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Twitter Online Assessment (HackerRank Challenge)

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pro 发表于 2014-11-5 13:21:44 | 显示全部楼层 |阅读模式

2014(10-12月) 码农类 硕士 全职@Twitter - 网上海投 - 在线笔试 |Pass

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1. Two Operations (Programming).鏈枃鍘熷垱鑷1point3acres璁哄潧
You are given only two operations, ADD_1 and MULTIPLY_2. You start from 0 and using the two operations reach a number N. Find the least number of operations needed to do this.
Input Format:
Given T, T lines follow. Each line contains
N - Number to be formed
Output Format
Print the minimum number of operations needed in each line corresponding to each test case.
Constraints
T <= 10000. From 1point 3acres bbs
N <= 1016
Sample Input
2. 1point3acres.com/bbs
5
3
Sample Output. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
4
3
Explanation
Case1: To reach 5 from 0. We do ADD1 then Multiply2 two times and then add 1 again , total 4 steps
Case2: To reach 3 from 0. We can either do Add 1 3 times or Add 1 , multiply 2 and again add 1. visit 1point3acres.com for more.


2. Rational Sum (Programming)
In mathematics, a rational number is any number that can be expressed in the form of a fraction p/q , where p & q are two integers, and the denominator q is not equal to zero. Hence, all integers are rational numbers  where denominator, in the most reduced form, is equal to 1.
You are given a list of N rational number, {a1/b1, a2/b2, ..., aN/bN}. Print the sum ( = a1/b1 + a2/b2 + ... + aN/bN = num/den) in the most reduced form.
Input
The first line of input contains an integer, N, the number of rational numbers.  N lines follow. ithline contains two space separated integers, ai bi, where aiis the numerator and bi is the denominator for the ith rational number.
Output. more info on 1point3acres.com
You have to print two space separated integers, num den, where num and den are numerator and denominator of the sum respectively.
Constraints.1point3acres缃
  • 1 <= N <= 15
  • 1 <= ai <= 10
  • 1 <= bi <= 10
Notes.1point3acres缃
  • Make sure the sum displayed as output is in the most reduced form.
  • If sum is an integer, you have to print 1 as denominator.
Sample Input. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
4
4 2. visit 1point3acres.com for more.
2 4
2 4
鏉ユ簮涓浜.涓夊垎鍦拌鍧. 2 3-google 1point3acres
Sample Output
11 3. From 1point 3acres bbs

Explanation
Sum is 4/2 + 2/4 + 2/4 + 2/3 = (24 + 6 + 6 + 8)/12 = 44/12 = 11/3. So you have to print "11 3", which is the most reduced form.

--------------------

有一点比较奇怪的就是,听说Twitter应该是做完笔试再通知phone interview的,但是我笔试都还没做phone interview的邮件就来了_(:з」∠)_
.1point3acres缃
(ps. HackerRank就是好用!) 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
(ps. 发帖的时候选公司那里居然没有Twitter_(:з」∠)_)
NdrZmansN 发表于 2015-1-23 06:36:54 | 显示全部楼层
关注一亩三分地公众号:
Warald_一亩三分地
谢谢分享.
楼主方便分享一下第二题的解法么? 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
谢谢,
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haiken 发表于 2015-1-24 02:00:44 | 显示全部楼层
关注一亩三分地微博:
Warald
NdrZmansN 发表于 2015-1-23 06:36
谢谢分享.
楼主方便分享一下第二题的解法么?
谢谢,
-google 1point3acres
主要是要写个最大公约数的func, 我想.
.鐣欏璁哄潧-涓浜-涓夊垎鍦
写个了C++的版本, 不是很简洁, 最大公约数也可以用C里自带的__gcd(). From 1point 3acres bbs

using Pair = pair<int, int>;

int _gcd(int a, int b){
        return b==0 ? a : _gcd(b, a%b);
}

Pair rationalSum(vector<Pair> &R){
        int gcd = _gcd(R[0].first, R[0].second);
        Pair res = {Pair(R[0].first/gcd, R[0].second/gcd)};
        for(int i=1; i<R.size(); ++i){.鐣欏璁哄潧-涓浜-涓夊垎鍦
                gcd = _gcd(R.first, R.second);
                R.first = R.first/gcd;
                R.second = R.second/gcd;
                int first = res.first * R.second + R.first  * res.second;
                int second = R.second * res.second;
                gcd = _gcd(first, second);
                res.first = first/gcd;
                res.second = second/gcd; 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
        }       
        return res;.鏈枃鍘熷垱鑷1point3acres璁哄潧
}
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douya 发表于 2015-2-2 04:46:02 | 显示全部楼层
谢谢分享,这个是online coding challenge?
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tyr034 发表于 2015-3-1 13:31:04 | 显示全部楼层
def rationalSum(lst):
    return functools.reduce(add,lst)
. From 1point 3acres bbs
#greatest common divider
def gcd(a,b):
    while a :
        a,b = b%a,a
    return b. from: 1point3acres.com/bbs

#lowest common multiplier
def lcm(a,b):
    return a*b//gcd(a,b)
       . 鍥磋鎴戜滑@1point 3 acres
def add(a,b):
   
    de = lcm(a[1],b[1])
    nu = a[0]*de//a[1] + b[0]*de//b[1]
    return(nu,de)
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manjusaka077 发表于 2015-3-11 04:22:21 | 显示全部楼层
第二题
Constraints
1 <= N <= 15
1 <= ai <= 10
1 <= bi <= 10.. Waral 鍗氬鏈夋洿澶氭枃绔,
竟然这么小?那可不可以直接把10以内的质数列出来放一个数组里,然后在分母里面找都出现过哪几个质数乘一下就是最终的分母了……
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nibuxing 发表于 2015-3-13 21:37:08 | 显示全部楼层
请问公司在Hackerrank上的test是不是都是考算法啊
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liuliuion1 发表于 2015-3-29 03:31:44 | 显示全部楼层
请问第一题该怎么做呢?
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