alfawise t1

ORA-00904: quot; T1 quot;. quot; AREA_ID quot;: Invalid identifier 1. Error description ORA-00904: "T1". "AREA_ID": Invalid identifier 00904. 00000-"% s: invalid identifier" * Cause * Action An error occurred while listing 45 columns and 8 columns in the row. 2. Error cause When splicing an SQL statement, "AREA_ID" is not queried, and "AREA_ID" is not found in group" 3. Solution When splicing

4033: [Haoi2015]t1 time limit: ten Sec Memory Limit: + MB Submit: 819 Solved: 375 [Submit] [Status] [Discuss] Description has a tree with a number of N points, and the tree edge has edge rights. Give you a positive integer k within 0~ N, you want to select K points in this tree, dye it black, and turn the other n-k Dianran into white. After all points are dyed, you will get the distance between the black dots 22 plus the whi

[HAOI2015] [bzoj4033] [T1], haoi2015bzoj40334033: [HAOI2015] T1 Time Limit: 10 Sec Memory Limit: 256 MBSubmit: 226 Solved: 111[Submit] [Status] [Discuss]Description A tree with N points has Edge Weight. You can get one from 0 ~ Positive Integers within NK, you need to select K points in the tree, dye them into black, andN-K points Dyed white. After dyeing all vertices, you will get the distance between the

Obtain kernel version number information from the hammer Smartisan T1 Mobile Official system EGL (provided by Cofface):I/ADRENO-EGL (816): The kernel source code addresses for the corresponding Qualcomm CAF are:https://www.codeaurora.org/cgit/quic/la/kernel/msm/tag/?id=AU_LINUX_ANDROID_LNX.LA.3.5.1.5.04.04.02.076.004 Hammer Smartisan T1 Phone official 4.4.2 System kernel version number information

Tags: Database repair uf database repair sql2005 RepairShanghai UF T1 agent, Customer hard drive bad word can not copy copyMdf/ldf Database sql2005 Repair Agents Encounter This problem, the credentials of customer data is very important can not be separated from the database can not be copied away, with MHDD detection is obvious discoveryThe disk has a bad block, with Diskgen and bad disk copy can not be copied away, we use disk mirroring tool The fil

1)First say the tuple: a data structure, separated by commas, that is used to pass a combination of values to a program or operating system.NET Framework directly supports tuples of one to seven elementsTupleT1> TupleT1, T2> TupleT1, T2, T3> TupleT1, T2, T3, T4> TupleT1, T2, T3, T4, T5> TupleT1, T2, T3, T4, T5, T6> TupleT1, T2, T3, T4, T5, T6, T7> Rest property by using the objects in the nested tuple Tuple T1, T2, T3, T4, T5, T6, T7, trest ; ob

); $Std::swap (b[j],b[num-1]); -b[i]-=tmp;b[j]=tmp; - } - returnRes; A } + LL INV (ll a) { the ll x, y; - EXGCD (a,mod,x,y); $ returnx; the } the voidSbpianfen () { the intK=dfs (a[1],a[2],a[3],a[4],a[5],n,0); theprintf"%d\n", ans); - } inll A (intNintm) { the return(Jc[n]*jcny[n-m])%Mod; the } About ll Pow (ll X,ll y) { thell res=1; the while(y) { the if(y%2) res= (res*x)%Mod; +x= (x*x)%Mod; -Y/=2; the }Bayi returnRes; the } the voidSxpianfen () { -ll

1#include 2#include 3#include 4#include 5#include 6 intf[257][257],n,type,v[257],g[257][257],ans,cnt;7 Charop[204];8 intRead () {9 intt=0, f=1;CharCh=GetChar ();Ten while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} One while('0''9') {t=t*Ten+ch-'0'; ch=GetChar ();} A returnt*F; - } - intFunintXinty) { the if(op[0]=='a')returnxy; - Else - if(op[0]=='o')returnX|y; - returnx^y; + } - voidAddintx) { + intA=x>>8,b=x255; Av[a]=1; at for(intI=0;i the; i+

; return ; } if(!h[k-1]!l[k-1]){ if(!a[dx[k-1]-1][dy[k-1]+1]) {a[dx[k-1]-1][dy[k-1]+1]=K; DX[K]=dx[k-1]-1; DY[K]=dy[k-1]+1; if(dx[k]==1) h[k]=1; if(dy[k]==n) l[k]=1; } Else{a[dx[k-1]+1][dy[k-1]]=K; DX[K]=dx[k-1]+1; DY[K]=dy[k-1]; if(dx[k]==1) h[k]=1; if(dy[k]==n) l[k]=1; } return ; }}intMain () {scanf ("%d",N); a[1][n/2+1]=1; dx[1]=1; dy[1]= (n+1)/2; h[1]=1; for(intI=2; i) {deal (i); } for(intI=1; i){ for(intj=1; j) {printf ("%d", A[i][j]); } Putchar ('\ n');

life""The right thing to do"1#include 2#include string.h>3#include 4#include 5#include 6#include 7#include 8#include 9 #defineIvorysiTen #defineMo 10007 One #defineSiji (i,x,y) for (int i= (x); i A #defineGongzi (j,x,y) for (int j= (x); j>= (y); j--) - #defineXiaosiji (i,x,y) for (int i= (x);i - #defineSigongzi (j,x,y) for (int j= (x);j> (y); j--) the #defineIvory (i,x) for (int i=head[x];i;i=edge[i].next) - #definePII pair - #defineFi first - #defineSe Second + #defineINF 10000000 - using name

is out: to fi for the first keyword ascending, subscript for the second keyword descending, a sequence, and then corresponding to fill in the 1-n these numbers just fine.1#include 2 using namespacestd;3 #defineN 1000054InlineintRead () {5 intx=0, f=1;CharA=GetChar ();6 while(a'0'|| A>'9') {if(a=='-') f=-1; A=GetChar ();}7 while(a>='0' a'9') x=x*Ten+a-'0', a=GetChar ();8 returnx*F;9 }Ten structdata{ One intNum,pos; A BOOL operatorConstdata W)Const{ - if(Num==w.n

He"Problem description"A length of?? The paper tape, we can be numbered from left to right 0??? (The left end of the tape is labeled as 0). Now there are?? , each time the tape is folded along a certain location, ask what the length of the tape is after all the operations."Input Format"First row two digits??,?? As described in test instructions. Next line?? An integer representing the position of each collapse."Output Format"An integer on one line represents the answer."Sample Input"5 2 3 5"Samp

]]) Break; $ if(pur0]!a[pur][1]) sum>>=1; -a[pur][id[i]]++; - } the if(!i) DFS (x+1, sum); - for(intj=first[x];j!=i;j=Next[j]) {Wuyi intPur=Go[j]; thea[pur][id[j]]--; - } Wu } - } About intMain () { $N=read (); M1=read (); m2=read (); - for(intI=1; i){ - intX=read (), y=read (); -tot++; Ago[tot]=y; +id[tot]=1; thenext[tot]=First[x]; -first[x]=tot; $ } the for(intI=1; i){ the intX=read (), y=read (); thetot

"Problem description"There is an unreasonable encryption method: randomly inserting characters anywhere in a string. corresponding to theThe unreasonable method of decryption is to delete exactly those characters that are randomly inserted from the string.Given the original text? And the encrypted string?, how many substrings can be decrypted to get the original text?"Input Format"Input the first line contains a string?, the second line contains a string?."Output Format"The output line contains

the calculationlCa(u,v) DP is only Considering that, so the complexity isO(n^2). #include #include#include#include#include#include#defineRep (i,s,t) for (int i=s;i#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i;i=next[i])using namespacestd;Const intBuffersize=1 -;Charbuffer[buffersize],*head,*Tail;inlineCharGetchar () {if(head==tail) { intL=fread (Buffer,1, Buffersize,stdin); Tail= (Head=buffer) +M; } return*head++;} InlineintRead () {intx=0, f=1;CharC=

. First talk about this problem, the problem of their own time to think, anyway, it is not all points, simply set the array smaller, get a sixty or seventy points. So began the violent screening method , the 1~m and 1~n of the prime numbers are screened out, and then subtract, just to note that N is the case of Prime, ans--; so he got 70.Positive solution: Such a large array must not, so need subscript (array) translation , the n~m and greater than sqrt (m) translation to the subscript 0~m-n. Th

//source code from Laekov for c0x17#definePrid "BXJL"#include#include#includeusing namespaceStd;typedefLong Longdint;Const intMAXN =100003;Const intMoD =998244353;CharA[MAXN];intN, M, NE[MAXN], SZ[MAXN];voidPrenext () {ne[1] =0; for(inti =2, j =0; I i) { for(; J A[i]! = a[j +1]; j =Ne[j]); if(A[i] = = A[j +1] J +1i) {Ne[i]= ++J; } Else{Ne[i]=0; }} memset (SZ,0,sizeof(SZ)); for(inti = n; I --i) {++Sz[i]; Sz[ne[i]]+=Sz[i]; }}intMainintargcChar*args[]) { if(ARGC 2|| strcmp (args[1],"-NF") {f

; the ifC=1 Then * begin $ Inc (SUM1);Panax Notoginsengeq[sum1].l:=D; -eq[sum1].r:=b; the End; + ifC1 Then A begin the Inc (SUM2); +neq[sum2].l:=D; -neq[sum2].r:=b; $ End; $ End; - fori:=1 toMax Doro[i]:=i; - fori:=1 toSum1 Do the ifRoot (EQ[I].L) ThenUnion (EQ[I].L,EQ[I].R); - fori:=1 toSum2 DoWuyi begin the ifRoot (NEQ[I].L) ThenContinue -

of the subtree, so we need to redefine the state.F[I][J] is a contribution to the answer, so we just need to multiply the number of black dots and white dots in the subtree, and then take the right value on the road.#include #include #include #define LL Long Longusing namespace STD;Const intn=2100;intn,k,tot=1, point[n],next[n*4],siz[n];Long LongF[n][n];structs{intSt,en,va;} aa[n*4];inline voidAddintXintYintZ) {tot+=1; next[tot]=point[x];p Oint[x]=tot; Aa[tot].st=x;aa[tot].en=y;aa[tot].va=z;

Background information:Previously, because the price of AWS EC2 more expensive and rented other foreign small vendors VPS, after using more than 3 years, found that the AWS EC2 Price is much cheaper than the VPS currently in use.The T1.MICRO model of the most ideal Japanese node of the global line and speed, the choice of 3-year long-term contract price of about 2500 yuan, less than 900 yuan a year, so it is intended to move the blog to AWS EC2.Prior