i=1, 2, 3,,, However, when (a-i)> 0 is calculated.

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haeyeon JI
haeyeon JI 2020 年 7 月 11 日
回答済み: madhan ravi 2020 年 7 月 11 日
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.)

採用された回答

madhan ravi
madhan ravi 2020 年 7 月 11 日
clear station
a = 4
ii = 1:a-1;
N = 2; % any nunber
station(ii+2) = (4-(4-ii))*N

その他の回答 (2 件)

Walter Roberson
Walter Roberson 2020 年 7 月 11 日
編集済み: Walter Roberson 2020 年 7 月 11 日
for i = 1 : a - 1
station(i+2) = (4-(4-i))*N;
end

John D'Errico
John D'Errico 2020 年 7 月 11 日
編集済み: John D'Errico 2020 年 7 月 11 日
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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