asmt 2115

Samba SMB Client Spoofing Vulnerability (CVE-2016-2115)Samba SMB Client Spoofing Vulnerability (CVE-2016-2115) Release date:Updated on:Affected Systems: Samba Samba 4.x-4.2.11Samba Samba 4.4.x-4.4.2Samba Samba 4.3.x-4.3.8Samba Samba 3.x Description: CVE (CAN) ID: CVE-2016-2115Samba is a free software that implements the SMB protocol on Linux and UNIX systems. It consists of servers and client programs.Samb

2115: [Wc2011] Xor time limit:10 Sec Memory limit:259 MBsubmit:2142 solved:893[Submit] [Status] [Discuss] DescriptionInputThe first line contains two integers n and M, indicating the number of points in the graph and the number of edges. The next M-line describes M-bar, with three integers per line si,ti, Di, which indicates that there is a non-forward edge between Si and Ti with a weighted value of Di. There may be heavy edges or self-loops

Test instructions: Given the initial value of the For loop, the end value and increment, and a modulo, the minimum number of cycles.Analysis:After reading the question should know is a concept of congruence, so it is to solve a one-time congruence equation, like the above problem with the expansion of Euclidean algorithm. The trick point of the problem is K Max is 32, then 2^32 out of int, to use a long long, so in 1Code:#include Copyright NOTICE: This article for Bo Master original article, wi

. Sample Input 3 3 2 163 7 2 167 3 2 163 4 2 160 0 0 0 Sample output 0232766FOREVER Or extend Euclidean... There is a for loop Statement (INT I = A; I! = B; I = (I + C) % 2 ^ K) sta; Given A, B, C, K; ask when the loop ends; Set the cycle to terminate X times, you can get (a + x * C) % 2 ^ K = B; that is, a + x * C-B = y * 2 ^ K; (y =, 2 ...) Sort C * X-2 ^ K * Y = B-A; make a = C; B =-2 ^ K; C = B-A; Obtain AX + by = C; The preceding solution means that the loop can be terminated, and the mini

not terminate.Sample Input3 3 2 163 7 2 167 3 2 163 4 2 160 0 0 0Sample Output0232766FOREVERSourceCTU Open 2004 Test Instructions: (A+BX)% (2^k) ==c, for x minimum; idea: the solution expands Euclid;#include #includestring>#include#include#includeusing namespacestd;#definell Long Long#defineESP 1E-13Const intn=1e4+Ten, m=1e6+50000, inf=1e9+Ten, mod=1000000007;voidExtend_euclid (ll A, ll B, ll x, LL y) { if(b = =0) {x=1; Y=0; return; } extend_euclid (b, a%b, x, y); LL TMP=x; X=y; Y= tmp-(A/b)

Computes a path of 1 to N to make the value XOR and maximum on the path.The XOR is calculated by taking one path at a time and then DFS is processed out of all the rings of XOR and, so for all the xor of the ring and for the linear base, the XOR and the greedy on the arbitrary path out.Correctness obviously, if the loop is coincident with the selected path, then the coincident part will be XOR two times, and there will be no XOR, which is equivalent to changing a part of the path. If the ring is

Approximate test instructions:For C's for (i=a; i!=b; I +=c) loop, ask how many loops in the K-bit storage system will end. If the end is within a finite number of times, the output cycle times.Otherwise the output is dead loop.Problem Solving Ideas:Test instructions is not difficult to understand, but the use of K-bit storage system data characteristics to cycle. For example , the int is 16 bits, then int can save 2^16 data, that is, the maximum number is 65535 (the default is unsigned), Whe

base, and finally there will be a set of bases ... The size is $o (log (n)), n is the size of the vector, which means that the highest bit of a base exists and no other base exists.Because of the closeness of the XOR operation, we just need to eliminate the value of the other elements in the same way as the Gaussian elimination element.Back to the subject ...Because the path of a 1~n can be made up of any single 1~n simple path plus any ring. (The path to the loop from the simple path and the p

I Love this GameTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 6357 Accepted Submission (s): 2175Problem Descriptiondo You like playing basketball? If you is, you may know the NBA Skills challenge. It is the content of the basketball skills. It include several parts, such as passing, shooting, and so on. After completion of the content, the player takes the shortest time would be the winner. Now give you their names and the time of finishing

C Looooops Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 19826 Accepted: 5299 DescriptionA Compiler mystery:we is given a c-language style for loop of typefor (variable = A; variable! = B; variable + = C) statement;i.e., a loop which starts by setting variable to value A and while variable are not equal to B, repeats statement followed By increasing the variable by C. We want to know what many times does the statement

Euclidean method (Euclid algorithm)Time complexity: Within O (LogMax (A, B))int gcd (int a, int b) {if (b = = 0) return A;return gcd (b, a% b);}　　Extended Euclidean algorithmThe complexity of time is the same as Euclid's algorithm.int extgcd (int a, int b, int x, int y) {int d = a;if (b! = 0) {d = EXTGCD (b, a% B, y, x); y-= (A/b) * x;} else {x = 1; y = 0;} return D;}For AX+BY=GCD (A, b) integer solution, XY returns an integer solution, and the return value of EXTGCD is the value of Ax+by.Title:

not terminate.Sample Input3 3 2 163 7 2 167 3 2 163 4 2 160 0 0 0Sample Output0232766FOREVERSourceCTU Open 2004Analysis:Extended Euclid:For ax+by=c, the minimum integer solution is calculatedD=GCD (A, B)1. Seek the x0,y0 of Ax0+by0=d first. Then both sides multiply D get a (X0*C/D) +b (Y0*C/D) =GCD (A, b) *c/d=c, find out X=X0*C/D,Y=Y0*C/D2. Find the minimum solution: The maximum solution x, the minimum solution = (x (b/d) +b/d)% (b/d)Reference Learning Website: http://www.cnblogs.com/comeon4my

3Davies 4Carter 5Iverson 6James 7Parker 8Kidd 9Jordan 10Author for JakeSourcehdu 2007-10 Programming Contest_warmup1#include 2#include 3#include 4#include 5#include 6#include 7#include string>8#include string.h>9#include Ten#include One#include A using namespacestd; - structDB - { the Charn[ -]; - Chart[ -]; - }; - BOOLCMP (db a,db b) { + if(a.t!=b.t) - returnstrcmp (A.T,B.T) 0; + Else A returnstrcmp (A.N,B.N) 0; at } -DB s[ -]; - intMain () - { - intn,i,j;

POJ 2115 C Looooops Extended Euclidean and pojlooooops It is not hard to understand the meaning of the question. The following formula can be obtained after reading the question: (A + C * x-B) mod (2 ^ k) = 0 (C * x) mod (2 ^ k) = (B-A) mod (2 ^ k) Use the modulus linear equation (linear homogeneous equation) to solve the problem. Template direct car access # Include

Winter vacation To do the problem, first paste at that time write code. POJ 1061 #include #includetypedefLong LongLL;using namespacestd;voidEXTEND_GCD (LL a,ll b,ll d,ll x,ll y) { if(b==0) {D=A; X=1, y=0; } Else{EXTEND_GCD (b,a%b,d,y,x); Y-=x* (A/b); }}intMain () {LL x,y,m,n,l; while(SCANF ("%i64d%i64d%i64d%i64d%i64d", x,y,m,n,l)! =EOF) {LL d,xx,yy; EXTEND_GCD (n-m,l,d,xx,yy); if(x-y)%d==0) {LL P=l/D; XX= (x-y)/d*xx; XX= (xx%p+p)%p; printf ("%i64d\n", XX); } Elseprintf"impo

]; All the current bits in the A array are 1 (a[j] and A[i] xor, A[j]=a[j] a[i], ext. 1). The final x guarantee must be the largest, with a time complexity of O (NB). (n is the size of a array, and B is bits number) Read the above analysis went to play, the results several times WA. Then, in contrast to the Hzwer code, it was found that the elements he had selected in step 2 were not considered at the back of Step 2 o'clock. Why didn't you say anything

not terminate.Sample Input3 3 2 163 7 2 167 3 2 163 4 2 160 0 0 0Sample Output0232766FOREVERSourceCTU Open 2004Analysis: a+cx%2^k=b, solving the smallest xa-b+cx%2^k=0cx= (b-a)%2^k @1: A typical congruence equation, solved by ext_gcd cx+ (2^k) y= (b-a) when and only if GCD (c,2^k) | (b-a), the equation has a solution. We obtained the solution b-a of cx+ (2^k) Y=GCD (X,Y,GCD) by EXT_GCD. The/gcd* of the two sides of the equation is B-a (@1). That is, x = x* (b-a)/gcd. Because we ask for the smal

Find a path + linear base.#include #include#include#include#defineMAXV 100050#defineMaxe 400500using namespacestd;structedge{Long LongV,W,NXT;} E[maxe];Long LongN,m,x,y,z,g[maxv],nume=0, dis[maxv],a[maxe],cnt=0, ins[ -];BOOLVIS[MAXV];voidAddedge (Long LongULong LongVLong LongW) {e[++nume].v=v; E[NUME].W=W; E[NUME].NXT=G[u]; G[u]=Nume;}voidDfsLong Longx) { for(Long LongI=g[x];i;i=e[i].nxt) { Long Longv=e[i].v; if(!Vis[v]) {Dis[v]=dis[x]^E[I].W; VIS[V]=true; DFS (v); } Els

does not terminate.Sample Input3 3 2 163 7 2 167 3 2 163 4 2 160 0 0 0Sample Output0232766FOREVERTest instructions is asking infor (variable = A; variable! = B; variable + C)In this case, the loop number of times.All of them have to be mod 2 for the K-th side. So the equation is (a+c*x)% (2^k) =b, Transformation is-c*x+ (2^k) *y=a-b. The smallest positive number x of the equation can be solved.It is also an extension of Euclid.Code:#include Copyright NOTICE: This article for Bo Master original

A simple extension of Euclid's questionHere 2^k can not be clever with 1Later honestly for (int i=1; i1#include 2#include 3 4 using namespacestd;5 #definell Long Long6ll EX_GCD (ll A, ll x, LL B, LL y)7 {8 if(b = =0){9x =1, y =0;Ten returnA; One } All ans = EX_GCD (b, X, a%b, y); - intt =x; -x = y, y = t-a/b*y; the returnans; - } - intMain () - { + //freopen ("a.in", "R", stdin); - intA, B, C, K; + while(SCANF ("%d%d%d%d", a, b, c, k)) { A if(A = =0 b