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School skipping class is happy, however, there are things that let er meow, that is, of course, the exam! This can be nasty, both skipping classes and want to play BC and prepare for the exam, uh meow. Uh meow in order to prepare for the exam no time to brush questions, want to play BC and do not want to fall points, had to resort to BCround92 Snowy_smile, let him say something ~~>_Enter a descriptionThe fi

Ollivanders: makers of fine wands since 382 BC. Time Limit: 2000/1000 MS (Java/others) memory limit: 131072/65536 K (Java/Others)Total submission (s): 964 accepted submission (s): 539 Problem descriptionin Diagon Alley, there is only one wand-seller, peeling gold letters over the door read ollivanders: makers of fine wands since 382 BC. A single wand lay on a faded purple cushion in the dusty window. A tin

Building BlocksTime limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)Total submission (s): Accepted submission (s): 6Problem Descriptionafter enjoying the Movie,lele went home alone. LeLe decided to build blocks.LeLe has already built n Piles. He wants to move some blocks to make W Consecutive piles with exactly the same height H .LeLe already put all of him blocks in these piles, which means he can not add any bl

; -Cin>>n>>m; + Forto (i,n) ACin>>a[i]>>B[i]; at CLR (c); - Forto (i,m) - { - intT; -Cin>>T; -t--; inc[t]++; - } to intTime=0; + intnres=0; - for(intI=0; i) the { *b[i]-=Time ; $time+=C[i];Panax Notoginseng if(a[i]==0) - { the while(!pq1.empty () pq1.top () B[i]) + { Anres++; the Pq1.pop (); + } - Pq0.push (B[i]); $ } $ Else - {

1#include 2#include 3#include 4 using namespacestd;5 intA,b,c,l,r;6 intFuncintx)7 {8 returnx*x*a+b*x+C;9 }Ten intMain () One { A intt,i,j; - intx, y; -Freopen ("In.txt","R", stdin); theCin>>T; - while(t--) - { -scanf"%d %d%d%d%d",a,b,c,l,R); + intmin=1000000, max=-1000000; - if(a==0) + { AMax=max (b*r+c,b*l+c); atMin=min (b*r+c,b*l+c); - } - Else - { - for(inti=l;i) - { in in

(two modules to the same pairing code) 3, At+uart = 115200,0,0 nbsp Change baud rate 1152004, At+role = 1 change from mode, 0-bit slave mode (default), 1-bit main Mode 5, At+role = 11 change main mode, 0-bit slave mode (default), 1-bit main mode 6, at+ ADDR reads the connection address of two modules (eg: if read from module 2013:123:000) 7, At+addr reads the connection address of two modules (eg: if the main module reads 2013:123:111) 8, At+bind = 2013,123,111 Send the address from the m

/************************************************* author:running_time* Created time:2015-9-5 19:48:28* File Nam E:b.cpp ************************************************/#include 　　BC 1002 (Put code)

/************************************************* author:running_time* Created time:2015-9-5 20:06:46* File Nam E:c.cpp ************************************************/#include 　　BC 1003 (Put code)

/*************************************************************************2 > File name:code/bc/#54/a.cpp3 > Author:111qqz4 > Email: [Email protected]5 > Created time:2015 September 05 Saturday 18:56 22 seconds6 ************************************************************************/7 8#include 9#include Ten#include One#include A#include -#include -#include string> the#include -#include Set> -#include -#include +#include -#include +#includ

Yesterday's BC was a hole in the pit = =Would have been able to rise rating opportunity and wasted ... Large abandoned numberA: The first reaction is high precision, the result template can not find the = =, and then now learn to sell patted a Java biginteger+ fast power, tune long time not say also tle. Looks like the problem is in the card JavaIn fact, the two sides take the log is not finished ...StrokeB:a problem tune for half a day, began to do B

final calculation is completed, cast to intOK paste the code offline tomorrow and then play the game again.1#include 2 using namespacestd;3 4 typedef __int64 LL;5 Const intSize =10000010;6 Const intMoD =1000000007;7 LL N;8 intf[1000010];9 intg[1000010];Ten One voidInit () A { -f[0] = g[0] =1; - intNow , next; thenow =1; - intsum, ans; -sum =1; - for(inti =2; i9; i++ ) + { -Next = (int) ((1LL * (now%mod) * i)%MoD); +Ans = (int) ((1LL * (sum%mod) * (next%mod))%MoD); Anow =Next;

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=4994Test instructions: there are n heap of stones, each heap has an AI stone, two people take a stone in turn, you can take a pile of one to many, but must take one of the piles to begin to take down a pile, who finally took the stone, who wonIdea: Simulation several times easy to find, who first made more than 1 of the gravel heap, who has the initiative, can win, in other words, when the encounter is 1 o'clock, then two people do not have t

ordering.Sample Input5 5 21 24 52 43 42 33 2 01 21 3Sample Output5 3 1 2 3 2Hintcase 1.Erase The Edge (2->3), (4->5). and the lexicographically largest topological ordering is (5,3,1,2,4).Sourcebestcoder Round #35Recommendhujie | We have carefully selected several similar problems for you:5197 5196 5193 5192 5189Test instructions: N points m Bar has a forward-and-no-loop graph, can delete up to the K-bar so that his topological sequence is the largest. Outputs the maximum topological order.Idea

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=1179Main topic:The title is too long, the simple meaning is: There are n wands, M wizards, wands have a number of matching wizards. But a magician.Only one wand can be matched. So the question is: how many magicians can get the magic wand at most.Ideas:Make a dichotomy, one side is a wand, the other is a magician. The corresponding match as the side of the binary graph. Using the Hungarian algorithm, find out twoWhat is the maximum matching o

preprocessing of things.By the way, the HDU is pressed. Shortest = = (space is too weak >_1#include 2#include 3 #definell Long Long4 using namespacestd;5 Constll modd=1000000007;6ll f[101],a[101];//the value of a array's factorial7 ll I,j;8 intn,x;9 BOOLu[101];Ten inline ll run () { Onell ans=0, now=0, MN; Amemset (U,0, n+1); - for(i=1; i){ -scanf"%d", x); u[x]=1; the for(j=1, mn=0; j) - if(!u[j]) ans+=f[n-i]+a[n-i]* (now+mn), ans%=modd,mn++; - for(j=1; jif(!u

using namespacestd;6 intl,r,x,n,m;7 intOkintAintBintc)8 {9 if(ab) Swap ( A, a);Ten if(ac) Swap (A,C); One if(B+c>a)return true; A Else return false; - } - Long LongCALC2 (intXinty) the { - Long Longtot=0; - for(intI=1; i) - if(OK (i,x-i,y)) tot++; + for(intI=1; i) - if(OK (i,y-i,x)) tot++; + returntot; A } at Long LongCalc1 (intN) - { - Long Longtot=0; - for(intI=1; i2; i++) - if(2*iN) - { intot+= (n1)/2-(n/2-i); -

PHP BC High Accuracy function library

1. The date of the command to display dates:Date May 24, 2015 Sunday 16:23:05 CSTDate +%y/%m/%d, 2015/05/24Date +%h:%m 16:262. Display the command CAL for the calendar:Cal,cal 2015,cal 3 20153. Calculator: BC:Scale=3 three digits after decimal pointQuit quitting4. Hotkey Tab:Complete the command after the first command of a sequence of commands;Complete the file after the second command of a sequence of commands.5.ctrl+c:Force exitCtrl+d:Keyboard input ends (equivalent to exit when you leave the

#include First to find the maximum value, and then use the maximum value to go down to try to be at least two number of approximate, the method is to sift up the 　　[BC] To find the number of any two number of conventions in the array, and to take maximum value in many cases

paradigm, and each non-primary attribute does not pass a candidate key that relies on R, then R is the pattern of the third paradigm. Take the student form (school number, name, course number, score) as an example, where the name of the student is not the same, so the table has two candidate code (school number, course number) and (name, course number), so there is a functional dependency: study number--name, (student number, course number)-- The only non-primary attribute score does not have