How to count and reduce values in matrix

Matrix input:

input = [
1  3  50  60
1  1  40  60
1  4  30  60
2  3  40  50
2  4  30  50
2  1  50  50
2  9  10  50
3  2  20  0
3  9  30  0
3  5  40  0
4  2  50  -20
4  2  60  -20
4  1  10  -20
4  1  25  -20
4  8  80  -20
];

% there are three 1, so 60/3, there are four 2, so 50/4, there are three 3, so 0/3 and there are five 4, so -20/5. 
% 50-60/3 = 30
% 40-60/3 = 20
% 30-60/3 = 10

Based on the similar arrays in the first column, I want to find for example how many 1 are there. Then divide the forth column by that (for example 1 is 3 here) and then reduce the amount from the third column. The first output should be:

output1 = [
1  3  30
1  1  20
1  4  10
2  3  27.5
2  4  17.5
2  1  37.5
2  9  -2.5
3  2  20
3  9  30
3  5  40
4  2  54
4  2  64
4  1  14
4  1  29
4  8  84
];

For the second output, I want the same process, but instead of counting all same numbers in the first column, this time just look at the second column and count if there is 3 and 4 (for every unique number in the first column).

output2 = [
1  3  20
1  1  40
1  4  0
2  3  15
2  4  5
2  1  50
2  9  10
3  2  20
3  9  30
3  5  40
4  2  50
4  2  60
4  1  10
4  1  25
4  8  80
];

% for "1", there are two 3&4, so 60/2
% 50-60/2 = 20
% 30-60/2 = 0

[Rik]

What have you tried? Have you looked at functions like unique? As someone said elsewhere: the community spends hours to help you, or seconds to ignore you. Show that you have put in some effort, and people are more likely to put in more effort helping.

[Tiffan]

This my effort so far:

KK = unique(input(:,1));
n  = histc(input(:,1),KK);
mm = n(input(:,1));
Dif = input(:,4)./mm;
output1 = [input(:,1) input(:,2) (input(:,3)-Dif)];

Return:

1  3  30
1  1  20
1  4  10
2  3  27.5000000000000
2  4  17.5000000000000
2  1  37.5000000000000
2  9  -2.50000000000000
3  2  20
3  9  30
3  5  40
4  2  54
4  2  64
4  1  14
4  1  29
4  8  84

But, I don't know how to count 3 & 4 for the second part in my question.

Answer 1

[Andrei Bobrov]

a = input;

ii = accumarray(a(:,1),1);
out1 = [a(:,1:2),a(:,3) - a(:,end)./ii(a(:,1))];

t = ismember(a(:,2),3:4);
i1 = accumarray(a(:,1),t);
out2 = a(:,1:3);
out2(t,3) = out2(t,3) - a(t,end)./i1(a(t,1));

[Tiffan]

Thank you @Andrei Bobrov