jj kk

standard threshold-dynamic range determination------------------------%-determining the Dynamic range---Calculate standard Thresholds---------------------------[Up-down mid mean]=signaldynamicrange (MX);[Up_level_std mid_level_std Down_level_std]=signalbiaozhunrange (Up,down,mean);%-14. Symbol synchronization algorithm-determine the best decision point, recorded as SYN---remove the DC component algorithm, remove F[Msigle I] = Synsampledingshi (Mx,mean);%-16. Threshold decision, inverse mapping-

" = "" Goto helpIf "% 1" = "? "Goto helpIf/I "% 1" = "H" Goto helpIf/I "% 1" = "T" if "% 3" = ""(Mstsc/v% 2: 3389) Else (If/I "% 1" = "T" mstsc/V % 2: % 3 goto help) Goto helpIf/I "% 1" = "add" Goto updateIf not exist jj.data.txt echo you have not created a chicken database, use JJ add to create a chicken database goto endFor/F % I in (jj.data.txt) do set/A count + = 1If/I "% 1" = "N" if "% 2" gtr "% count %" Echo: The number you entered is greater

Topic linkshttp://poj.org/problem?id=3070Just learned the matrix, I wrote a Fibonacci, the problem looked at the input and output, directly affixed to the code, on AC. Matrices are a great way to do this.Code#include #include #define MOD 10000__int64 A[5];__int64 B[3][3];__int64 Quick_mod (__int64 N){__int64 i,j,k;A[1]=1; A[2]=1;b[1][1]=0; B[1][2]=1;B[2][1]=1; B[2][2]=1;while (n){if (n1){__int64 S[3];Memset (s,0,sizeof (s));for (i=1;i{for (j=1;j{S[i]= (S[i]+a[j]*b[j][i])%mod;}}for (i=1;iA[i]=s[i

Oracle10g manually create a database stored in ASMProject Process:0. Add two SCSI hard disks to the computer.1. Install ASM2. Create two ASM disk groups: GROUP1 and GROUP23. Create a database. First, create a parameter file and save it as $ ORACLE_HOME/dbs/initnestling. ora.4. Create a password file $ ORACLE_HOME/dbs/orapwdnestling5. Create a path for tracking and log files6. Start the database to the nomount status.7. Create and run the database creation script8. Create an erp tablespace9. Crea

', 'C: C', 'd: d', 'E: E', 'f: f', 'G: G', 'H: H', 'I: I', 'J: J ', 'K: K', 'L: l', 'M: M', 'n': N', 'O: O', 'P: p', 'Q: q', 'R: R','s: S', 'T: t', 'U: U ', 'V: V', 'W: w', 'X: X', 'Y: Y', 'Z: Z', 'AA: A', 'BB: B ', 'CC: C', 'dd: d', 'ee: E ', 'FF: f', 'GG: G', 'hh: H', 'II: I ', 'JJ: J', 'KK: K','ll: l', 'Mm: M', 'nn: N ', 'Oo: O', 'pp: p', 'qq: q', 'rr: R', 'SS: S', 'TT: t', 'uu: u', 'vv: V', 'WW: w ', 'X

Recently in the study of Python, easy to make some records, to facilitate later review#字符串是不可再改变的序列Aa= ' ABCD '#aa [2:]= ' FF ' #报错, not directly assignable#字符串格式化: Use the format operator, or percent%, to implementprint ' Price of Aggs: $%d '%42mm= ' Hello 'Nn= ' World 'Print '%s go%s '% (MM,NN)#String template string Formatting: Replace the $foo in the string with the passed keyword argument (foo)S=template (' $x is $x ')Print S.substitute (x= "Lyq")#字段宽度: The minimum number of characters reta

(); + } A while('0''9') at { -ans=ans*Ten+s-'0'; -s=GetChar (); - } - returnans*ff; - } in Long LongQuick_mod (ll A,ll b)/*slow multiplication prevents overflow*/ - { toa%=q;b%=Q; +ll ans=0; - while(b) the { * if(b1) $ {Panax Notoginsengans+=A; -ans%=q;// the } +b>>=1; Aa1; thea%=Q; + } - returnans; $ } $ Jz martax (Jz x,jz y) - { - Jz ans; theAns.line=X.line; -Ans.cal=y.cal;Wuyi for(intI=1; ii) the for(intj=1;

; - while(~SCANF ("%d%d%d",n,k,m)) { -scanf"%s%s", S1,S2); the intCnt=0; - for(intI=0; ii) { - if(S1[i]!=s2[i]) + +CNT; - } +memset (D,0,sizeof(d)); -d[0][cnt]=1; + for(intI=0; ii) { A for(intj=0; jj) { at if(d[i][j]==0)Continue; - for(intk=0; kk) { - if(J (K-(m-k)) 0|| n-j0|| m-k0)Continue;

Title DescriptionFor Fibonacci series: 1,1,2,3,5,8,13 ... everyone should be familiar with it. But now there is a very simple question: what is the greatest common divisor of the nth and the M items?Update: Added a set of data.input/output formatInput format:Two positive integers n and M. (n,mNote: The data is very largeOutput format:FN and FM greatest common divisor.As a result of looking at the large number of dizziness, so as long as the output of the last 8 digits can be.input/Output sampleI

[i].to]; Q.push (TMP); P[e[i].to]=E[i].st; }Else if(d[e[i].to]==d[e[i].st]+TT) {P[e[i].to]=min (p[e[i].to],e[i].st); } } }}voidPrintintu) { if(p[u]==-1) {printf ("%c\n", rec (u)); return; } printf ("%c-", rec (u)); Print (P[u]);}intMain () {Charst1[Ten],st2[Ten]; intu,v; intTc=0; while(SCANF ("%d", m)!=eofm!=-1) {init (); for(intI=0; i) {scanf ("%s%s", St1,st2); U= Getnum (st1[0]); V= Getnum (st2[0]); Addedege (U,V); Addedege (V,u); } scanf ("%lld%

(1///if the previous line and the current line are horizontally tiled, you must appear in a two-1 if(i+111)) (y (11)))++i; Else return 0; } ///if the current line is placed vertically, the condition is met without any further judgment } Else{///If the previous line is vertical brick, the current row must correspond to 1 if(! (y (1return 0; } } return 1;}intMain () {inti,j,x; while(SCANF ("%d%d", n,m)!=eof (n| |m)) {MST (DP,0); if(n///Th

returned), and then merging the two, then the length, bounds, and string of the combination before and after the enumeration, directly returning, Two strings, with multiplication sign in the middle, with parentheses on both sides to return*/2#include 3 using namespacestd;4#include 5#include 6vectorstring> Dfs (string*SS,intBeginintend)7 {8vectorstring>ret;9 if(Begin>end)returnret;Ten if(begin==end) One { A Ret.push_back (Ss[begin]); - returnret; - } the if(begin+1==e

It is easy to think of the state Dp[n][s][m] (S is the set of numbers appearing), indicating that the first n bits use the number set S and the modulo K remainder is the scheme number of M.using (XY) Base% K = (x*base+y)% K = ((x%k ) * base + y)% K, the third dimension of the state is transferred. However D[16][216][20] has more than 2000 watts of state number, and does not say the problem of timeouts, memory has long exceeded the limit.It can be found that this dimension of s actually contain

Topic Portal1 /*2 BFS: This is very interesting, like the underground city, the figure is three-dimensional, can be from the previous map to the next map of the corresponding position, then the three-dimensional search, more Z axis3 */4#include 5#include 6#include 7#include 8 using namespacestd;9 Ten Const intMAXN = -; One Const intINF =0x3f3f3f3f; A structPoint { - intx, y, z, step; - }; the CharMAZE[MAXN][MAXN][MAXN]; - intdx[6] = {-1,1,0,0,0,0}; - intdy[6] = {0,0, -1,1,0,0}; - intdz[6] =

#include#include#include#include#include#includeusing namespaceStd;typedef pairint,int>Pii;typedefLong Longll;#defineLson l,m,rt#definePi ACOs (-1.0)#defineRson m+1,r,rt#defineAll 1,n,1#defineRead Freopen ("In.txt", "R", stdin)Constll infll =0x3f3f3f3f3f3f3f3fll;Const intinf=0x7ffffff;Const intMoD =1000000007;Doublea[1010],b[1010],p[ -];intn,k1,k2,k3,a1,b1,c1;voidsolve () {Doubletp=1.0/(k1*k2*K3); Memset (P,0,sizeof(p)); Memset (A,0,sizeof(a)); memset (b,0,sizeof(b)); for(intI=1; ii) for(intj=1

structMatrix9 {Ten intr,c; OneLL m[3][3]; AMatrixintArrint_c): R (_r), C (_c) {} - }; - voidInit (Matrix m,intAintb) the { -m.m[1][1]=a,m.m[1][2]=-b; -m.m[2][1]=1, m.m[2][2]=0; - } + matrix Multiply (matrix A,matrix b) - { + Matrix M (A.R,B.C); A for(intI=1; ii) { at for(intj=1; jj) { -m.m[i][j]=0; - for(intk=1; kk) -m.m[i][j]+=a.m[i][k]*B.m[k][j]; - } - } in

State compression +bfs, once AC.1 /*1885*/2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 9 #defineMAXN 105Ten One BOOLvisit[maxn][maxn][ -]; A -typedefstructnode_t { - intx, Y, T, K; the node_t () {} -node_t (intXxintYyintTtintKK) { -x = XX; y = yy; t = TT; K =KK; - } + } node_t; - + intN, M; A node_t Beg; at CharMAP[MAXN][MAXN]; - intdir[4][2] = { --1,0,1,0,0,-1,0,1 - }; - -InlineBOOLCheckintXinty) { in retur

3822 Domination + intT scanf ("%d", T); A while(t--) { the intN, M; scanf ("%d%d", n, m); +Memset (DP,0,sizeof(DP)); -dp[1][1][1] =1.0;intTOT = n *m; $ for(intI=1; ii) { $ for(intj=1; jj) { - for(intk=1; kk) { - if(Dp[i][j][k] >0) { thedp[i+1][j+1][k+1] + = Dp[i][j][k]/(tot-i) * (n-j) * (M-k); -dp[i+1][j+1][k] + = Dp[i][j][k]/(t

=TEMP.R; TC=temp.c; if(Judge (tl-1, TR,TC)) Que.push (State (TL-1, tr,tc,temp.num+1)); if(Judge (tl+1, TR,TC)) Que.push (State (TL+1, tr,tc,temp.num+1)); if(Judge (tl,tr-1, TC)) Que.push (State (Tl,tr-1, tc,temp.num+1)); if(Judge (tl,tr+1, TC)) Que.push (State (Tl,tr+1, tc,temp.num+1)); if(Judge (tl,tr,tc-1)) Que.push (state (TL,TR,TC-1, temp.num+1)); if(Judge (tl,tr,tc+1)) Que.push (state (TL,TR,TC+1, temp.num+1)); } return-1;}intMain () {Chars[ -]; intans; Ios::sync_with_stdio (false); whi

Topic Portal1 /*2 violence: Violence too, no words. No-map, both sides of the point to add degrees and points3 */4#include 5#include 6#include 7#include 8#include 9 using namespacestd;Ten OnetypedefLong Longll; A Const intMAXN = 2e2 +Ten; - Const intINF =0x3f3f3f3f; - ll A[MAXN]; thevectorint>G[MAXN]; - intDEGREE[MAXN]; - - intMainvoid)//Hihocoder 1179 Timeless Games + { - //freopen ("c.in", "R", stdin); + A intN, M; at while(SCANF ("%d%d", n, m) = =2) - { - for(intI