Design an algorithm and write the program for a machine that must give the current amount of change from rs500 for any amount of change costing less than rs 500. The program must specify the number and all notes and coins as possible. Assume purc

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beepo pony 2016-3-17

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编辑： Stephen23 2016-3-17

采纳的回答： Stephen23

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Answer 1

Stephen23 2016-3-17

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编辑：Stephen23 2016-3-17

在 MATLAB Online 中打开

Here is a simple greedy algorithm, based on Geoff's answer:

function chg = change(num)
den = [200,100,50,20,10,5,2,1,0.5,0.2,0.1,0.05];
vec = int32(100*den);
lft = int32(100*num);
tmp = zeros(size(vec),'int32');
for v = 1:numel(vec)
    tmp(v) = idivide(lft, vec(v));
    lft = mod(lft, vec(v));
end
chg(:,2) = den;
chg(:,1) = double(tmp);
end

You should be able to adapt its output to suit your homework. You will have to change the denominations to suit your requirements, and also consider how to display the output (hint: use fprintf). Currently the functions simply returns a matrix showing how many of each denomination are used:

>> change(499)
ans =
            2          200
            0          100
            1           50
            2           20
            0           10
            1            5
            2            2
            0            1
            0          0.5
            0          0.2
            0          0.1
            0         0.05
>> change(99)
ans =
            0          200
            0          100
            1           50
            2           20
            0           10
            1            5
            2            2
            0            1
            0          0.5
            0          0.2
            0          0.1
            0         0.05
>> change(1.75)
ans =
            0          200
            0          100
            0           50
            0           20
            0           10
            0            5
            0            2
            1            1
            1          0.5
            1          0.2
            0          0.1
            1         0.05