fixlock 340

Provincial Urban database download, for the general version of the provinces and cities to do the processing.General Edition download provincial and municipal database access version download (Universal version, without any processing) Access file format, without any processing.Due to business requirements, the following processing is done on the basis of the General edition: treated by the municipality. to Shanghai, Tianjin, Beijing, Chongqing to do a special treatment, will not distingui

Ultraviolet A 340 Master-Mind Hints (tips for guessing digital games)Question: Guess the number. The game calculates the number. The number is correct. The number is correct. The number appears in the answer. However, the position is incorrect. Each character can only match once. Direct match per Digit #include #include #include using namespace std; const int N = 1005; int a[N], b[N], c, d; int main() { int n, cas = 0; whi

#include int main (){int m,i,x,n=0;int a[1010],b[1010];while (scanf ("%d", m) m){printf ("Game%d:\n", ++n);for (i=0; iscanf ("%d", a[i]);while (1){int k=0;int b=0;int a=0;for (i=0; i{scanf ("%d", b[i]);if (b[i]==0)k++;}if (k==m)Breakfor (i=0; i{if (A[i]==b[i])a++;}for (x=1; x{int c1=0;int c2=0;for (i=0; i{if (a[i]==x)c1++;if (b[i]==x)c2++;}C1}printf ("(%d,%d) \ n", a,b-a);}}return 0;}340-master-mind Hints

) {returna.idb.id;}intMain () {CIN>>n>>m>>K; Rep (I,1, N) { intnum; scanf ("%d",num); Seq[i]=seq[i-1]^num; } Rep (I,1, M) {scanf ("%d%d",q[i].l,Q[I].R); Q[i].id=i; Q[I].L--; } sort (Q+1, q+1+m,cmp); intl=q[1].l,r=q[1].R; ll ans=0; Rep (I,l,r) {ans+=exist[k^Seq[i]]; Exist[seq[i]]++; } q[1].ans=ans; Rep (I,2, M) { while(rQ[I].R) {R++; Ans+=exist[k^Seq[r]]; EXIST[SEQ[R]]++; } while(r>Q[I].R) {Exist[seq[r]]--; Ans-=exist[k^Seq[r]]; R--; } while(l>Q[I].L) {L--; Ans+=exi

Test instructions: 01 strings divided into segments each segment can only have one 1 ask how many ways a string is dividedIdea: There is a 0 between two 1, there are two ways to divide, and then according to the principle of step multiplication,(However, there is a set of 0 1 0 This data can not be processed directly, so encountered the first 1 to start marking)1#include 2 using namespacestd;3 intMain ()4 {5 Long LongL=0;6 intN;7Cin>>N;8 intA;9 intj=0;Ten ints=0; One for

hint per line. Each hint should is represented as a pair of integers enclosed in parentheses and separated by a comma. The entire list of hints for each game should is prefixed by a heading indicating the game number; Games is numbered sequentially starting with 1. Look at the samples below for the exact format.Sample Input41 3 5 51 1 2 34 3 3 56 5 5 16 1 3 51 3 5 50 0 0 0101 2 2 2 4 5 6 6 6 91 2 3 4 5 6 7 8 9 11 1 2 2 3 3 4 4 5 51 2 6 1 91 2 2 5 5 5 6 6 6 70 0 0 0 0 0 0 0 0 00Sample OutputGame

maximum value for N would be 1000.OutputThe output for each game should list the hints that would is generated for each guess, in order, one hint per line. Each hint should is represented as a pair of integers enclosed in parentheses and separated by a comma. The entire list of hints for each game should is prefixed by a heading indicating the game number; Games is numbered sequentially starting with 1. Look at the samples below for the exact format.Sample Input41 3 5 51 1 2 34 3 3 56 5 5 16 1

single of these pairs are suitable for the second query. In the second, the sample XOR equals 1 for any subarrays of an odd length.Test instructions: Find out how many pairs of xor and the logarithm of k are in the interval;Idea: Mo team algorithm, mainly how O (1) update, a[l]^a[l+1].....^a[r]=pre[r]^pre[l-1]=k;pre[i] means a[1]^a[2]^....^a[i];Pos[r]^pre[l-1]=k; pre[r]=pre[l-1]^k; update the pre;#include using namespacestd;#definell Long Long#definePi (4*atan (1.0))Const intn=1e5+Ten,

any subarrays of an odd length.Test instructions　　Give you a sequence of length nM inquiries, every time you ask how many pairs of l,r between the sub-range are equal to KExercisesGodless, a team of Mo.#include using namespacestd;Const intN = 1e6+Ten, M = 5e6+Ten, INF = 1e9+7, mod = 1e9+7;intH[m];intSum[n],n,m,k,a[n],pos[n];Long LongAn[n],ans;structss{intL,r,id;} Q[n];intCMP (ss S1,ss S2) {if(POS[S1.L]==POS[S2.L])returns1.rElse returns1.lS2.L;}intCmpl (ss S1,ss S2) {returns1.ids2.id;}voidUpdate

Link:Http://codeforces.com/contest/617/problem/ECode: to structNode { + intL, R, id; - }Q[MAXN]; the * intN, M, K; $ intA[MAXN];Panax Notoginseng intPOS[MAXN]; - ll ANS[MAXN]; thell sum[2000007]; + A BOOLCMP (Node A, Node B) { the if(POS[A.L] = = POS[B.L])returnA.R B.R; + returnPOS[A.L] POS[B.L]; - } $ $ intL =1, R =0; -ll Ans =0; - the voidAddintx) { -Ans + = Sum[a[x] ^K];Wuyisum[a[x]]++; the } - Wu voidDelintx) { -sum[a[x]]--; AboutAns-= sum[a[x] ^K]; $ } - - intMain () { -Io

", a[i]); theA[i] = a[i]^a[i-1];//computes the prefix XOR or +Pos[i] = I/sz;// chunking A } the for(inti =1; I //read-in Query +scanf"%d%d", AMP;Q[I].L, Q[I].R); -Q[i].id =i; $ } $Sort (q+1, q+1+M, CMP); -flag[0] =1; - for(inti =1; I ){ the while(L //The current left node is smaller than the query node. -Del (L-1);Wuyil++; the } - while(L > Q[I].L) {//The current left node is larger than the left node of the query. Wul--; -Add (l1); About } $

Http://www.iteye.com/news/27727-jinnianshilongnianSome of the specifications in Jave EE 7 have been formally approved, including the JSR340 Java Servlet 3.1 specification, which was translated last year and shared here, hoping to help some friends, please correct me.Click Direct downloadOnline catalogueServlet3.1 specification Translation--preface Servlet3.1 Standard Translation--Overview Servlet3.1 Specification translation--servlet ContextServlet3.1 specification translation--responseservlet3.

Designer, Selects a secret code. The other,Breaker, Tries to break it. A code is no more than a row of colored dots. At the beginning of a game, the players agree upon the lengthNThat a code must have and upon the colors that may occur in a code.

The English description of this question is too redundant. Finally, after reading the question described in one sentence, I understood http://blog.163.com/kakarrot@yeah/blog/static/12011592520106241155854 /. In fact, it is to count the number of

13.8.5 instance attributes and class attributes of functions The javascirp function is not only a function, but also a ray. According to the way variables are declared in the function, there are 3 variables in the function. Local variables:

It takes a long time to understand the problem description. The question itself is very simple. The designer provides a string of code with a length of N, and breaker uses guess to decrypt the code. For two strings of numbers, if the same column has