help solving string question
显示 更早的评论
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
采纳的回答
Star Strider
2014-5-7
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 个评论
Joseph Pauwels
2014-5-7
Star Strider
2014-5-7
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
2014-5-8
fix, log10, rand, and rem are functions. The way i read the question is that no function is allowed.
更多回答（4 个）
Carlos
2014-5-7
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
2014-5-7
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 个评论
Wesley Ooms
2014-5-7
ok, i see i used 2 functions (floor and numel) you can get around the numel by placing the part of the for loop in the while loop, but i don't kno how to get around floor. I think Carlos answer is not what your teacher meant.
Wesley Ooms
2014-5-7
you can get around floor by using this:
clear a b
a=132385
i=1
c=0
d=1
e=0
while a
c=-1:(a/10)
a1=c(end)
b(i)=a-a1*10
a=a1
i=i+1
end
for c=0:i-2
e=e+d*b(end-c)
c=c+1
d=d*10;
end
Joseph Pauwels
2014-5-7
Wesley Ooms
2014-5-8
编辑：Wesley Ooms
2014-5-8
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
2014-5-8
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
2014-5-7
编辑：Sagar Damle
2014-5-8
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
Joshua Amuga
2016-11-2
0 个投票
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
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