i9 alienware

Summary of 80 typical Python documents (tutorials + source code + tools)-download directory. Hello everyone, the 51cto download center collects 80 Python documents based on the popularity and praise of the materials, share it with Python developers. The content includes 1 Python topic, 66 learning materials, 7 Python source codes, and 6 related software. There are many attachments that cannot be shared one by one. Here we will organize a download Directory, which can be saved for future use and

Class has some special attributes, so that we can obtain some additional information. 1 >>> class class1 (object): 2 "class1 Doc. "3 def _ init _ (Self): 4 self. I = 1234 5 6 >>> class1. _ Doc _ # type help information 7' class1 Doc. '8 >>> class1. _ name _ # type name 9 'class1' 10 >>>> class1. _ module _ # type module 11' _ main __' 12 >>> class1. _ bases _ # base class 13 inherited by the type ( Member Python class also contains two types of members: type and instance. 1 >>> class Class1:2

). http: // www. fx998.cn4 Q8 q) _ # E9 t In the consolidation of the market, 60% people is making money. http://www.fx998.cn! O) r-u/U (E8 x0 I0 P + n o + S Most people do not lose money in the Consolidation market? When the market starts, is most people's judgment correct? Indeed. http: // www. fx998.cn5 y # F "K; I9 W1 L 'G (K However, in the entire 30% phase of the market, 80% of these people chose to close their positions and leave the market. M

. Forced type conversionThe mandatory conversions available in ECMAScript are as follows:Boolean (value) -- converts a given value to a Boolean value.Number (value) -- converts a given value to a Number (which can be an integer or floating point Number)String (value) -- converts a given value to a String.Example:Copy codeCode: var b1 = Boolean (""); // false-empty stringVar b2 = Boolean ("hi"); // true-not empty stringVar b3 = Boolean (100); // true-not zero numberVar b4 = Boolean (null); // fal

] ww; -Memset (G,0,sizeof(g)); +Memset (F,0,sizeof(f)); f[n]=1; - for(intI=0; i9; i++) g[n][i]=1; + for(inti=n-1; i;i--){ A for(intj=0; j9; j + +) at for(intk=0; k9; k++) - if(ABS (J-K) >=2) g[i][j]+=g[i+1][k]; - for(intj=0; j1];j++) - if(ABS (A[X][I]-J) >=2) f[i]+=g[i+1][j]; - if(ABS (a[x][i]-a[x][i+1]) >=2) f[i]+=f[i+1]; - } in for(inti=ww+1; i) - for(int

: 250px; margin:0 auto;}input{ text-align: center; font: 40px/60px 'Microsoft YaHei'; width: 30%; float:left; box-sizing:border-box}button{ position: absolute; bottom:-30px; left: 50%; margin-left: -30px; width: 40px;} 2. next we will focus on the js part.I have already mentioned the js implementation method. here is the relevant code. Var oBtn = document. getElementsByTagName ('Button ') [0], aInp = document. getElementsByTagName ('input'); function isNum () {var aTemp = []; // cre

,p+1, now+v[s[p][(F-DEP +1)]]); }Else { for(intI=0; i9; i++) { if(Num[i])Continue; if(F-DEP +1==1i==0)Continue; Num[i]=1; v[s[p][(f-Dep +1)]] =i; DFS (Dep,last,p+1, now+i); v[s[p][(f-Dep +1)]] = -1; Num[i]=0; } } }ElseDFS (dep,last,p+1, now); }Else { if(v[s[p][(F-DEP +1)]]!=-1) { if((f-dep+1==1) v[s[p][(F-DEP +1)]]==0)return ; if((now+last)%Ten! = v[s[p][(F-DEP +1)]])return ; DFS (

Describe http://www.lydsy.com/JudgeOnline/problem.php?id=1833Statistics \ (a~b\) in number \ (0,1,2,..., 9\) how many times respectively.Analysis Digital DP is really the details and more error-prone, I do not bother to read the problem, so also lazy to write the problem ...Pay attention to the details, or ...1#include 2 using namespacestd;3 4typedefLong Longll;5 ll A, B;6ll a[Ten],b[Ten],num[ the];7ll f[ -][Ten][Ten];8 9 voidSolve (ll X,ll *a) {Ten if(x==0)return; One if(xTe

BZOJ1295[SCOI2009] the longest distanceTest instructionsN*m Block ground, if two pieces of land are not obstacles, then each other can reach. If the two pieces of land can reach each other (can go through the other) then the distance between them is the Euclidean distance of their central point, the maximum distance between the land can be removed if it is not greater than the T obstacle. N,m≤30ExercisesConsider an obstacle as a side length of 1, find out the length of the edge to be crossed bet

l1-003. Single-digit statistical time limit of MS memory limit 65536 KB code length limit 8000 B procedure StandardAuthor ChenGiven a k-bit integer n = dk-1*10k-1 + ... + d1*101 + d0 (0Input format:Each input contains 1 test cases, a positive integer n of no more than 1000 bits.Output format:For each of the different single digit numbers in n, output the digit D and the number of occurrences of M in N in a row in d:m format. Required to be output in ascending order of D.Input Sample:100311Sample

][x1][y1][x2][y2]=t*T; returnt*T; } for(a=x1;a) {C=calsum (A +1, y1,x2,y2);//The and of the Matrix on the rightE=calsum (X1,Y1,A,Y2);//The and of the left MatrixT=min (N-1, X1,y1,a,y2) +c*c,fun (n1, A +1, Y1,x2,y2) +e*e); if(min>t) min=T; } for(b=y1;b) {C=calsum (x1,b+1, x2,y2);//The following matrix and theE=calsum (X1,Y1,X2,B);//The and of the matrix aboveT=min (N-1, x1,y1,x2,b) +c*c,fun (n1, x1,b+1, X2,y2) +e*e); if(min>t) min=T; } Res[n][x1][y1][x2][y2]=MIN; returnMIN;}intMain () {me

Dfstime limit:5000/2000ms (java/other) Memory limit:65536/32768k (Java/other) total submission (s): Accepted SubMission (s): 32font:times New Roman |Verdana | Georgiafont Size:←→problem Descriptiona DFS (digital factorial sum) number is found by summing the factorial of every dig It's a positive integer.For example, consider the positive integer 145 = 1!+4!+5!, so it ' s a DFS number.Now you should find out all the DFS numbers in the range of int ([1, 2147483647]).There is no input for this prob

return, and the next time you enter this function again. But the addition of yield is not so, and each time yield is encountered, it returns a value to the caller, and then the current function pauses (rather than exits), and then the control is returned to the caller.Yield is the equivalent of return, except that the return ends the current function, but the yield is paused, which saves the scene.So the next time this function is called, it continues from the last saved state, not from the beg

Title Link: http://vjudge.net/problem/UVA-1225#include#include#includeusing namespacestd; intf[10000][Ten]; intMain () {memset (F,0,sizeof(f)); for(inti =1; I 10000; ++i) { for(intj =0; J Ten; ++j) F[i][j]= f[i-1][j]; intleft =i; while(left) {F[i][left%Ten] ++; Left/=Ten; } } intT,n; while(~SCANF ("%d",t)) while(T--) {scanf ("%d",N); for(inti =0; I 9; ++i) printf ("%d", F[n][i]); printf ("%d\n", f[n][9]); } return 0; } View Codeand wrote a similar code, but WA1#include

Train of thought: Dp,f[i][j] indicates that the new arrangement state is I (that is, which number has been selected for the new arrangement), and then the remainder of modulo d is the number of schemes of J.But given that there may be several occurrences, assuming a number x appears cnt[x] times, then for a workable answer, apparently including cnt[x] x, then the answer will be counted multiple times, Because if the state I first joins the first X and then joins the second X and the state I firs

C-language string to note that the last one is the '/0 ' problem by default. This is a simple error point.Strlen () calculates the length without considering the end of '// Example 1 void charstring[tenchar"0123456789" string,str1);} The length defined by string is the last of the 10,STR1 by default, so the length of the str1 is actually 11. In the strcpy () function, if the second string is longer than the first string, there is no error, the length of the first string is incremented, and the

(order[1],order[3]); Adis[6] = cal (order[1],order[4]); thedis[7] = cal (order[2],order[4]); +dis[8] = cal (order[2],order[5]); -dis[9] = cal (order[3],order[5]); $ intFlag2 =1; $ for(intI=5;i9; i++) - { - if(Fabs (dis[i]-dis[9]) >EPS) the { -Flag2 =0;Wuyi Break; the } - } Wu returnFlag1 Flag2; - } About $ BOOLDfsintPOS) - { - if(pos = =6) - { A if(isOK ()) + retur

Test instructionsTo a positive even N of not more than 10, the number of n number of digits before the string of N/2 and the second N/2 bit and the number of equal.Ideas:DP[I][J] The number of number strings consisting of an I-digit and J.DP[I][J]+=DP[I-1][J-K];Finally, arrange a combination.#include Long Longdp[6][ -];intMain () {intN; scanf ("%d",N); for(intI=0; i9; i++) {dp[1][i]=1; } for(intI=2; i2; i++) { for(intj=n/2*9; j>=0; j--

(intI=1; i1; i++){ Constuchar* previous = src.ptr1); Constuchar* current = src.ptr(i); Constuchar* next = src.ptr1); for(intj=1; j1; j + +) { for(intk=0;k3; k++) {Arr[k]= previous[j+k-1]; Arr[k+3] = current[j+k-1]; Arr[k+6] = next[j+k-1]; } l= Nc*i+j;//calculate the location in the table of current pixellmaskoperation (table_lap,arr,l); TABLE_ORIG[L]= arr[4]; }}//Pixels Scaleuchar* la_scaled =NewUchar[n];table_scale (table_lap,la_scaled,n);//padding ValuesMat Laresult_own; Laresul

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=2089Qualifying questions encountered when the number of DP will not, first find a way to write a simple orz, after all do not know how to live, anyway first pass up OrzDP[I][J] means that the number of digits is I, the highest bit is J when the number does not contain 4 and 62, learn from Baidu Library: http://wenku.baidu.com/view/9de41d51168884868662d623.html#include #include#includeusing namespacestd;intf[Ten][ A],digit[9];voiddp () {f[0][0