atex class 1 div 2

. Instead He gave you an array a1, A2, ..., Aksuch that In other words, n was multiplication of all elements of the given array. Because of precision difficulties, Barney asked you to tell him the answer as an irreducible fraction. In all words need to find it as a fraction p/q such that, where is the greatest common divisor. Since p and Q can be extremely large, you have need to find the remainders of dividing each of the them by 109 + 7. Please note that we want the P and Q to is

, for example, the rule of "balance" (The winning player have to is at least, points ahead to win a set) have no power within the present problem.Test instructions: Test instructions good fan, do the problem by enumerating test instructions.The title says two people are playing, each game has a lot of rounds, each round of winning players can get a point. If someone's score is k in a match, the game ends and a game is re-opened.Now give the two people a total score, a, B, ask you whether the sco

,sizeof(c)); for(i=1; i1); for(i=1; i) {scanf ("%d",x); X++; P2[i]= Getsum (X-1); Modify (x,-1); } memset (P3,0,sizeof(p3)); for(i=n;i>=1; i--) {P3[i]+ = p1[i]+P2[i]; if(P3[i] >= (n-i+1) ) {P3[i]= p3[i]-(n-i+

read his blog, and tell it in detail.1#include 2 3 using namespacestd;4 5 #defineMP Make_pair6 #definePB push_back7typedefLong LongLL;8typedef pairint,int>PII;9 Const Doubleeps=1e-8;Ten Const DoublePi=acos (-1.0); One Const intk=2e5+7; A Const intmod=1e9+7; - - intn,mx; thevectorint>Mp[k]; - - intDfsintXintf) - { + inta[3],num=0; - for(intI=0; I) + if(mp[x][i]!=f) A { at intv=Mp[

case of the root of 1 and O (1) Move the root#include #includeusing namespaceStd;inlineintRead () {intXCharC; while((C=getchar ()) '0'|| C>'9'); for(x=c-'0';(C=getchar ()) >='0'c'9';) x= (x3) + (x1) +c-'0'; returnx;}#defineMN 200000structedge{intNx,t;} e[mn*2+5];inth[mn+5],en,d[mn+5],d1[mn+5],d2[mn+5],c[mn+5],g;inlinevoidInsintXinty) {e[++en]= (Edge) {h[x],y};

. If There is multiple possible solutions, print any of them.Sample Test (s) input20 11 0Output2 1Input50 2 2) 1 22 0 4) 1 32 4 0) 1 31 1 1) 0 12 3 3)

Map[i][j]=min (A[i],a[j]), that is, the value is the smaller of the two, and array A is a sequence that allows you to restore the array A through this array of valuesIdea: In each row if the number of different elements equals n, then this sequence except 0 think the other value is the number of series in the array A, the remaining as long as the 0 is changed to N.Another way of thinking (being taught by others) is that the maximum value in each row is actually the value of the array a[i], but

diamonds.AC Code:1#include 2 #defineINF 0x3f3f3f3f3 #definell Long Long int4 using namespacestd;5 6 Const intMAXN = 1e5+Ten;7 8 intT_1[MAXN], T_2[MAXN];9 intN, Num_c, num_d;Ten One intLowbit (intx) A { - returnx (-x); - } the voidAddintNointStintvalue) - { - for(inti = st; I lowbit (i)) - { + if(no) t_1[i] =Max (T_1[i], value); - ElseT_2[i] =Max (T_2[i], value); + } A } at

); - } About Else $ { - if(p>d)Continue; - Dd.push_back (Make_pair (p,b)); -dd_max=Max (dd_max,b); A } + } the if(cc_max!=0 dd_max!=0) Ans=cc_max+dd_max;//first Case -Ans=max (Ans,solve (cc,c));//Two of them are coins . $Ans=max (Ans,solve (dd,d));//Two of them are diamonds . theprintf"%d\n", ans); the } the return 0; the}In addition, the tree array can also be done, you can refer to http://blog.csdn.net/ssimpl

home or contestSolution: Since the departure from home will return, then we just have to compare the number of times the odd-and-even sex is on the line, the numbers are at home, odd is not back1#include 2 using namespacestd;3 intN;4 Chars1[ $],s2[ $];5 intMain ()6 {7 intans=0;8Cin>>N;9scanf"%s", S1);Ten for(intI=0; i) One { Ascanf"%s", S2); - if(STRSTR (S2,S1)! =NULL) - { theans++; - } - } - if(ans%

, longer lengths, i.e. larger areas After that, you can select another sequence in two, use a double loop to get the sub-columns, and then divide the limit by this sum to get a target value. After that, in the array of initial preprocessing to find the target value, return is greater than its first value, that is, with Upper_bound, to find the length minus one, so that the longest length. Finally, by multiplying the length of the lookup with the length of the current child column, the result is

". ExamplesinputAbacabaOutputYESinputJinotegaOutputNONoteIn the first sample case, one possible list of identifiers would is "numberstring number character number string numb Er ". Here are Kostya would obfuscate the program: Replace all occurences of number with A, the result would is "a string a character a string a", Replace all occurences of string with B, the result would is "a B a character a b a", Replace all occurences of character with C, the result would is "a B

Div 2 A simulation. Simulation. Simulation. I have read a wrong sentence. I have always done this kind of thing recently... According to his experience, he thought that Freda always said "Lala." at the end of her sentences, while rainbow always said "Miao ."At thebeginning of his sentences. Cause wawawa ---- find problem --- solve ----- AC #include B proportional sequence infinite summation formula A + AP ^

Codeforces Round #513 by Barcelona bootcamp (rated, div. 1 + div. 2) Solved:2 out of 8 ... rank:2730 Unrated A. Phone Numbers Difficulty: Universal group. Simulation can be: Summary: Do not waste too much time on simple questions, to improve the

Tag: INF pair scanf has root tree oid test i++ form ClassF-uniformly branched Trees#include #defineLL Long Long#defineFi first#defineSe Second#defineMk Make_pair#definePII Pair#definePLI Pair#defineull unsigned long Longusing namespacestd;Const intN = 1e3 +7;Const intINF =0x3f3f3f3f;ConstLL INF =0x3f3f3f3f3f3f3f3f; LL dp[n][ One][n], Inv[n], f[n], finv[n], g[n][ One];intN, D, mod;voidinit () {inv[1] = f[0] = finv[0] =

print mnumbers h(1),?...,? H(m). If There is several correct answers, you could output any of the them. It is guaranteed that if a valid answer exists and then there was an answer satisfying the above restrictions.Examplesinput31 2 3Output31 2 31 2 3input32 2 2Output

Div2 A: Water question Div2 B: change each number to 1. the required number of steps is the F value of this number (the odd number minus one and the even number is divided by 2) Div1 A: record the highest current height. Div1 B: Sb question, there will be many points with the same X coordinate, but there may be some key points, so multiply the CNT! /(2 ^ c). B

", u, v, W); Edge[u].push_back (Mk (W, v)); Edge[v].push_back (Mk (W, u)); } LL ans=0; for(inti =1; I ) { if(!Vis[i]) { Base. Init (); Id.clear (); DFS (i,0); for(intj =0; J +; J + +) {num[0] =0, num[1] =0; for(Auto x:id) {num[(d[x]GT;GT;J) 1]++; } BOOLFlag =false; for(intK =0; K the; k++) {

　　A question, water problem, judge all points is not into degrees and out of the same can be.Question B, test instructions understand is water problem.Question C. The other party must win at least one of the points after the score mod K. This method can be used to judge.Problem d, construction problem, not = =.E, test instructions is the choice of a node each time, the two sons of the same length of the chain to merge into a new chain of the same length. Ask if the end can be a straight chain, i

such then All Stars lie on that Line.OutputPrint three distinct integers on a single line-the indices of the three points so form a triangle that satisfies the C Onditions stated in the problem.If There is multiple possible answers, you could print any of the them.Sample Test (s) input30 11 01 1Output1 2 3Input50 00 22 02 21 1Output1 3 5NoteIn the first sample, we can print the three indices on any order.In the second sample, we had the following pic