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sum times comparaison à un max
roronoa
Jul 17th, 2019
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m
=
readline
(
)
N
=
readline
(
)
t
=
[
3600
,
60
,
1
]
s
=
x
=>
{
k
=
0
;
return
x.
split
(
':'
)
.
reduce
(
(
a
,
b
)
=>
a
+
t
[
k
++
]
*
b
,
0
)
}
m
=
s
(
m
)
z
=
0
for
(
i
=
0
;
i
<
N
;
i
++
)
z
+=
s
(
readline
(
)
)
print
(
z
<=
m
?
true
:
false
)
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