t1 wiring

); $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

Determines whether the T1 tree contains all the topologies of the T2 tree public class containsubtree{the public static class node{the public int value; Public Node left; public Node right; public Node (int data) {this.value=data; }//To determine whether to include public static Boolean Iscontainsubtree (Node h1,node h2) {if (h2==null) { return true; } if (H1==null) {return false; ///Three case matching

August 15, 2017 raise group T1 string Description There are two string s,t of length n and consist only of lowercase letters, satisfying s and T happen to have K-bit difference. Q in all strings that happen to have K-bit different from S, T is sorted by dictionary order. Because the answer can be large, the die 10^9+7 output. Input The first row of two integers n,k.The second line is a string s.The third line is a string T. Output An integer line rep

want to set the response data, you can first respond to the assertion there is not filled out, run the entire project this time is displayed all through, so you then the if judgment is quite empty, then everything is right. This time you only have to look at the response data, the comparison will be able to know that the data only successful login, you choose to modify the response assertion there.Response assertion is empty hereThe result tree is all through, the greenThe response data is not

number, the quality of transmission has been improved and the price has been improved. The following is a simple generalization: CAT 1: A traditional telephone line. Suitable for analog voice and data. CAT 2: Originally the IBM 3 class. Suitable for 4 Mbps token network LAN and T1, less used. Cat 3: Designed for Mbps Ethernet, it is widely used in digital PBX systems but tends to decline due to cat 5. CAT 4: Designed for Mbps Ethernet and the ten M

First, the introduction Today, most of the voice, data and image transmission in office buildings, buildings, university campuses, government departments and even residential communities in intelligent buildings are based on structured cabling systems on their physical structure. We know that in the Iso/osi protocol, the physical layer is the basis of the network system, all network communications rely on the physical layer of the cable to the voice, data to the destination. With the populariza

It is worth learning about the Integrated Wiring System. This article comprehensively introduces the Integrated Wiring System from two aspects: the characteristics of the Integrated Wiring System and the system introduction, I also hope to teach you more knowledge. Overview The Integrated Wiring System is a modular an