Need to create a for loop for this problem

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Md. Atiqur Rahman
Md. Atiqur Rahman 2022-8-23
回答： KSSV 2022-8-23
clear all;
clc;
Key = randi([0,1],[1,2048]);
%Key(randperm(size(Key,2)))
Plaintext = randi([0,1],[1,256])
%Dividing PLaintext into Left and Right part
L = Plaintext(1:128);
R = Plaintext(129:256);
%Round one
F1 = xor(R,Key(1:128));
C1 = xor(F1,L);
%Round two
F2 = xor(C1,Key(129:256));
C2 = xor(F2,R);
%Round three
F3 = xor(C2,Key(257:384));
C3 = xor(F3,C1);
%Round four
F4 = xor(C3,Key(385:512));
C4 = xor(F4,C2);
%Round five
F5 = xor(C4,Key(513:640));
C5 = xor(F5,C3);
%Round six
F6 = xor(C5,Key(641:768));
C6 = xor(F6,C4);
%Round seven
F7 = xor(C6,Key(769:896));
C7 = xor(F7,C5);
%Round eight
F8 = xor(C7,Key(897:1024));
C8 = xor(F8,C6);
%Round nine
F9 = xor(C8,Key(1025:1152));
C9 = xor(F9,C7);
%Round ten
F10 = xor(C9,Key(1153:1280));
C10 = xor(F10,C8);
%Round eleven
F11 = xor(C10,Key(1281:1408));
C11 = xor(F11,C9);
%Round twelve
F12 = xor(C11,Key(1409:1536));
C12 = xor(F12,C10);
%Round thirteen
F13 = xor(C12,Key(1537:1664));
C13 = xor(F13,C11);
%Round fourteen
F14 = xor(C13,Key(1665:1792));
C14 = xor(F14,C12);
%Round fifteen
F15 = xor(C14,Key(1793:1920));
C15 = xor(F15,C13);
%Round sixteen
F16 = xor(C15,Key(1921:2048));
C16 = xor(F16,C14);
%Output
Ciphertext = [C15,C16]

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回答（1 个）

KSSV
KSSV 2022-8-23
Ran in:
idx = (0:128:2048)' ;
I = [idx(1:end-1)+1 idx(2:end)] ;
for i = 1:size(I,1)
I(i,:)
end
ans = 1×2
1 128
ans = 1×2
129 256
ans = 1×2
257 384
ans = 1×2
385 512
ans = 1×2
513 640
ans = 1×2
641 768
ans = 1×2
769 896
ans = 1×2
897 1024
ans = 1×2
1025 1152
ans = 1×2
1153 1280
ans = 1×2
1281 1408
ans = 1×2
1409 1536
ans = 1×2
1537 1664
ans = 1×2
1665 1792
ans = 1×2
1793 1920
ans = 1×2
1921 2048

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