i=1, 2, 3,,, However, when (a-i)> 0 is calculated.
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haeyeon JI
am 11 Jul. 2020
Beantwortet: madhan ravi
am 11 Jul. 2020
i=1, 2, 3,,,
However, when (a-i)> 0 is calculated.
N, a is any positive integer,
station(2+i)=(a-(a-i))*N
I am trying to make a do statement, but I get an error
I am wondering how I should write the code to run it.
for example.
if a=4
i=1 station(3)=(4-(4-1))*N
i=2 station(4)=(4-(4-2))*N
i=3 station(5)=(4-(4-3))*N
i=4 station(6)=(4-(4-4))*N %->This case is not considered. Because (a-i) <= 0 (when (a-i)> 0 is calculated.)
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madhan ravi
am 11 Jul. 2020
clear station
a = 4
ii = 1:a-1;
N = 2; % any nunber
station(ii+2) = (4-(4-ii))*N
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Weitere Antworten (2)
Walter Roberson
am 11 Jul. 2020
Bearbeitet: Walter Roberson
am 11 Jul. 2020
for i = 1 : a - 1
station(i+2) = (4-(4-i))*N;
end
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John D'Errico
am 11 Jul. 2020
Bearbeitet: John D'Errico
am 11 Jul. 2020
Or, with no loop at all, and no test required. Just
i = 1:a-1;
station(i+2) = (4-(4-i))*N;
This will define the entire vector at once, leaving station(1) and station(2) as zero. But then you never said what they were anyway.
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