abacus counting

1000000007#defineINF 999999999#definePi 4*atan (1)//#pragma COMMENT (linker, "/stack:102400000,102400000")Const intn=1e5+Ten, m=2*1e5+Ten;intA[n],b[n];ll Ans[n];intL[m];intR[m];intMain () {intx,y,z,i,t; intT; while(~SCANF ("%d",x) {memset (ans,0,sizeof(ans)); for(i=1; i) scanf ("%d",A[i]); for(i=1; i) { for(t=1; t) { if(a[t]==A[i]) b[t]=0; ElseB[t]=a[t]>a[i]?1:-1; } intflag=8000; for(t=i;t>0; t--) Flag+=b[t],l[flag]++; Flag=8000; for(

note is that the recursive DP when the three rectangles are in the lower right corner (because the number of bars is unchanged because of the downward or right translation of a unit), so that if the recursion is to consider from this point to (I,J) this point of the situation can be, while the recursive ans, The rectangle is the upper left side, then, as long as the entire large rectangle inside to (i,j) this point of the situation can be. Of course, purely personal understanding.UVA 1393 Highw

Starting from all of the one-digit contained 1, the recursion pushes all the K-digits contained in 1, recursively:AK = ak-1 * + POW (10,k-1);AC Code:#include #includeusing namespacestd;intMain () {intN; Vectorint>Rec; scanf ("%d",N); intNC (n); while(n! =0) {rec.push_back (n%Ten); N/=Ten; } Vectorint>R (Rec.size ()); Vectorint>REM; intRadix1); for(inti =0; i ) {rem.push_back (radix); if(i = =0) r[0] =1; Else{R[i]= R[i-1] *Ten+Radix; } Radix*=Ten; } intRet0); for(inti = rec.size ()-1; I >=0

Some of the important role nodes are over-counted:node100 node101 (LSF Job scheduling system node, and PAC Site page management)node108 node105 (NIS Service Master node)node166 (parastor Parallel storage MGR Node)Backup policy: Backups can be backed up by scheduled task/etc/crontab, backed up to the/home directory, or backed up to a storage directory;1 Back up important directories such as the/etc/opt/root of the node108 node:* 3 * * 6 root tar-zcpf/home/public/node108. ' data+\%y\%m\%d '. ETC.T

Simple question. Statistics can be a bit.#include #include#include#include#include#include#include#includeusing namespacestd;Const intmaxn=100000+Ten;structnode{intLeft ; intRight ; intVal; intDEP;} S[MAXN];intN;intA[MAXN];intmax_dep,n1,n2;voidDfsintXintDEP) {MAX_DEP=Max (MAX_DEP,DEP); S[X].DEP=DEP; if(s[x].left!=-1) DFS (s[x].left,dep+1); if(s[x].right!=-1) DFS (s[x].right,dep+1);}voidDFS (intx) { if(S[X].DEP==MAX_DEP) n1++; Else if(s[x].dep==max_dep-1) n2++; if(s[x].left!=-1) DFS (s[x].left

(s1[mark+1] =='-') {cout"-"0"."; value2--; while(value2) {cout0; value2--; } for(n =1; n ){ if(n! =2) {coutS1[n]; } } } Else if(s1[mark+1] =='+') {cout"-"; if(value2 >= Mark-3) { for(n =1; n ){ if(n! =2) {coutS1[n]; } } for(M =0; M 3-Mark; m++) {cout0; } } Else if(Value2 3){ for(n =1; n ){ if(n! =2){

#pragma once#includeA smart pointer based on reference counting

; - + intNumislands = 0; A introws =grid.length; at intcols = grid[0].length; - for(inti=0; i ) - { - for(intj=0; j ) - { - if(Grid[i][j] = = ' x ')Continue; in - if(Grid[i][j] = = ' 1 ' ) to { + Discoverisland (grid, rows, cols, I, j); -numislands++; the } * Else $ {Panax NotoginsengGRID[I][J] = ' x '; - } the } + } A the

: 1≤sThe 2nd behavior has a W lowercase string, which is a required jam number.The data given is correct and does not have to be verified.output FormatThe output is up to 5 rows, which is the 5 jam numbers immediately following the input jam number, and if there are not so many jam numbers behind, then a few will output. Each line only outputs one jam number, which is a string of W lowercase letters and no extra spaces.Sample input:2 10 5BdfijSample output:BdghiBdghjBdgijBdhijBefghAlgorithm anal

original can be simplified to n (n-1) * (2n-4)/3 + (m-n+1) *n* (n-1);In addition to the diagonal condition, the same as above, then d = * (n (n-1) * (2n-4)/3 + (m-n+1) *n* (n-1));　　　　The answer is a+b+d;The code is as follows1#include 2#include 3#include 4 using namespacestd;5 6typedef unsignedLong LongULL;7 8 intMain ()9 {Ten ULL m, N; One while(Cin >> M >>N) A { - if(!m !n) Break; - if(mN) Swap (m, n); theULL A = m*n* (n1), B = n*m* (M-1); -ULL D =2* (n1)*(2*n-4)/3+ (

module is a separate relationship so to use the else if instead of the original if to represent. However, after running the result is still error, the output is correct and the number of errors is always reversed, and found that in the main function, the return of the pointer is wrong, resulting in a result error. This is the problem encountered, these problems are caused by carelessness, there is a logic error. In the final analysis is the foundation is too poor, lack of practice!Arithmetic im

, Len; the intans;111 the #ifndef Online_judge113Freopen ("data.in","R", stdin); theFreopen ("Data.out","W", stdout); the #endif the 117 while(SCANF ("%s", s)!=eof (s[0]!='#')) {118 //printf ("%s\n", s);119 for(i=0; S[i]; ++i) -Str[i] = s[i]-'a'+1;121Str[i] =0;122Len =i;123Da (str, SA, len+1, MAXM);124 calheight (str, SA, Len); the #ifndef Online_judge126 Printsa (len);127 Printrank (len); - Printheight (len);129 #endif the for(ans=0, i=1; i2; ++i) {131Ans +

Given a number n, the calculation from 1 to n can be composed of several different triangles.The range of n is 10^6, which is probably recursion. From f[i-1] to f[i] can be calculated linearly. Be aware that the result exceeds int.#include #include#includeusing namespacestd;Long Longdp[1000100];intN;intMain () {dp[3] =0; dp[4] =1; dp[5] =3; for(intI=6;i1000010; i++) { Long LongK = i3; if(k1) Dp[i]= dp[i-1]+ (k +1) * (k +1)/4; ElseDp[i]= dp[i-1]+ k* (k +2)/4; } while(~SCANF ("%d",

Push_back#defineFi first#defineSe Second#defineMP Make_pairConst intn=5005;Constll INF = 1ll A;Const intINF =1 to;Const intMod=1000000007;Const intM =1000000;structss{intx, y;} A[n],p[n*N];intCNT;intCMP (ss S1,ss S2) {if(s1.x = = s2.x)returns1.ys2.y; returns1.xs2.x;}voidinit () {CNT=0; mem (p);} ll solve () {ll ans=0; for(inti =0; I CNT;) { intnow = i+1; while(p[i].x = = p[now].x p[i].y = = p[now].y) now++; ll c= Now-i; if(c>=2) ans + = c* (C-1)/2; I=Now ; } returnans;}intMain () {intT,n

; - } - voidEX_GCD (ll A, ll B, ll x, LL y) the { - if(b)Wuyi { theEX_GCD (b, a%b, x, y); -LL tmp=x; Wux=y; -Y=tmp-(A/b) *y; About } $ Else - { -x=1, y=0; - return ; A } + } the ll C (ll N, ll M) - { $ if(N==m | | m==0) the return 1; the if(m==1|| m==n-1) the returnN; the //LL tmp=1, Ans=1; - //For (LL i=min (n, m); i>=1;i--) in // { the //tmp= (tmp*i)%mod; the //ans= (ans* (n+1-i))%mod; About // } the LL x, y; theEX_GCD (jc[m]*jc[n-m

SGU-117Counting Time Limit: 250MS Memory Limit: 4096KB 64bit IO Format: %i64d %i64u Submit StatusDescriptionFind amount of numbers for given sequence of integers numbers such that after raising them to the M-th Power they Would be divided by K.InputInput consists of lines. There is three integer numbers N, M, K (0first line. There is N positive integer numbers? Given sequence (each number isn't more than 10001)? On the second line.OutputWrite answ

points by twice times, adding the polar angle of all points plus the PI to the extended array.#include #include #include #include using namespace STD;Const intN =1205;Const DoublePI =4*Atan(1.0);Const DoubleEPS =1e-9;intNDoubleS, r[2*n];structPoint {Doublex, y;} P[n];DoubleCount (intD) {intc =0, MV =0; for(inti =0; I if(i = = d)Continue;DoubleA =atan2(P[i].y-p[d].y, p[i].x-p[d].x); R[C] = A; r[c+n-1] = a +2*pi; C + +; } C =2N2; Sort (R, R + C);DoubleAns =0; for(inti =0; I 1; i++) {Dou

Divided into Queens in the same row, the same column, the same diagonal count of the sum can be added separately.1#include 2#include 3#include 4#include 5 using namespacestd;6 7typedefLong Longll;8 9 ll A (ll N, ll M)Ten { One returnn * M * (M-1 ); A } - - ll B (ll N, ll M) the { - returnA (M, n); - } - + ll D (ll N, ll M) - { + returnn * (N-1) * (N-2) /3*4+2* (M-n +1) * N * (n-1 ); A } at - intMain () - { - ll N, m; - while(Cin >> N >>m) - { in if(n = =0 m = =0)

Test instructions: Gives N, writes the first n integers sequentially, and counts the number of occurrences of each number.The stupidest way to do it--direct statistics--and later found that the online solution had to be done first.1#include 2#include 3#include 4#include 5 using namespacestd;6 7 Chars[10005];8 inta[ the];9 Ten intMain () One { A intncase,n,i; -scanf"%d",ncase); - while(ncase--) the { -scanf"%d",n); -Memset (A,0,sizeof(a)); - for(i=1; i) + { -

Objective-c class type, if Nil. Represents an object in Objective-c. NSNull An empty object that itself represents a class type.5. What is the difference between a deep copy and a shallow copy? A shallow copy of its essence, just copy the object pointer, that is, memory or the original memory. Deep copy: Re-applied to the system memory, and the original data are all copied in a deep copy of a copy, shallow copy in the manual management of memory mode, equivalent to the reference counter added 1