Codility-task 1-Tape Equilibrium
来源:互联网 发布:淘宝喵喵 官换机 编辑:程序博客网 时间:2024/06/12 09:58
Codility是个great的OJ,用起来非常爽,所以我要来CSDN上安利一下!
网址是 https://codility.com/programmers/challenges/
Task Description
A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3
We can split this tape in four places:
- P = 1, difference = |3 − 10| = 7
- P = 2, difference = |4 − 9| = 5
- P = 3, difference = |6 − 7| = 1
- P = 4, difference = |10 − 3| = 7
Write a function:
int solution(vector<int> &A);
that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3
the function should return 1, as explained above.
Assume that:
- N is an integer within the range [2..100,000];
- each element of array A is an integer within the range [−1,000..1,000].
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
题目的等级是easy,无需多言,下边是我的解答。
//Codility //Zhang Wenjian 07-06-2015int solution(vector<int> &A) { // write your code in C++11 long long sum; long long th; for(vector<int>::size_type it=0;it<A.size();it++) sum+=A[it]; th=abs(sum-2*A[0]); if(A.size()==2) return abs(A[0]-A[1]); for(vector<int>::size_type it=0;it<A.size()-1;it++) { sum-=2*A[it]; if(abs(sum)<th) th=abs(sum); } return th;}
- Codility-task 1-Tape Equilibrium
- [codility]Tape-Equilibrium
- [codility]Task description
- [codility]Task description
- [codility]Task description
- codility
- codility
- UVa10878 - Decode the tape-字符串-难度1
- codility上的练习 (1)
- Task 1
- HP StorageWorks 1/8 G2 Tape Autoloader - privilege escalation, DOS
- Ticker Tape
- TGT + TAPE
- UVa 12166 - Equilibrium Mobile
- uva12166 - Equilibrium Mobile
- UVa 12166 - Equilibrium Mobile
- 12166 - Equilibrium Mobile
- UVa 12166 Equilibrium Mobile
- 87.计算n!
- Leetcode|Divide Two Integers
- IOS沙盒详解
- HBase学习总结(3):HBase的数据模型及工作机制
- linux下mysql安装
- Codility-task 1-Tape Equilibrium
- LR vs LWLR
- nodejs之SVG转图片下载方案
- 输入一行字符,统计有多少个单词
- HDU 3681 Prison Break(bfs+二分+状压DP)
- 1005. Spell It Right (20)
- 黑马程序员—网络编程
- 线程级并发
- 代理练习 两个UIView界面之间的传值--Delegate