mathematica试除法解素数问题

DivPrime[n_]:=
Block[
{primeset={2,3}},
For[k=4,k<=n,k++,

Block[{i,flag=True},
For[i=1,primeset[[i]]<=Sqrt[k]&&i<=Length[primeset],i++,

If[Mod[k,primeset[[i]]]==0,flag=False;Break[]]
];

If[flag,AppendTo[primeset,k]];
primeset

];
primeset

]

]

(*重要说明，while为当型循环，执行无穷多次，if为判断，只执行一次*)

(*程序员最忌讳用循环，效率太低了*)
DivPrime[n_] :=
Block[{primeset = {2}, i, j, flag},
If[n == 1, primeset = {}];
For[i = 3, i <= n, i++, flag = True;
For[j = 1, primeset[[j]] <= Sqrt[i] && j <= Length[primeset], j++,
If[Mod[i, primeset[[j]]] == 0, flag = False; Break[]]];
If[flag == True, AppendTo[primeset, i]]];
primeset]

(*第一次改进*)
DivPrime1[n_] := Block[{primeset = {2, 3}, i}, For[i = 4, i <= n, i++,
If[! MemberQ[Mod[i, primeset], 0], AppendTo[primeset, i]]];
primeset]

(*第二次改进，改进后速度大大提高，n取100000没几秒就出来了*)
DivPrime2[n_] :=
Fold[If[! MemberQ[Mod[#2, #1], 0], Append[#1, #2], #1] &, {2, 3},
Range[4, n]]

(*总结：能不用循环尽量不用循环，尽可能使用内置函数，多次用到的量要先存起来*)

function isp = primeornot(X)
isp = false(size(X));

if ~isempty(X)
X = X(:);
if ~isreal(X) || any(X < 0) || any(floor(X) ~= X) || ...
any(isinf(X))
error(message('MATLAB:isprime:InputNotPosInt'));
end

n = max(X);
if isinteger(X) || n <= flintmax(class(X))
if (isa(X,'uint64') || isa(X,'int64')) && n > flintmax
p = primes(2.^(nextpow2(n)/2));
else
p = primes(cast(sqrt(double(n)),class(X)));
end
for k = 1:numel(isp)
Xk = X(k);
isp(k) = (Xk>1) && all(rem(Xk, p(p<Xk)));
end
else
fm = flintmax(class(X));
p = primes(sqrt(fm));
for k = 1:numel(isp)
Xk = X(k);
isp(k) = (Xk<fm) && (Xk>1) && all(rem(Xk, p(p<Xk)));
end
end
end