (8) = f (4) = F (2) = F (1) If it is an odd number, add 1 after dividing by 2 directly, for example, F (1) = f (0) + 1 = 0 + 1 = 1 It can be proved that the Left and Right numbers can be continuously divided by 2, and the formula is obtained: F (0) + x = x. So what is X? First, we can see through observation that as l

Tags: Io for SP on HTML amp as HTM C #include Cf 479d long jumps

Input n a B k has n floors. The starting point of n floors cannot be reached at layer a and layer B. Assuming that the current position on layer x can reach layer y at each time. | x-y | Dp [I] [j] is the number of solutions that reach layer j for the I-th dp [I] [j] = sum (dp [i-1] [k]) | k-j | K that meets the condition is a continuous segment with prefix and optimization# Include CF 479E Riding in a Lift prefix and DP