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
标签： no-crashes-allowed development followed software
上传时间： 2016-04-14
上传用户：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.
标签： dimensions arbitrary function reverse
上传时间： 2016-04-16
上传用户：waitingfy
用宏和高级汇编技术实现类似高级语言中的条件分支语句IF功能。同时，编写一个程序证明所编写宏的正确性。要求如下： (1).iff后的条件为“x1,op,x2”形式，其中x1和x2为操作数，op为关系比较符，用g(>),l(<),e(=), ge(>=), le(<=)表示。另外，x1和x2必须是字。 (2).elsee语句（相当于else语句）是可选项，即iff后可以不跟elsee。 (3)条件分支iff的结束用.ifend表示.
上传时间： 2013-12-31
上传用户：风之骄子
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
标签： DESProcess ttbitdiff xorRes bdiff
上传时间： 2016-07-02
上传用户：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 {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：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 {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：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 {
标签： AOrigin APoint Point PointToAngle
上传时间： 2013-12-18
上传用户：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.
标签： enumeration utility entire devic
上传时间： 2013-12-17
上传用户：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 }
标签： insert_sort int return void
上传时间： 2014-01-22
上传用户：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.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
