brute shoes

Topic Links:AbsTime limit:2000/1000 MS (java/others)Memory limit:131072/131072 K (java/others)Problem Descriptiongiven A number x, ask positive integer y≥2, that satisfy the following conditions:1. The absolute value of y-x is minimal2. To prime factors decomposition of Y, every element factor appears. Inputthe first line of input was an integer T (1≤T≤ )For each test case,the a single line contains, an integer x ( 1≤x≤ten)Outputfor each testcase print the absolute value of y-xSample Input 5

) -cout " "; thecout Endl; * } $ Panax Notoginseng voidGet_nexts (Char* p,intPlint*nexts) - { the intJ0), K (-1); +nexts[0] = -1; A while(J PL) the if(P[j] = = P[k] | | k = =-1) +nexts[++ J] = + +K; - Else $K =Nexts[k]; $ } - - intKmp_search (Char* S,intSlChar* p,intPL) the { - int* Nexts =New int[PL +1];Wuyi get_nexts (p, PL, nexts); the - intI0), J (0); Wu intIndexbegin (0); - while(J SL) About if(S[i] = =P[j]) $(+ + I, + +j); - Else

Test instructions: Given a n*m chess board, put some pieces, ask you can put a few guns (Chinese chess gun).Analysis: Actually very simple, because the chessboard is 5*5 biggest, then the direct violence on the line, can be regarded as a line, very water, time is very short, only 62ms.The code is as follows:#include HDU 4499 Cannon (brute force solution)

; J ) the { the intScore = v[i] * -;94 intRound1 =Divide (P[j], a[i]); the intRound2 =Divide (H[i], b[j]); the if(Round1 > Round2) score =-score; the if(i = = j) score++;98W[I][J] =score; About } - 101 KM ();102 103 intAns =0;104 for(inti =1; I W[lft[i]][i]; the 106 if(Ans 0)107 {108Puts"Oh, I lose my dear seaco!");109 Continue; the }111 th

, in the case of no optimal positive solution, can also be written in accordance with the time complexity of the violent solutionSimilar topics, more from factorization. In fact, the prime number itself is less, the quality factor is less, so efficiency will be more efficientCode#include #include#include#includeusing namespacestd;structNode {Long LongDataintzs,nt,mp; BOOL operatorConstNode A)Const {returndataA.data;} }now,tmp;priority_queueQ;Long LongN,x;intk,j;intprime[50]={0,2,3,5,7, O

illegal link to shield off. and block out the spam, only keep normal mail delivery to the server, greatly reducing the malicious connection and virus mail caused by the security risks, see below the relevant schematic diagram.650) this.width=650; "src=" Http://www.unioncyber.net/images/20160418-3.jpg "height=" 534 "width=" 793 "alt=" 20160418-3.jpg "/>650) this.width=650; "src=" Http://www.unioncyber.net/images/20160418-3.jpg "height=" 534 "width=" 793 "alt=" 20160418-3.jpg "/>Can be clearly se

Sometimes it's lazy, and I can't see anyone else trying to get the SSH port.Try to use the next fail2ban this software, simple rough is the effect we want.yum installfail2ban -y //epelcp /etc/fail2ban/jail.conf /etc/fail2ban/jail.localvim /etc/fail2ban/jail.local# "IGNOREIP" can be a IP address, a CIDR mask or a DNS host. Fail2ban would not# Ban a host which matches an address on this list. Several addresses can be# Defined using space separator. Multiple IPs are separated by spaces, and this is

Test instructions: Given a sequence of DNA of length n of M, a shortest sequence of DNA is obtained, which minimizes the total hamming distance.Hamming the distance is equal to the number of different positions of characters.Analysis: See this problem, my first feeling is to calculate the time complexity, good small, nothing, completely can violence, as long as the same position on each string,The choice appears most, if has the same choice ASIIC code small (because requires the dictionary order

I can finally say that this is my own independent completion of the topic, did not see the problem, not read the notes, although the time to write, there is a sense of accomplishment, last night wrote a probably, a little bug, because too late, and a little sleepy, went to bed, then really himself seriously thought, and very deep thinking, Use the bit vectors that have just been learned in the book to judge by their own generation. In the future, we must strive to think, solve their own, focus .

E.... Rice still did not read the question .... T_t .....E.... This is the legendary violence .... It's bloody .... It's too violent ... Kokonoe for loop .... It's such a naked ac ....Water is a bit of water. But.. I didn't think it was possible. Because each operation can only be done at most three times or it is repeated. So 、、、 attached code:#include #include #include #define for (x) for (x=0; xusing namespace Std;int main (){int a[10], b[10], c[10];for (int i=0; iCIN >> A[i];for (B[0])for (B

) = (s);(i) >= (t);--(i))#define REPOK (i, S, T, O) for (int (i) = (s);(i) #define MEM0 (addr) memset ((addr), 0, sizeof ((addr)))#define MP (x, y) make_pair (x, y)#define REV (S, e) reverse (S, e )#define SET (P) memset (pair,-1, sizeof (p))#define CLR (P) memset (p, 0, sizeof (p))#define MEM (P, v) memset (P, V, sizeof (p))#define CPY (d, s) memcpy (d, S, sizeof (s))#define READ (f) freopen (F, "R", stdin)#define WRITE (f) freopen (F, "w", stdout)#define SZ (c) (int) c.size ()#define PB (x) pu

) in{//interval Update, - if(L==TREE[ROOT].L tree[root].r==R) to { +TREE[ROOT].C =C; - return ; the } * if(Tree[root].c! =-1) $ {Panax Notoginsengtree[2*root+1].C = tree[2*root+2].C =tree[root].c; -TREE[ROOT].C =-1; the } + if(L >=Tree[root]. Mid ()) AUpdate (2*root+2, L, R, c); the Else if(R Tree[root]. Mid ()) +Update (2*root+1, L, R, c); - Else $ { $Update (2*root+1, L, Tree[root]. Mid (), c); -Update (2*root+2, Tree[root]. Mid (), R, c); -

11.07.48.png "alt=" Wkiom1uiodtdxntdaaabt-z6kjs787.jpg "/>650) this.width=650;" Src= "http://s3.51cto.com/wyfs02/M01/5D/58/ Wkiol1uioytCkvtaaaofgc0eahw959.jpg "title=" screenshot 2015-04-06 pm 11.07.59.png "alt=" wkiol1uioytckvtaaaofgc0eahw959.jpg "/>650) this.width=650; "src=" Http://s3.51cto.com/wyfs02/M02/5D/58/wKioL1Uioy3hbtPkAADrz8AdrpQ928.jpg "title=" screenshot 2015-04-06 pm 11.08.10.png "alt=" Wkiol1uioy3hbtpkaadrz8adrpq928.jpg "/>This article from "Lee Chenguang original Technology blo