prom programmer

computer systems and provide the most basic input/output (I/O) programs, such as the CMOS chip storing BIOS parameters. Rom is characterized by the existence of data in the memory after the computer is powered off. Prom (programmable read-only memory)-programmable read-only memory, also called one-time programmable (OTP) Rom "one-time programmable read-only memory", is a read-only memory that can be operated by a program. The primary feature is that

John's N (2≤n≤10000) cows are very excited because it's prom night! They wear dresses and new shoes, not flowers, they want to perform a round dance. Only cows can perform this round dance. The round dance needs some ropes and a round pool. The cows were standing around the pool, numbered 1 to n in the clockwise order. Every cow is faced with a pool so she can see every other cow. In order to dance this round, they found the M (2≤m≤50000) rope. On

DescriptionThe N (2 John's N (2≤n≤10000) cows are very excited because it's prom night!They wear dresses and new shoes, not flowers, they want to perform a round dance. Only cows can perform this round dance. The round dance needs some ropes and a round pool. The cows were standing around the pool, numbered 1 to n in the clockwise order. Every cow is faced with a pool so she can see every other cow. In order to dance this round, they found the M (2≤m≤

namespacestd;Const intMAXN =505;intN,A[MAXN][MAXN],F[MAXN][MAXN],ANS,SUM[MAXN];intMain () {CIN>>N; Charcmd; sum[1] =1; for(inti =1; I ){ for(intL =1; l 2* (n-i) +1; l++){ intj = i + L-1; scanf ("%c",cmd); while(cmd! ='-' cmd! ='#') scanf ("%c",cmd); if(cmd = ='#') A[i][j] =1; ElseA[I][J] =2; } } for(inti =3; I 2N1; i + =2) Sum[i] = sum[i-2] +i; for(inti =1; I ){ for(intL =1; l 2* (n-i) +1; l++){ intj = i + L-1; if(A[i][j] = =2) F[i][j] =1; if(L 1)

"Thinking": The rhythm of the brush water problem."AC Code": #include Blue Bridge Cup ALGO-94 freshman Prom (structural)

Water problem, direct sticker code.POJ 3180//sep9#include POJ 3180 The Cow Prom strong connectivity Component

Topic: Given two sequence, how many matches satisfy a[i]See http://blog.csdn.net/popoqqq/article/details/44514113High precision has been wasted ...#include Bzoj 2024 SHOI2009 Prom Dynamic planning + principle + high precision

I can't understand the question, and the Chinese and English letters are too bad... So orz itwiiioi is huge! The result is finally understood: calculate the number of strongly connected components of a digraph without a single point. Tarjan... (Board * 1 get √) 1 /************************************************************** 2 Problem: 1654 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:32 ms 7 Memory:1476 kb 8 ************************************************

1654: [Usaco2006 jan]the Cow prom Dairy BallTime Limit:5 Sec Memory limit:64 MB submit:287 solved:215 [Submit] [Status] [Discuss] Description The N (2 Input * Line 1:two space-separated integers:n and M * Lines 2..m+1:each line contains, space-separated integers a and B that describe a rope from cow A to cow B in the CL Ockwise direction. Line 1th enters N and M, and the next M-line is two integers a and B, indicating a traction B. Output * Line 1:

1380 Prom without a bosstime limit: 1 sspace limit: 128000 KBtitle level: Diamonds Diamond SolvingTitle DescriptionDescriptionUral University has a staff of N and is numbered 1~n. They have affiliation, which means that their relationship is like a tree rooted in the headmaster, and the parent node is the direct boss of the child node. Each employee has a happiness index. There is now an anniversary party, which asks the staff to have the most happine

the sum of the funny values of all staff on the platform.Input:71 1 1 1 1 1 1115144Output:5Without the boss's prom question.#include #include#includeusing namespacestd;Const intmaxn=5005; typedefLong Longll;intN;intV[maxn];vectorint>tree[maxn];ll dp[maxn][2];voidDfsintu) {dp[u][1]+=V[u]; for(intI=0; I) { intv=Tree[u][i]; DFS (v); dp[u][0]+=max (dp[v][1],dp[v][0]); dp[u][1]+=dp[v][0]; }}intMain () {CIN>>N; for(intI=1; i) {cin>>V[i];

algorithm training freshman Promtime limit: 1.0s memory limit: 512.0MBProblem DescriptionThe freshman prom began. Each of the N freshmen has three attributes: Name, student number, gender. wherein, the name with a length of not more than 20 is composed of only uppercase and lowercase letters of the string representation, the study number is not more than 10 of the length of the number-only string representation, the gender with an uppercase character

Simple Tree DP1#include 2#include 3#include 4#include 5 using namespacestd;6 intn,w[6001],ne=0;7 BOOL inch[6001];8 intf[6001][2];9 inthd[6001];TenInlineintR () { One intx=0, f=1; A CharCh=GetChar (); - while(ch>'9'|| ch'0') { - if(ch=='-') f=-1; theCh=GetChar (); - } - while(ch>='0'ch'9') { -x=x*Ten+ch-'0'; +Ch=GetChar (); - } + returnx*F; A } at - structData { - intV,next; -} e[6001]; - voidAddintVintu) { -ne++; ine[ne].v=v; -e[ne].next=Hd[u]; tohd[u]=

happiness value when the subtree state is s with the root of I, and s==0 indicates that there is no s==1 present. Then there is the transfer equation: d[i][1]=sum{d[chi][0]}D[I][0]=SUM{MAX{D[CHI][0],D[CHI][1]}}It is important to note that when s==0, the subtree status is not necessarily 1.Code1#include 2#include 3#include 4#include 5 using namespacestd;6 7 Const intMAXN =6000+Ten;8 intr[maxn],d[maxn][2],CNT[MAXN];9vectorint>G[MAXN];Ten intN; One A voiddpintu) { - if(d[u][0])return ; -

F[I,0] indicates that the first person does not attend the dancef[i,1] indicates that the first person attends the danceF[i,1]=sigma (f[j,0]) +v[i] J for Children of IF[i,1]=sigma (Max (f[j,0],f[j,1)) J for Children of IAns=max (f[root,0],f[root,1])Program CODEVS1380;typeArr=RecordU,v,next:longint; End;Constmaxn=10000;varegArray[0.. MAXN] ofarr; Last:Array[0.. MAXN] ofLongint; A:Array[0.. MAXN] ofLongint; F:Array[0.. MAXN,0..1] ofLongint; FA:Array[0.. MAXN] ofLongint; I,n,u,v,j,root:longint;proc

appears two times, that is, CH has brothers, so next is the brother of CHfc[fa]=cur;}voiddpintX//0 indicates no participation, 1 indicates participation{f[x][0]=0; f[x][1]=V[x]; for(inti=fc[x];i!=0; i=Poi[i].next) { intNe=poi[i].ch; DP (NE); f[x][1]+=f[ne][0];//subordinates are not involved .f[x][0]=f[x][0]+max (f[ne][1],f[ne][0]); }}intMain () {CIN>>N; for(intI=1; iV[i]; while(Cin>>ch>>fa ch) {L[ch]++; Insert (CH,FA); } for(intI=1; i) { if(l[i]==0) {DP (i); co

Basic tree-type DP. DP[I][J] I means someone, J says the man is coming. Transfer equation See procedure.Note Pre-processing: First start all dp[i][1] for the given weight. First the people who have no subordinates into the team.#include Codevs 1380 no boss's prom

ifs[x]=0 Then beginF[x,0]:=0; F[x,1]:=A[X]; ExitEnd; - fori:=1 toN Do ifFa[i]=x Then - begin the DP (i); -Inc (F[X,0],max (F[i,1],f[i,0])); -Inc (F[X,1],f[i,0]); - End; + End; - begin + READLN (n); A fori:=1 toN Doreadln (A[i]); at fori:=1 toN-1 Do - begin - readln (u,v); -fa[u]:=v; - Inc (S[v]); - End; in readln (u,v); - fori:=1 toN Do iffa[i]=0 Thenk:=i; to fori:=1 toN DoF[i,1]:=A[i]; + DP (k); -Writeln (Max (F[k,1],f[k,0])); the End.Discla

The main topic: There are n cattle in a piece, M single rope, there is m link relationship, ask how many groups inside any two cows can reach each otherThe thinking of solving problems: the map of strongly connected component modeThe code is as follows:#include #include#include#includeSet>#includeusing namespaceStd;typedefLong Longll;Const intN =10003; Vectorint>G[n], DQ;intLow[n], dfn[n], tot;BOOLMk[n];intN, m, ans;voidinit () {ans= Tot =0; Dq.clear (); for(intI=1; ii) {g[i].clear (); Low[i

(i) instruction set Format: ok> Instruction 1.banner Display current machine configuration status, CPU, memory, hostid,ethernet 2.PROBE-SCSI displays the device attached to the internal SCSI channel 3.probe-scsi-all displays all SCSI channels and attached devices 4.probe-ide displays all IDE channels and attached devices (for U10,U5) Probe-fcal-all 5.devalias display device alias, such as CDROM,DISK,DISK0,DISK1, etc. 6.printenv no parameters, display environment variables or parameter i