Soluție HackerRank pentru Game Of Rotation, subdomeniul Algebra, în Java 8. Include cerința formatată, exemple, explicația pașilor și cod sursă.
- Problemă: Game Of Rotation
- Domeniu: Algebra
- Limbaj: Java 8
Challenge: Game Of Rotation
Subdomeniu: Algebra (algebra)
Scor cont: 30.0 / 30
Submission status: Accepted
Submission score: 1.0
Submission ID: 464733999
Limbaj: java8
Link challenge: https://www.hackerrank.com/challenges/game-of-rotation/problem
Cerință
Mark is an undergraduate student and he is interested in rotation. A conveyor belt competition is going on in the town which Mark wants to win. In the competition, there's A conveyor belt which can be represented as a strip of _1xN_ blocks. Each block has a number written on it. The belt keeps rotating in such a way that after each rotation, each block is shifted to left of it and the first block goes to last position.
There is a switch near the conveyer belt which can stop the belt. Each participant would be given a single chance to stop the belt and his *PMEAN* would be calculated.
*PMEAN* is calculated using the sequence which is there on the belt when it stops. The participant having highest *PMEAN* is the winner. There can be multiple winners.
Mark wants to be among the winners. What *PMEAN* he should try to get which guarantees him to be the winner.
PMEAN = ∑_i = 1^n i × a[i]
where a represents the configuration of conveyor belt when it is stopped. Indexing starts from 1.
Input Format
First line contains N denoting the number of elements on the belt.
Second line contains N space separated integers.
Output Format
Output the required *PMEAN*
Constraints
1 ≤ N ≤ 10<sup>6</sup>
-10<sup>9</sup> ≤ each number ≤ 10<sup>9</sup>
For any rotation, *PMEAN* will always lie within the range of 64-bit signed integer.
Sample Input
3
20 30 10
Sample Output
140
Explanation
Number on top can be written in these manners.
Initial numbers on belt, 20 30 10 *PMEAN* = 1x20 + 2x30 + 3x10 = 110
After first rotation, 30 10 20 *PMEAN* = 1x30 + 2x10 + 3x20 = 110
After second rotation, 10 20 30 *PMEAN* = 1x10 + 2x20 + 3x30 = 140
So maximum possible value will be 140.
Cod sursă
//https://www.hackerrank.com/challenges/game-of-rotation
import java.io.*;
public class Solution{
public static void main(String[] args) throws IOException{
//Input
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(br.readLine());
int[] A = new int[N];
N = 0;
for(String s : br.readLine().split(" ")){
A[N++] = Integer.parseInt(s);
}
//Find sum
long sum = 0;
for(int v : A){
sum += v;
}
//Find cur pmean
long curPmean = 0;
for(int i = 0; i < N; ++i){
curPmean += ((long)A[i])*(i+1);
}
//Cycle through
//Keep track of max pmean
long maxPmean = curPmean;
for(int i = 1; i < N; ++i){
curPmean = curPmean - sum + ((long)A[i-1])*N;
if (curPmean > maxPmean){
maxPmean = curPmean;
}
}
//Output
System.out.print(maxPmean);
}
}