computer game competition

: Press the number in array B to 0->a, 1->b, 2->c, ..., 24->y, 25->z, convert the rules to ciphertext For example: Encryption July, Matrix A=[11 8;3 7] (semicolon denotes another line) Step1: Convert July to an array a={9, one, one, and one}; STEP2: Divide a into {9, 20},{11, 24} two parts Step3: Because [9,] * [one 8;3 7] = [159, 212] [11, 24] * [11 8;3 7]=[19 3, []] b=[159%, 212%, 193%, [3, 4, one,] Step4: Ciphertext is DELW Your task is: give you ciphertext and matrix A, the ori

Character Recognition? Time Limit: 1 sec memory limit: 128 MB Submit: 156 solved: 98 [Submit] [Status] [web board] Description Your task is to write a program for character recognition. Don't worry, you only need to identify 1, 2, 3, as follows: .*.****** .*...*..* .*.****** .*.*....* .*.******Input The input contains only one group of data, which consists of six rows. The first row contains N (1 Output The output should contain one line, that is, each character recognized.Sample Input 3.*..***

Question link: http://acm.csu.edu.cn/OnlineJudge/problem.php? Id = 1112 Simple question simulation. Command of the csuoj robot (Hunan's eighth competition for computer programming for College Students)

Contrast MoeTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 0 Accepted Submission (s): 0Problem description has 2N individuals, each with an Moe mi (1They are required to be divided into n pairs, so that the sum of the contrast values is maximal.The meaning of the contrast value is as follows: if x and Y are a pair, they contribute | mx-my| point contrast value. The first line of input is a number T, which indicates the number of test instances

- #defineMAX 0x7f7f7f7f to #definePI 3.14159265358979323 + #defineN 100004 - using namespacestd; thetypedefLong LongLL; * intCas,cass; $ intN,m,lll,ans;Panax Notoginseng LL Aans; - LL E[n],sum[n],l[n],r[n]; the LL A; + CharS[n]; A intMain () the { + #ifndef Online_judge - //freopen ("1.txt", "R", stdin); $ //freopen ("2.txt", "w", stdout); $ #endif - inti,j,k; - LL x, y; the //for (scanf ("%d", cass); cass;cass--) - //for (scanf ("%d", cas), cass=1;cassWuyi //while (~scanf ("%s", s)) th

voidWorkintXintc[]) + { A inti; thec[0]=x/ .; + for(i=1; i .); i++) c[i]=c[0]+1; - for(I= (x .)+1;i .; i++) c[i]=c[0]; $ } $ intMain () - { - #ifndef Online_judge the //freopen ("1.txt", "R", stdin); - //freopen ("2.txt", "w", stdout);Wuyi #endif the inti,j,k; - Wu //for (scanf ("%d", cass); cass;cass--) - //for (scanf ("%d", cas), cass=1;cass About //while (~scanf ("%s", s+1)) $ while(~SCANF ("%d",N)) - { -scanf"%d",m); -aans=0; A Work (n,a); + Work (m,b); t

If a * b% 2016 = = 0If a = 1, and a * b% 2016 = = 0Consider a = 2017.* B = (+ 1) * b% 2016 = = 0 must be establishedSo that means the b,2017 in 1 can be paired with the same.Same: 4033 * b = (+ + 1) * b% 2016 = = 0 must be establishedSo, I can enumerate [1,2016] in [1,2016], I * j% 2016 = = 0 of the logarithm, and then multiply on the corresponding [1,n] in the number of I, instead of the number is also counted, the substitution number is those equivalent number, 1------4033#include #include#inc

#include #include#include#includeusing namespacestd;Const intmaxn= -;Long Longx,k;Long LongBASEX[MAXN];Long LongBASEK[MAXN];intTOTX,TOTK;intMain () {intT; scanf ("%d",T); while(t--) {scanf ("%lld%lld",x,j); Totx=totk=0; memset (Basex,0,sizeofbasex); memset (Basek,0,sizeofBasek); while(X) {Basex[totx++]=x%2; X=x/2; } while(K) {BASEK[TOTK++]=k%2; K=k/2; } intnow=0; for(intI=0; i) { for(intJ=now;; J + +) { if(basex[j]==0) {Basex

#include #include#include#includestring>#includeusing namespacestd;intT;Chars[ -+Ten];Charr[ -+Ten];mapstring,int>m;structdan{Chars[ -+Ten]; intnum;} d[1000000+Ten];intsum;inttot;BOOLcmpConstDana,Constdanb) { if(A.num==b.num)returnstrcmp (A.S,B.S) 0; returnA.num>B.num;}//Turn lowercasevoidF () { for(intI=0; s[i];i++) if(s[i]>='A's[i]'Z') S[i]=s[i]-'A'+'a';}voidWork () {intlen=strlen (s); Tot=0; for(intI=0; i) { if(s[i]>='a's[i]'Z') r[tot++]=S[i]; Else{R[tot]=' /'; if(strlen

(classmate of H) 10 Right of the H (H right side of the classmate) InputInput has only one set of data. The first line is the number of students N (1Each name is either a no more than 3 English letters, or a question mark.At least one student's name is not a question mark. The next line is the number of queries Q (1OutputFor each of the inquiries. The output is termed. Note "Middle of X and Y" only have when the caller has two recent known students X and Y, and X is

Original title Link: http://acm.hunnu.edu.cn/online/?action=problemtype=showid=11654courseid=0A large number of comparisons, decisive use of Java, but also pay attention to the details.Note: The comparison can not use equals, if used 0.0 and 0.00 is not equal, a bit like HDU2054, the problem has a detailed explanationhttp://blog.csdn.net/hurmishine/article/details/51382141AC Code:Import Java.math.bigdecimal;import Java.util.scanner;public class Main {public static void main (string[] args) {in

each Tianma, and the number of I indicates the height of the Pegasus. The n number is separated from each other by a space.The third row has n positive real numbers, each representing the height of each day, and the number of I for the first day. The n number is separated from each other by a space.There is only one row in the output output file Horse.out, and the line has only a positive integer, which is the longest queue length that meets the criteria. According to the height of the horse, i