boy it is a good thing this was not an English class. that spelling and miss used work are horrible. I am glad yall knew what I meant
help solving string question
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Joseph Pauwels
am 7 Mai 2014
Beantwortet: Joshua Amuga
am 2 Nov. 2016
Durring my last semester, we were given a bonus quiz to write a function the received and numberical input and produced the revise as an output.
example a=123
b=321
we were not allowed to use any string variable or function. Any ideas now that the semester is over Id like to know what you think.
My original thought was to divide the input but the got harder with larger number as the function should receive any number
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Star Strider
am 7 Mai 2014
This works, but obviously only for integers:
a = 123;
% a = fix(rand*1E6) % Test Integer
La = fix(log10(a));
x = a;
for k1 = La:-1:0
d(k1+1) = fix(x/(10^k1));
x = rem(x,10^k1);
end
v10 = 10.^(La:-1:0)';
Flipped_a = d*v10
The Flipped_a variable is the result. I tested it on other random integers as well.
3 Kommentare
Star Strider
am 7 Mai 2014
The fix and rem functions are not string functions.
- The fix function rounds toward zero.
- The rem function (similar to mod) returns the remainder after division.
Wesley Ooms
am 8 Mai 2014
fix, log10, rand, and rem are functions. The way i read the question is that no function is allowed.
Weitere Antworten (4)
Carlos
am 7 Mai 2014
a=[1 2 3 4 5];
>> b=zeros(1,length(a));
count=1;
l=length(a);
while(count<=l)
b(count)=a(l+1-count);
count=count+1;
end
Wesley Ooms
am 7 Mai 2014
totally not optimized, but this will do the trick:
clear a b
a=32385
i=1
while floor(a/10)
b(i)=a-floor(a/10)*10
i=i+1
a =floor(a/10)
end
b(i)=a
d=1
e=0
for c=0:numel(b)-1
e=e+d*b(end-c)
d=d*10
end
5 Kommentare
Wesley Ooms
am 8 Mai 2014
Bearbeitet: Wesley Ooms
am 8 Mai 2014
that is because you
do c=-1:(a/1); instead of c=-1:(a/10);
the following must work: (but i admit it is not optimized for speed)
function b=swapnumber(a)
b = 0;
while a
c = -1:a/10;
c = c(end);
b = (b+a-c*10)*10;
a = c;
end
b = b/10;
Wesley Ooms
am 8 Mai 2014
the following code also works but is faster for large numbers since it predetermines the size of the number
function b=swapnumber(a)
b=0;d=0;while a>1;a=.1*a;d=d+1;end
for i=0:d;c=0:a*10;c=c(end);b=b+c*10^i;a=a*10-c;end
Sagar Damle
am 7 Mai 2014
Bearbeitet: Sagar Damle
am 8 Mai 2014
I think the code which I am going to put here is the standard code to reverse a number.(This code is used in C language,of course,according to its own syntax!)Also,it is easy to understand.Remember this code,I think it is very helpful!
a = 126986;
b = a; % Save value of "a" in new variable "b".
reverse = 0;
while b > 0 % OR while b ~= 0 (Both 'while' statements are same.)
r = rem(b,10);
reverse = reverse * 10 + r;
b = floor(b/10);
end
a
reverse
0 Kommentare
Joshua Amuga
am 2 Nov. 2016
IVP :=ode({y''(x)+4*y'(x)+3*y,y(0)=3,y'(0)=4},y(x)); Undefined function or variable 'IVP'. this is what i got
0 Kommentare
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