The standard does not specify a limit, it is implementation defined. The practical limit on my system is 30 dimensions. You can easily determine the upper limit by instantiating a static array of char (the smallest data type), where each dimension is 2 (the minimum size for any dimension). static char ch[2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2] = {0}; // ok static char no[2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2][2] = {0}; // no can do! Note that if you allocate on the stack (non-static allocation), the limit drops to just 19 dimensions. Note also that a dimension of size 1 isn't an array. You can easily have hundreds of dimensions all of size 1 in an array, but there would only ever be 1 element in the entire array. Thus any dimension of 1 in any array (no matter which dimension) becomes redundant in terms of the array's size. For example, the following definitions are all exactly the same size: char ch[3][2][1] = {0}; char ch[3][1][2] = {0}; char ch[1][3][2] = {0}; char ch[3][2] = {0};
if (i==2) if (j==2) System.out.println ("i==2 and j==2"); else System.out.println ("i==2 and j<>2"); else System.out.println ("i<>2");
It is not very difficult if you can image how it looks. In your case what you need is rectangular Cuboid with height N, width N and depth 2. Here is example in C: /* There N = 2 */ int matrix[2][2][2] = 1, 2}, {1, 2}}, {{1, 2}, {1, 2; /* There N = 3 */ int matrix[3][3][2] = 1, 2}, {1, 2}, {1, 2}}, {{1, 2}, {1, 2}, {1, 2}}, {{1, 2}, {1, 2}, {1, 2; These two are statical, it is possible to create dynamical cuboid, but it is more complex and requires knowledge of pointer and memory management.
2
Floor division is division where the answer is rounded down. For example, 5/2 in floor division is not 2.5, but 2. In Python 2, floor division is the default. In Python 3, the floor division operator is //. Python 2: >>> 5/2 2 >>> 5.0/2 2.5 Python 3: >>> 5/2 2.5 >>> 5//2 2
dalawa ang balya ng isang nota o pahinga at dalawa ang laman ng isang hati
magbigay ng tatlong awitin na nasa kumpasang 2/4 3/4 4/4
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(2+2)/(2+2)=1 (2/2)+(2/2)=2 (2*2)-(2/2)=3 (2+2+2)-2=4 (2+2)+(2/2)=5 (2*2*2)-2=6 ((2-.2)/.2)+2=7 2+2+2+2=8 (22/2)+2=9 ((2+2)*2)+2=10 ((2-.2)/.2)+2=11 (2+2+2)*2+=12 (22/2)+2=13 (2/.2)+(2+2)=14 2+(2/2)/.2=15 (2+2)*(2+2)=16 ...??????????=17 ((2+2)/.2)-2=18 ((2+2)-.2)/.2=19 22- √(2+2)=20
230.2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 = 1073741824
21
As a fraction it is: 2/9
2^18 = 262,144(2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2)
2,222,2222 + 2 + 2 + 2 + 2 + 2 + 2 = 7(2) = 142 - 2 - 2 - 2 - 2 - 2 - 2 = 2 - 6(2) = -102*2*2*2*2*2*2 = 27 = 1282/2/2/2/2/2/2 = 2/26 = 1/32Otherwise, it's just a bunch of 2's.
256 2, 128 2, 2, 64 2, 2, 2, 32 2, 2, 2, 2, 16 2, 2, 2, 2, 2, 8 2, 2, 2, 2, 2, 2, 4 2, 2, 2, 2, 2, 2, 2, 2
8192
32