oracle srm

at the border, so only multiply $n -1$The other is to put the $i$ ball in the middle of a position, assuming that the left side of the $j$ ball, then the transfer is necessarily$ $DP [i]=\sum\limits_{j=1}^{i-2}c_{i-1}^j*dp[j]*dp[i-j-1]* (n-2) $$So the total shift is$ $DP [i]= (\sum\limits_{j=1}^{i-2}c_{i-1}^j*dp[j]*dp[i-j-1]* (n-2)) +2*dp[i-1]* (n-1) $$The answer is obviously $dp[n]$.Long LongDp[max_n];Long LongMod=1000000007;Long LongC[max_n][max_n];classsumoverpermutations{ Public: int

Enumerates the number of cases in which Alice throws the dice to get results.Enumeration Bob throws the dice to get the result of the extreme number of each case.Then enumeration Alice throws the dice to get the number of results and Bob throws the dice lower than Alice's number.The results are then computed.#include tc-srm-626-div1-250

to dividing all y into y1 parts, C (y-1,y1-1)For each of the multi-plug y, we fortress a Z to prevent y adjacent, then after the end of our z is z2=z-z1-y2, and then insert these z y1 sequence of two ends, the scheme is C (2*Y1,Z2)At this point, the assignment ends#include #include#include#include#includeusing namespaceStd;typedefLong Longll;intn=1000000;intmo=1000000007;intjc[1000011],fc[1000011];intans,tmp,ts,x,y,z,xl,yl,req,zl,dt,tj,m,r,g,b,j,i;intMiintXintz) { intl; L=1; while(z) {if(z%

Objective:As I have been in the DIV2 of the weak dish. I don't know what to say.A: It must be judged that there are 8 ' R ', only one per column per rowQuestion B: Probably int e,int em,int m,int mh,int HThen em can give value to E,M,MH can give value to h,m;I'm doing two points and then judging.C: When encountering number theory, kneel.Seek a*x^2+b*x+c=0 (mod p); p=10^9;Not satisfied with the output-1 a,b,c,x are in "0,999999999";First of all, p=10^9 is very special, so consider it from here.p=

#defineEPS 1e-9#defineAll (x) X.begin (), X.end ()#defineINS (x) Inserter (X,x.begin ())#definefor (i,j,k) for (int i=j;i#defineMAXN 1005#defineMAXM 40005#defineINF 0X3FFFFFFFusing namespaceStd;typedefLong LongLL; LL I,j,k,n,m,x,y,t,big,cas,num;BOOLFlag; LL Cur,ans;BOOLprim[2000005]; LL ver[2000005]; voidGetprim (ll size) {ll m=SQRT (size+0.5); memset (Prim,0,sizeof(prim));//can be emptied according to the situationnum=0;//put the found prime number in the Ver array, num is the length of the ver

2333 ...Because the TC is too few participants. Plus constantly FST I dropped div2 that.Fortunately, the finished back Div1 the.250The problem of water500The problem of water.Direct BFS expansion.Pay attention to the weight. I also started with Cantor. It's really superfluous.1000The question test instructions the wrong understanding.I said, look at the code, it's not right.It's just a very easy question.Binary enumeration of which leaf nodes to burnAnd then for each method of burningFind the sh

); } }returnAns }};1000 points:Dp[i][sta][k] Represents the enumeration to the I card, /************************************************************************* > File name:1000.cpp > Author: ALex > Mail: [email protected] > Created time:2015 May 08 Friday 21:25 26 seconds *********************************** *************************************/#include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespac

= =-1) || (Case = =4)) Test_case_4 ();if(Case = =-1) || (Case = =5) ) test_case_5 (); } the Private: theTemplate stringPrint_array (ConstVector"{ "; for(TypeName Vector'\"'"\","; OS " }";returnos.str ();} the voidVerify_case (intCase,Const intexpected,Const intreceived) {Cerr "Test Case #""...";if(expected = = Received) Cerr "PASSED"Else{Cerr "FAILED""\texpected: \ ""'\"'"\treceived: \ ""'\"'Endl;} } - voidTest_case_0 () {intARG0 =5;intARG1 =3;intARG2 = A;intARG3 =1; Verify_case (0,

= 1000000007;const int N = 55; ll dp[n][n];void Add (ll a, ll b) {A + = B;a = (a% mod + MoD)% MoD;} ll Mut (ll A, ll b) {a = a * b% Mod;a = (A + MOD)% Mod;return A;} LL gao_up (vectorDIV1 250PTExtract all the l,r, first infer whether it is consistent.And then compare the positionDIV1 PTThe XOR result is less than or equal to limit, the case of equality is processed first, and the result is different or results each bit is equal to the limit, and the column equation Group is evaluated for rank.o

must lose, we can use the memory of the search to achieve the above process ~ ~ ~Code:1vectorint>Card;2 intdp[ -][555], N;3 intDfsintThintmask)4 {5 if(Mask = =511)return 1;6 if(th = = N)return 0;7 if(~dp[th][mask])returnDp[th][mask];8 intCNT =0;9 for(inti =0; I if((Card[i] | mask) = = mask) cnt++;Ten if(cnt > th !dfs (th +1, mask))returnDp[th][mask] =1; One for(inti =0; I if((Card[i] | mask)! =mask) A { - if(!dfs (th +1, Mask | Card[i]))returnDp

].length (); j1++) { for(intI2 = I1; I2 for(intJ2 = J1; J2 0].length (); j2++) {inta1 = A[i2][j2];if(I1 >0) A1-= a[i1-1][J2];if(J1 >0) A1-= a[i2][j1-1];if(I1 >0 J1 >0) A1 + = a[i1-1][j1-1];if(A1 = = k) {intT1, t2, anst =0; T1 = i1, t2 = (int) b.size ()-I2-1; Anst + =2* min (t1, T2) + max (t1, T2); T1 = J1, t2 = (int) b[0].length ()-J2-1; Anst + =2* min (t1, T2) + max (t1, T2); ans = min (ans, anst); } } } } }if(ans = =1e

{ Public:stringShortestcommon (stringS1,stringS2) {intT1 =0, t2 =0; for(inti =0; I if(S1[i] = =' * ') T1 = i; for(inti =0; I if(S2[i] = =' * ') t2 = i;if(T1 > T2) swap (S1, S2);stringANS1 ="", ANS3 =""; for(inti =0; I if(S1[i] = =' * ') ans1 = S1.substr (0, i); for(inti =1; I if(s2[i-1] ==' * ') Ans3 = S2.substr (i, s2.length ()-i); while(s1[0] !=' * ' s2[0] !=' * ') {if(s1[0]! = s2[0])return "Impossible"; S1.erase (0,1), S2.erase (0,1); } while(S1.back ()! =' * ' s2.back ()! =' * ')

) {memset (vis,0,sizeof (VIS)); node kiss; kiss.a[0]=a,kiss.a[1]=b,kiss.a [2]= C; int sum=kiss.a[0]+kiss.a[1]+kiss.a[2 ]; if (sum%3!=0 ) return "impossible" ; sort (KISS.A, Kiss.a+3 ); Vis[kiss.a[0]][kiss.a[1]]=1 ; queue Q; Q.push (Kiss); while (! Q.empty ()) {node now= q.front (); if (NOW.A[0]==SUM/3NOW.A[1]==SUM/3 ) return "possible" ; Q.pop (); for (int i=0;i) {A (int j=i+1;j) {if (now.a[i]!= now.a[j]) {node next= now; next.a[j]=next.a[j]- next.a[i]; next.a[i]=next.a[i]+ next.a[i]; sor

#include #includestring.h>#include#includeusing namespacestd;classbearcheats{ Public: stringEyesight (intAintB) { Chart[ the]; stringS1; sprintf (T,"%d", A); S1=T; stringS2; Chart2[ the]; sprintf (T2,"%d", B); S2=T2; if(S1.size ()! =s2.size ())return "glasses"; intflag=0; for(intI=0; I) if(s1[i]!=S2[i]) flag++; if(flag>1) return "glasses"; return "Happy"; }};#include #includestring.h>#include#include#includeusing namespacestd;classbearplaysdiv2{ Public: intvis

Test instructions: Give a n*m chess board, choose a position (x, y), place a horse, the horse can go to (x-1,y-1), (x-1,y-2), (x-1,y+1), (x-1,y+2), (x+1,y-1), (x+1,y-2), (x+1,y+1), (x +1,Y+2) Eight positions, provided that the board cannot be walked out. A horse can walk forever. Q. What is the maximum number of different positions the horse can walk to?Solution: If N>m,swap (n,m), if N=1,ans=1, if n=2,ans= (m+1) >>1; if n=3 and m=3,ans=8; other conditions ans=n*m;Specifically why, not yet prove

From A.M. To the lab, I flipped through my mailbox to see the TC email... SRM 546, because I went to review last night and didn't come to the lab, I don't know how to play again. T_t took an hour out in the morning and flipped through... 250pt Question 550pt Difficult to think about. My thinking is complicated and hard to write. Then we can see other people's writing, just a few lines without wordsCodeDone. I have to admire it. Class Tworec

Test instructions: Give two large integers to determine which one is larger. Large integers are given in "AB" form, "a" is an integer without a leading 0 (greater than 0, not more than 1e9), and "B" is a number (possibly empty) factorial symbol ("!"). For example: 3!! =6!=720Solution: Set two large integer form a part of the A,b;b section respectively has n1,n2 a symbol. Assuming N1 > n2, then we just need to determine the size of AA and B, where A is (N1-N2) a factorial symbol. N1 = N2 or N1 My

[j]*x[K] + b[j]*y[K] + c)sQ (d)* (sQ (a[j])+sQ (b[j]))) ans2++; } ans = max (ans, ans2); } } for(inti =0; I x. Size (); i++) { for(intj = i +1; J x. Size (); J + +) {Double X1 = (x[I] +x[j]) *0.5, y1 = (y[I] +y[j]) *0.5; for(intK =0; K 4; k++) {Double c =-(a[k] * x1 + b[k] * y1);intAns2 =0; for(intL =0; L x. Size (); l++) {if(sQ (a[k]*x[l] + b[k]*y[l] + c)sQ (d)* (sQ (a[k])+sQ (b[k]))) ans2++; } ans = max (ans, ans2); } } }returnAns }}tt; Copyri

Topcoder srm 623 solution report We recommend that you use the plug-in greed 2.0, which is very useful. But I don't know how to add test data by myself. I will study it next time.Greed 2.0 https://github.com/shivawu/topcoder-greed250ptThere are n trees on the ring runway, marked as 1 -- n. Some trees are recorded when Alice is running. The number is used to calculate the minimum number of laps.Question: If you think about it, you will know that the n

3) {20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1} 14 returns:105