Index exeeds matrix dimensions
Mostra commenti meno recenti
for n = 1:k
%HS or LE
prompt = 'Enter individual tasks memory location \n High Speed = 1 \n Low Energy = 0 \n';
fprintf('M(%d) = \n',n);
M(n) = input (prompt);
if M(n) == 1
h = h+1;
prompt = 'Enter the hiding value of high speed tasks \n';
fprintf('V(%d) = \n',h);
Vh(h) = input (prompt);
hst(h) = n;
else
l = l+1;
prompt = 'Enter the hiding value of low speed tasks \n';
fprintf('V(%d) = \n',l);
Vl(l) = input (prompt);
let(l) = n;
end
end
for n = 1:k
if M(n) == 1
a = 'High speed';
fprintf('M(%d) = %s \n',n,a);
else
b = 'Low Energy';
fprintf('M(%d) = %s \n',n,b);
end
end
fprintf('Number of High speed tasks = %d \n',h);
fprintf('Number of Low Energy tasks = %d \n',l);
for n = hst(1):hst(h)
fprintf('Vh(%d) = %d \n',n,Vh(n));
end
for n = let(1):let(l)
fprintf('Vl(%d) = %d \n',n,Vl(n));
end
Output
M(1) = High speed
M(2) = High speed
M(3) = High speed
M(4) = High speed
M(5) = Low Energy
M(6) = Low Energy
M(7) = Low Energy
Number of High speed tasks = 4
Number of Low Energy tasks = 3
Vh(1) = -100
Vh(2) = -1
Vh(3) = -2
Vh(4) = -3
Index exceeds matrix dimensions.
3 Commenti
KSSV
il 17 Feb 2017
We cannot help you unless variables inside are known to us..
hariharan ilango
il 17 Feb 2017
hariharan ilango
il 17 Feb 2017
Risposta accettata
Più risposte (0)
Categorie
Scopri di più su Matrices and Arrays in Centro assistenza e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!