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MessagePosté le: Mer 10 Jan - 19:46 (2018)    Sujet du message: Facebook Hacker Cup Checkpoint Répondre en citant




Facebook Hacker Cup Checkpoint
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There are two ways to get from the origin to the checkpoint depending on whether you move to the right first or later. You will be given the length N of the original sequence, and the debug sequence. 2014 (94) 12 (4) 11 (4) 10 (3) 9 (10) 8 (3) 7 (8) 6 (11) 5 (6) 4 (17) 3 (9) 2 (13) 1 (6) . This time the original sequence is 2 1. In the second example, N is 2 and the debug sequence is 2. R. The first line of each test case contains the length of the original sequence, N. BackSign UpCreate a Page for a celebrity, band or business.English (US)NederlandsFryskPolskiTrkeDeutschFranais (France)EspaolPortugus (Brasil)ItalianoSign UpLog InMessengerFacebook LiteMobileFind FriendsPeoplePagesPlacesGamesLocationsCelebritiesMarketplaceGroupsRecipesSportsLookMomentsInstagramLocalAboutCreate AdCreate PageDevelopersCareersPrivacyCookiesAd ChoicesTermsHelpSettingsActivity Log Facebook 2018. .. In the first example, N is 2 and the debug sequence is 1. InicioAbout Google Code JamBolivia Questions ? Buscar: Categoras ACM ICPC (27) Algorithm (22) Bipartite Graph (1) Sweep Line (1) Bolivia (17) C++ (6) CodeChef (3) Codeforces (17) Facebook Hacker Cup (5) Games (6) Google CodeJam (11) IOI (1) Java (1) Javascript (1) Music (1) Other (1) Topcoder (7) UI Design (1) Uncategorized (1) Web Development (3) Suscripcin por correo electrnico Escribe tu direccin de correo electrnico para suscribirte a este blog, y recibir notificaciones de nuevos mensajes por correo. To avoid having to upload the entire original sequence, output an integer checksum of the original sequence, calculated by the following algorithm:. Error U heeft geprobeerd om een pagina te bekijken die niet bestaat. Jump toSections of this pageAccessibility HelpPress alt + / to open this menuRemoveTo help personalize content, tailor and measure ads, and provide a safer experience, we use cookies. Het moet gemakkelijk en opbeurend zijn. The server has a key k1 and the client has a key k2 where: Sigue leyendo &rarr; 3 febrero, 2013vudduu Bipartite Graph, facebook hacker cup 2013, Java, Min Cost Max Flow Deja un comentario Facebook Hacker Cup 2012 Round 1 SquishedStatus Dynamic Programming O(n^2) #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define FOR(i,a,b) for(int i=(a),##i=(b);i >r; return r; } LL solve(int m, string text){ F(i, text.S){ string aux = ""; for(int j=i; j>=0 ;j--){ aux = text[j] + aux; if(aux.size() >= 4) break; if(text[j] != '0'){ if(stringtoint(aux) >m>>text; printf("Case #%d: %lldn", cas+1, solve(m, text)); } } 30 enero, 2012vudduu Algorithm, dp, dynamic programming, facebook hacker cup 2012 Deja un comentario Facebook Hacker Cup 2012 Round 1 Recover theSequence #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define FOR(i,a,b) for(int i=(a),##i=(b);i merge2(vector arr1, vector arr2){ vector result; int i=0, j=0; while(i reversesort(vector arr){ int k = arr.size(); if(k firsthalf, secondhalf; F(i, mid) firsthalf.PB(arr[i]); FOR(i, mid, arr.S) secondhalf.PB(arr[i]); firsthalf = reversesort(firsthalf); secondhalf = reversesort(secondhalf); return merge2(firsthalf, secondhalf); } int checksum(vector arr) { int result = 1; for(int i=0; i solve(){ vector v(n), r(n); F(i, n) v[i] = i; v = reversesort(v); F(i, n) r[ v[i] ] = i+1; return r; } int main(){ //freopen("a.in", "r", stdin); //freopen("recoverthesequence.txt", "r", stdin); freopen("a.out", "w", stdout); int t; scanf("%d", &t); F(cas, t){ cin>>n>>s; its = 0; printf("Case #%d: %dn", cas+1, checksum(solve())); } } 30 enero, 2012vudduu C++, facebook hacker cup 2012 Deja un comentario Facebook Hacker Cup 2012 Round 1 Checkpoint Dynamic Programming O( 160000000 , sqrt(n) ) #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define FOR(i,a,b) for(int i=(a),##i=(b);i 10000000LL) break; if(DP[ DP[i][j] ][0]) DP[ DP[i][j] ][0] = min(DP[ DP[i][j] ][0], LL(i)+LL(j)); else DP[ DP[i][j] ][0] = LL(i)+LL(j); } } } int cuad(int x) { int r = int(DP[x][0]); if(r == 0) return x; return r; } int solve(int x){ int m = sqrt(x), r = cuad(x)+1; for(int i=2; i<=m ;i++) { if(x % i == 0){ int j = x/i; r = min(r, cuad(i)+cuad(j)); } } return r; } int main(){ //freopen("a.in", "r", stdin); //freopen("checkpoint.txt", "r", stdin); freopen("checkpoint.out", "w", stdout); generate(); int r, s; scanf("%d", &r); F(cas, r){ scanf("%d", &s); printf("Case #%d: %dn", cas+1, solve(s)); } } 30 enero, 2012vudduu Algorithm, dp, dynamic programming, facebook hacker cup 2012 Deja un comentario . This gives a total of 4 distinct shortest paths, and takes T = 2 + 3 = 5 steps. The second line contains a string of 1s and 2s, the debug sequence produced by merge sort while sorting the original sequence. The debug sequence tells us that the first number was smaller than the second so we know the sequence was 1 2. Output. Similarly, there are two ways to get to the goal, which gives a total of 4 distinct shortest paths. .. In this problem merge sort is used to sort an array of integers in ascending order. Output. The exact behavior is given by the following pseudo-code:. 5 &le; T &le; 20. Het spijt ons dat u een error heeft ervaren. Examples. Once you are at the checkpoint, there is only one way to reach the goal with minimal number of steps. Input. S. Dit is ook dezelfde ervaring die onze medewerkers hebben. function checksum(arr): result = 1 for i=0 to arr.length()-1: result = (31 * result + arr[i]) mod 1000003 return result. Luckily, that sequence had been sorted by the above algorithm and the debug sequence of 1s and 2s was recorded on a different disk 5a02188284
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