Swfdec still is development software, but has also followed a rigid no-crashes-allowed policy. I believe it s stable enough now to be installed as a default plugin for people that can live with occasional crashes of their browser. But don t blame me if it does crash. File a bug at https://bugs.freedesktop.org/enter_bug.cgi?product=swfdec
標(biāo)簽： no-crashes-allowed development followed software
上傳時間： 2016-04-14
上傳用戶：franktu
1) Write a function reverse(A) which takes a matrix A of arbitrary dimensions as input and returns a matrix B consisting of the columns of A in reverse order. Thus for example, if A = 1 2 3 then B = 3 2 1 4 5 6 6 5 4 7 8 9 9 8 7 Write a main program to call reverse(A) for the matrix A = magic(5). Print to the screen both A and reverse(A). 2) Write a program which accepts an input k from the keyboard, and which prints out the smallest fibonacci number that is at least as large as k. The program should also print out its position in the fibonacci sequence. Here is a sample of input and output: Enter k>0: 100 144 is the smallest fibonacci number greater than or equal to 100. It is the 12th fibonacci number.
標(biāo)簽： dimensions arbitrary function reverse
上傳時間： 2016-04-16
上傳用戶：waitingfy
用宏和高級匯編技術(shù)實(shí)現(xiàn)類似高級語言中的條件分支語句IF功能。同時，編寫一個程序證明所編寫宏的正確性。要求如下： (1).iff后的條件為“x1,op,x2”形式，其中x1和x2為操作數(shù)，op為關(guān)系比較符，用g(>),l(<),e(=), ge(>=), le(<=)表示。另外，x1和x2必須是字。 (2).elsee語句（相當(dāng)于else語句）是可選項(xiàng)，即iff后可以不跟elsee。 (3)條件分支iff的結(jié)束用.ifend表示.
標(biāo)簽： 匯編 技術(shù)實(shí)現(xiàn) 分支 高級語言
上傳時間： 2013-12-31
上傳用戶：風(fēng)之驕子
DESProcess FILE *mfp,*cfp int ttch=0,xorRes,ttbitdiff=0 char mch,cch float bdiff=0 if((mfp=fopen(mfile,"r"))==NULL) {cout<<"Cannot open the file to compare"<<endl } if((cfp=fopen(cfile,"r"))==NULL) {cout<<"Cannot open the file to compare"<<endl } else { while(!feof(mfp)&!feof(cfp)) { ttch++ mch=fgetc(mfp) cch=fgetc(cfp) xorRes=mch^cch ttbitdiff+=(xorRes&1)+(xorRes&2)/2+(xorRes&4)/4+(xorRes&8)/8+(xorRes&16)/16+(xorRes&32)/32+(xorRes&64)/64+(xorRes&128)/128 } bdiff=float(ttbitdiff)*100/float(ttch*8) } *bitdiff8byte=float(ttbitdiff)*100/float(8*8) return bdiff
標(biāo)簽： DESProcess ttbitdiff xorRes bdiff
上傳時間： 2016-07-02
上傳用戶：1079836864
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
標(biāo)簽： AOrigin APoint Point PointToAngle
上傳時間： 2016-10-31
上傳用戶：zhyiroy
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
標(biāo)簽： AOrigin APoint Point PointToAngle
上傳時間： 2016-10-31
上傳用戶：sunjet
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
標(biāo)簽： AOrigin APoint Point PointToAngle
上傳時間： 2013-12-18
上傳用戶：rocketrevenge
This utility has two views: (a) one view that will show you the entire PnP enumeration tree of device objects, including relationships among objects and all the device s reported PnP characteristics, and (b) a second view that shows you the device objects created, sorted by driver name. There is nothing like this utility available anywhere else.
標(biāo)簽： enumeration utility entire devic
上傳時間： 2013-12-17
上傳用戶：zjf3110
void insert_sort(int *a,int n) { if(n==1) return insert_sort(a,n-1) int temp=a[n-1] for(int i=n-2 i>=0 i--) { if(temp<a[i]) a[i+1]=a[i] else break } a[i+1]=temp }
標(biāo)簽： insert_sort int return void
上傳時間： 2014-01-22
上傳用戶：banyou
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標(biāo)簽： the subsequence determine Instead
上傳時間： 2013-12-17
上傳用戶：evil
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