I have a function that uses fix and rem functions but if the entered number is big it does not give the right number.

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Adnan Fino
Adnan Fino 2021-12-12
评论： Adnan Fino 2021-12-13
double d;
function denominationsCounter(d)
double d50;
double d20;
double d10;
double d5;
double d2;
double d1;
double d50p;
double d20p;
double d10;
double d5p;
double d2p;
double d1p;
d50 = fix(d/50);
d = rem(d,50);
d20 = fix(d/20);
d = rem(d,20);
d10 = fix(d/10);
d = rem(d,10);
d5 = fix(d/5);
d = rem(d,5);
d2 = fix(d/2);
d = rem(d,2);
d1 = fix(d/1);
d = rem(d,1);
d50p = fix(d/0.5);
d = rem(d,0.5);
d20p = fix(d/0.2);
d = rem(d,0.2);
d10p = fix(d/0.1);
d = rem(d,0.1);
d5p = fix(d/0.05);
d = rem(d,0.05);
d2p = fix(d/0.02);
d = rem(d,0.02);
d1p = fix(d/0.01);
d = rem(d,0.01);
disp("Number of Notes: 50:"+d50 +",20:"+ d20 +",10:"+d10+",5:"+d5)
disp("Number of Coins: 2:" +d2 + ",1:"+ d1 + ",50p:" +d50p + ",20p:" +d20p + ",10p:" +d10p + ",5p:" + d5p + ",2p:" +d2p+ ",1p:"+d1p)
disp(d)
This function will calculate the least number of denominations required for an entered amount
for example if I entered 9.33, it show this 10p:0,5p:0,2p:1,1p:1 which means that it will take 1 of 2p coins and 1 of 1p coins for the required amount. while if I entered 15.33 or 20.33 it will show 2p:1,1p:0. It will not add the 1p coin. If the reminder is 1 it is not showing 1p:1 it is showing as zero. Any ideas please to make it suitable for all numbers?
  1 个评论
Walter Roberson
Walter Roberson 2021-12-12
Note: MATLAB does not use declarations of variables like that.
double d50;
would be interpreted as
double('d50');
which would look up the codes used to encode 'd' '5' '0' and convert those uint16 to double precision, and then throw away the result because of the semi-colon.

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采纳的回答

Walter Roberson
Walter Roberson 2021-12-12
function denominationsCounter(D)
F = 100;
d = round(D * F);
f = round(50*F);
d50 = fix(d/f);
d = rem(d,f);
f = round(20*F);
d20 = fix(d/f);
d = rem(d,f);
f = round(10*F);
d10 = fix(d/f);
d = rem(d,f);
f = round(5*F);
d5 = fix(d/f);
d = rem(d,f);
f = round(2*F);
d2 = fix(d/f);
d = rem(d,f);
f = round(1*F);
d1 = fix(d/f);
d = rem(d,f);
f = round(0.5*F);
d50p = fix(d/f);
d = rem(d,f);
f = round(0.2*F);
d20p = fix(d/f);
d = rem(d,f);
f = round(0.1*F);
d10p = fix(d/f);
d = rem(d,f);
f = round(0.05*F);
d5p = fix(d/f);
d = rem(d,f);
f = round(0.02*F);
d2p = fix(d/f);
d = rem(d,f);
f = round(0.01*F);
d1p = fix(d/f);
d = rem(d,f);
disp("Number of Notes: 50:"+d50 +",20:"+ d20 +",10:"+d10+",5:"+d5)
disp("Number of Coins: 2:" +d2 + ",1:"+ d1 + ",50p:" +d50p + ",20p:" +d20p + ",10p:" +d10p + ",5p:" + d5p + ",2p:" +d2p+ ",1p:"+d1p)
disp(d/F)

更多回答（1 个）

Voss
Voss 2021-12-12
This problem is due to the nature of finite-precision floating-point numbers.
In this case, since you know (presumably) that any number input into the program will be a multiple of 0.01 (i.e., no fractional cents/pence allowed), a solution is to round d to 2 decimal places each time it's calculated, e.g.:
d = round(rem(d,50),2); % and so on every time d is updated
Aternatively, you can redefine d to be in units of cents/pence rather than dollars/pounds/euros/whatever and then just round to the nearest integer.
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