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iax0583
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Commit
9c5f51bd
authored
Oct 18, 2017
by
videid
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lab6 & 7
parent
fefe20b5
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2 changed files
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80 additions
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0 deletions
lab6/lab6.py
lab7/lab7.cpp
lab6/lab6.py
0 → 100644
View file @
9c5f51bd
##TASK1
i
=
int
j
=
int
x
=
0
y
=
0
som
=
[]
def
algorithm
()
:
print
(
"select n"
)
n
=
int
(
input
())
if
(
0
<=
n
<=
10
)
:
print
(
"select m"
)
m
=
int
(
input
())
if
(
0
<=
m
<=
10
):
A
=
[[
0
for
x
in
range
(
n
)]
for
y
in
range
(
m
)]
for
i
in
range
(
0
,
n
):
for
j
in
range
(
0
,
m
):
print
(
"enter element"
,
"["
,
i
,
"]"
,
"["
,
j
,
"]"
)
A
[
i
][
j
]
=
int
(
input
())
for
h
in
range
(
0
,
min
(
m
,
7
)):
som
.
append
(
0
)
for
g
in
range
(
0
,
n
):
som
[
h
]
=
som
[
h
]
+
A
[
g
][
h
]
return
A
,
som
## i don't really get the second question
##,if it is asked to return only the negative
##sums here is the code :
##if som[h] < 0 listofnegativesums.append(som[h]):
## return listofnegativesums
##maybe it is about returning a sum
##of only negative elemtents from the rows
##in which case here is the solution:
##for h in range (0,min(m,7)):
## som.append(0)
## for g in range (0 , n):
## if A[g][h] < 0 :
## som[h] = som[h] + A[g][h]
## return A,som
##TASK2
def
banknotes
():
print
(
"enter S"
)
S
=
int
(
input
())
print
(
"enter number of different Bij"
)
NbBij
=
int
(
input
())
B
=
[]
for
l
in
range
(
NbBij
):
print
(
"enter Bij size"
,
l
)
Bijsize
=
int
(
input
())
B
.
append
([
Bijsize
])
print
(
"enter number of Bij"
,
Bijsize
)
NbofBij
=
int
(
input
())
B
[
l
]
.
append
(
NbofBij
)
return
B
## the question 3 is a complex mathematical problem i have no idea
##how to resolve,the python implementation would be easier though
\ No newline at end of file
lab7/lab7.cpp
0 → 100644
View file @
9c5f51bd
#include <stdio.h>
#define ARRAY_SIZE 25
int
function
(
int
input
);
int
main
()
{
int
array
[
ARRAY_SIZE
]
=
{
0
};
for
(
int
i
=
0
;
i
<
ARRAY_SIZE
;
i
++
)
{
array
[
i
]
=
function
(
i
);
}
for
(
int
j
=
0
;
j
<
ARRAY_SIZE
;
j
++
)
{
int
b
=
array
[
j
];
printf
(
""
,
array
[
j
]);
}
return
0
;
}
int
function
(
int
input
)
{
return
(
input
*
(
input
-
1
));
}
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