s*3
2 楼
best buy??
bless
bless
j*y
3 楼
Bless! take it easy.
c*d
4 楼
bless
B*a
8 楼
bless
l*e
9 楼
updates:
老印面试:
1.why bloomberg, why software engineering
2. two to find square root for doubles instead of int
3, maximum sum of a subset from a array, two cases: 1) continuous sequence 2
)not continuous
4. find a 6 digits: first 3 digits sum= last 3 digits sum
老印面试:
1.why bloomberg, why software engineering
2. two to find square root for doubles instead of int
3, maximum sum of a subset from a array, two cases: 1) continuous sequence 2
)not continuous
4. find a 6 digits: first 3 digits sum= last 3 digits sum
b*g
10 楼
Newton ?
2: )not continuous
1. maxsofar maxendinghere
2. all positive
dont'know what it means...
【在 l****e 的大作中提到】
: updates:
: 老印面试:
: 1.why bloomberg, why software engineering
: 2. two to find square root for doubles instead of int
: 3, maximum sum of a subset from a array, two cases: 1) continuous sequence 2
: )not continuous
: 4. find a 6 digits: first 3 digits sum= last 3 digits sum
y*e
11 楼
bless
d*o
12 楼
bless,典型的bb面试
d*u
13 楼
bless
q*h
15 楼
big Bless
c*o
16 楼
maximum subset in an array有时间O(n)的解么?
j*9
17 楼
take it easy. bless.
p*p
18 楼
bless you!
z*8
20 楼
第三题not continuous是怎么回事? 把所有的正数挑出来不就行了?
第四题能否详细点?
2
【在 l****e 的大作中提到】
: updates:
: 老印面试:
: 1.why bloomberg, why software engineering
: 2. two to find square root for doubles instead of int
: 3, maximum sum of a subset from a array, two cases: 1) continuous sequence 2
: )not continuous
: 4. find a 6 digits: first 3 digits sum= last 3 digits sum
第四题能否详细点?
2
【在 l****e 的大作中提到】
: updates:
: 老印面试:
: 1.why bloomberg, why software engineering
: 2. two to find square root for doubles instead of int
: 3, maximum sum of a subset from a array, two cases: 1) continuous sequence 2
: )not continuous
: 4. find a 6 digits: first 3 digits sum= last 3 digits sum
h*0
22 楼
bless
b*e
23 楼
bless
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