rfc 1035

Title Address: http://ac.jobdu.com/problem.php?pid=1035Topic 1035: Identifying immediate family memberstime limit:1 seconds memory limit:32 trillion special sentence: no submission:2388 Resolution:939 Title Description: If a, B is the parent of C, then a A A, B is the parent,c of C is a A, a A, if a is a, C is the (outside) grandfather, grandmother, then a B is the grandparent,c of C is a A, B is the grandchild, if A is C

1#include 2#include 3#include string>4#include 5 using namespacestd;6 7 BOOLcmpstringAstringb);8 BOOLMatchstringAstringb);9 Ten One intMainvoid) { A inttestsize, couple; - stringstr[101]; - intlen[101]; the BOOLused[101]; - - //For each test case: (t -CIN >>testsize; + while(testsize--) { - //Scan and Store n strings (n +CIN >>couple; A for(inti =0; i i) { atCIN >>Str[i]; - } - - //Sort by length -Sort (str, str +couple, CMP); -

; while(1) { if(PAGT;=LENAAMP;AMP;PBGT;=LENB) Break; if(a[pa]==B[PB]) {PA++; PB++; } Else if(!flag) {Flag=1; PA++; } Else return 0; } if(flag==1)return 1; } if(lena==lenb-1) { intPa=0. P =0, flag=0; while(1) { if(PAGT;=LENAAMP;AMP;PBGT;=LENB) Break; if(a[pa]==B[PB]) {PA++; PB++; } Else if(!flag) {Flag=1; PB++; } Else return 0; } if(flag==1)return 1; } retu

Merge sort does not understand, first do the problem, then look at the back, and now there is no solution is to seek to merge the order of the middle sequence of the next sequence PackageCom.company;Importjava.util.Arrays;ImportJava.util.Scanner; Public classMain { Public Static voidMain (string[] args) {Scanner sc=NewScanner (system.in); intnum=Sc.nextint (); int[] s=New int[num]; int[] s0=New int[num]; int[] s1=New int[num]; for(inti=0;i) {S[i]=sc.nextint ();//3 1 2 8 7 5 9 4 6 0s0[i]=S[i]; }

sequence lengths previously recordedcd = nol[0]/Factor_two (nol[0]);//Because the sequence length generated by the merge must be a multiple of 2, this extracts the maximum number of 2 contained in the CD for(inti =0; I 1 CD >1; i++)if(NOL[I]%CD! =0) CD/=2, i--;//Because the entire sequence can be divided into equal length (except the last one) ascending sequence, the subsequence length of the current step must be an approximate number of all ascending sequence lengths for(i =0; i+2*CD 2

]) {Tmp.str.erase (--tmp.str.end ()), tmp.cnt--; //tmp.str.push_back (' 1 ');Q.push (TMP); Tmp.cnt=2; Tmp.str.clear (); Tmp.str.push_back (A[i]); if(i==len-1) Q.push (TMP); } Else{tmp.str.push_back (a[i]), tmp.cnt++; if(i==len-1) Q.push (TMP); } } } while(!Q.empty ()) {tmp=Q.front (); Q.pop (); if(tmp.str.size () = =1tmp.cnt!=1) printf ("%d%c", tmp.cnt,tmp.str[0]); Else{cout"1"; string:: Iterator it=Tmp.str

of rows in the grid and the number of columns in the GRI D, and the number of the column in which the robot enters from the north. The possible entry columns is numbered starting with one in the left. Then come the rows of the direction instructions. Each grid would has at least one and at the most rows and columns of instructions. The lines of instructions contain only is the characters N, S, E, or W with no blanks. The end of input is indicated by a row containing 0 0 0.Outputfor each grid wi

the very 10000 words in the dictionary.The next part of the "file contains all words" is checked. Each word occupies it own line. This was also finished by the single character ' # ' on a separate line. There'll is at the most of words that is to be checked.All words in the input file (words from the dictionary and words to being checked) consist only of small alphabetic character s and each one contains the characters at the most.OutputWrite to the output file exactly one line for every checke

account is modified" instead.Sample Input 1:3team000002 rlsp0dfateam000003 perfectpwdTeam000001 R1spodfaSample Output 1:2team000002 rlsp%dfateam000001 [email protected]Sample Input 2:1team110 abcdefg332Sample Output 2:There is 1 account and no account is modifiedSample Input 3:2team110 abcdefg222team220 abcdefg333Sample Output 3:There is 2 accounts and No. is modifiedIn order to output in the order of input, we use the vector to accommodate each record, each record contains the name and passwor

At the beginning of this topic I thought it was a collection, and later found more simple than the collection, is a family tree, and then find the depth difference between two nodes.#include Nine degrees 1035-tree-Find immediate family

famous double helix structure. this pairing occurs because of a mutual attraction, call hydrogen bonding, that exists between AS and Ts, and between GS and CS. hence a/t and G/C are called complementary base pairs. In the molecular biology experiments dealing with DNA, one important process is to match two complementary single strands, and make a DNA Double Strand. here we give the constraint that two complementary single strands must have equal length, and the nucleus otides in the same positi

Despite this, I am very disappointed !!! Come on !!! Come on !!! "《《《《《《《《《《《《《 " A very simple question, step by step. Going out of the matrix indicates that there is no ring. If you go to a place that has already passed, it indicates that there is a ring. You can output the result in the format !!! # Include # Include Int n, m, temp;Int ans [1010] [1010];Char map [1010] [1010]; Void DFS (INT Sx, int SY){ While (SX> = 0 SX {If (Map [SX] [sy] ='s '){// Temp ++;Map [SX] [sy] = 'O ';Ans [SX] [

can write a program to help yourself. but you must know that you have your other assignments to finish, and you shoshould not waste too much time here, so, hurry up please!InputInput may contain multiple test cases. the first line is a positive integer T (t OutputFor each test case, the output is one line containing a single integer, the maximum number of double strands that can be formed using those given single strands.Sample Input23AtcgTagcTagg2AattAttaSample output10D In fact, it seems tha

1 second Memory limit:32 MB Special Judgment:No Submit:1892 Solution:764 Description: If A and B are the parents of C, A and B are the parents of C, and C is the child of a and B. If a and B are the grandfather and grandmother of C, then A and B are the grandparent of C, and C is the grandchild of A and B. If a and B are the grandfathers of C, then, B is the great-grandparent of C, and C is the great-grandchild of A and B. If there are more than one generation, a great-is added to th

characters [Unicode]. an object is an unordered collection of zero or more name/value pairs, where a name is a string and a value is a string, number, Boolean, null, object, or array. an array is an ordered sequence of zero or more values. the terms "object" and "array" come from the conventions of JavaScript. JSON's design goals were for it to be minimal, portable, textual, and a subset of JavaScript. crockford informational [page 1] RFC 4627 JSON

. Similarly, to reduce a letter, the premise is that the word is better than the word in the dictionary. One more letter and the rest of the letters are the same. Changing a word requires that the words to be judged have the same length and only one letter. Use strcmp () to retrieve input words in the dictionary (). Feelings: note the writing of input. I am not familiar with using new dynamic memory allocation .... Thanks ~!!! Remember to delete and release space .. It is said that this

Click to view the topicAt first look at this problem, there is really no train of thought (really ashamed).And then I saw this paper: PortalSuppose a cyclic section that requires a positive integer reciprocal is actually the last to solve a minimum x satisfying 10x=1 (mod C)Tenx≡1(moDC)If GCD (10,c)!=1, it is clear that there is no solution. If there is a solution, according to Euler's formula, then this solution X|phi (C), so the direct violent enumeration of x is good.#include using namespaces

; } for(inti =0; I 1; ++i) { if(Arrsorted[i+t*gap] > arrsorted[i+1+t*Gap]) {Flag=false; Break; } } } if(!flag) { Break; } Gap*=2; } for(intt =0; T t) {Sort (arrsorted+t*gap, arrsorted+ (t+1)*Gap, CMP); } if(n% Gap! =0) {sort (arrsorted+n/gap*gap, arrsorted+N, CMP); } for(inti =0; I i) { if(i = =0) {printf ("%d", Arrsorted[i]); } Else{printf ("%d", Arrsorted[i]); }

((c== ' X ') | | (c== ' \ \ ')){a[ul][side]++;a[dr][side]++;if (!judge (UL,DR)) P[P[UL]]=P[DR];}if ((c== ' X ') | | (c== '/')){a[ur][side]++;a[dl][side]++;if (!judge (UR,DL)) P[P[UR]]=P[DL];}}}int main (){scanf ("%d%d", n,m);int i,j,ans=0;For (i=0;iMake (0);Make (1);For (i=0;i{if (a[i][0]| | A[I][1]) sum[find (i)]+=abs (a[i][0]-a[i][1]);}For (i=0;i{if (a[i][0]| | A[I][1]) (p[i]==i)) ans+=sum[i]?sum[i]/2:1;}printf ("%d\n", ans);return 0;}In fact, it's better to use DFS,,, to merge the situation

the single character ' # ' on a separate line. There'll is at the most of words that is to be checked.All words in the input file (words from the dictionary and words to being checked) consist only of small alphabetic character s and each one contains the characters at the most.OutputWrite to the output file exactly one line for every checked word in the order of their appearance in the second part of th e input file. If the word is correct (i.e. it exists in the dictionary) write the message: