cad computer design

Tags: data type entity algorithm limit logo card conversion logout further before the first time the computer room charges, the database used is someone else's. Cognition can only be built on the basis of others, such as the "database system principle" after learning the book, then to see the database, found that the database really need further optimization. Here are some insights into my database design.

,backcardcash,allgetcash,username,thedate)Thinking in the study:1, register need not be associated student entity. 2. Whether the association between SystemUser and card entities is 1:n or N. 3, Systemuserlog can be considered an entity exists. Thinking Harvest:1, in the process of this design, began to like the design. Always thought the code is full of infinite fun, now can coordinate the overall, from th

1505: Cool WordsTime limit:1 Sec Memory limit:128 MBsubmit:237 solved:88[Submit] [Status] [Web Board]DescriptionEnter some words that consist only of lowercase letters. Your task is to count how many words are "cool", that is, each letter appears in a different number of times.Ada is cool, for example, because a appears 2 times, D appears 1 times, and 1 and 2 are different. For example, banana is also cool, because a appears 3 times, N appears 2 times, B appears 1 times. However, BBACCCD is not

: 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

(scanf ("%d", cass); cass;cass--)111 //for (scanf ("%d", cas), cass=1;cass the //while (~scanf ("%s", s))113 while(~SCANF ("%d",N)) the { theaans=0; thex=l=n+1, y=r=0;117 for(i=1; i"%d", A +i);118 for(i=1; i)119 { - if(!a[i])Continue;121 if(ABS (A[I]-I) >1) {Work1 (i); Break;}122 if(a[i]-i==1) X=min (x,i), y=Max (y,i);123 if(a[i]-i==-1) L=min (l,i), r=Max (y,i);124 } the if(iContinue;126 i

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

Test instructionsHave n jewels m box Jewel value a box value bEach jewel is placed in a box and costs ABS (A-B)The box can put jewels in infinity.Ask for the minimum costWater problemPre-treatment of each value of jewelry placed in the box O (n)Find the closest left box from left to right L find the closest right box from right to left RTake min#include #include#includestring>#include#include#includeusing namespacestd;Const intn=20005;intn,k,m;intL[n],r[n];intv[Ten*N],s[n];intMain () {inti,j; in

#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

-type Domino.The Fillboard () function is a function that overrides the chessboard with some kind of L-shaped domino. Fillchessboard () is a recursive function that solves the whole problem, and is entered as a checkerboard pointer. The participation level is also K. Blue_row,blue_col is the position of the special lattice in the coordinate system of the sub-chessboard, Baserow and Basecol are the coordinates of the sub-chessboard (0,0) points throughout the large checkerboard.Keep up with the n

11th Hunan University student Computer Programming Competition "Blue Fox network Cup" B. Hunan Design CompetitionB-big or small?Time Limit:5000 MSMemory Limit:65535KB64bit IO Format: Description Input two real numbers to determine whether the first number is large or not. The format of each number is: [integer]. [decimal part] For simplicity, both the integer part and the decimal part must be non-null, And

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

of the recursive solution, assuming that the four sub-problems have been solved, we should use a specific L-shaped dominoes cover three false special lattice, so that the entire large board has been solved. we will think of the entire solution process, first of all to simplify the problem, simplified to directly see the point of the answer, and then analyze a slightly more complex situation, summarize the law, using the idea of divided treatment , the problem is resolved into four sub-problems,

Single responsibility: The computer has just been repaired. Obviously, the memory is broken and should not be the reason for CPU replacement. Their Respective responsibilities are clear;Open-closed Principle: The memory is not enough, as long as the slot is enough to be added, the hard disk is not enough to use mobile hard disks, etc. The PC interface is limited, so the expansion is limited, and the software system is well designed, but it can be expa

In February this year, Microsoft launched its next-generation Windows Phone 7 operating system for mobile devices. This is a brand new system. Its development mode and user interface are completely different from those before. To help developers better understand the interface design and Human-Computer Interaction Features of Windows Phone 7, Microsoft specially released a documentUI