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codeforces D. Ice Sculptures 题解

2024-07-02 12:53:17   227人阅读

The Berland University is preparing to celebrate the 256-th anniversary of its founding! A specially appointed Vice Rector for the celebration prepares to decorate the campus. In the center of the campus n ice sculptures were erected. The sculptures are arranged in a circle at equal distances from each other, so they form a regular n-gon. They are numbered in clockwise order with numbers from 1 to n.

The site of the University has already conducted a voting that estimated each sculpture‘s characteristic of ti — the degree of the sculpture‘s attractiveness. The values of ti can be positive, negative or zero.

When the university rector came to evaluate the work, he said that this might be not the perfect arrangement. He suggested to melt some of the sculptures so that:

  • the remaining sculptures form a regular polygon (the number of vertices should be between 3 and n),
  • the sum of the ti values of the remaining sculptures is maximized.

Help the Vice Rector to analyze the criticism — find the maximum value of ti sum which can be obtained in this way. It is allowed not to melt any sculptures at all. The sculptures can not be moved.

Input

The first input line contains an integer n (3?≤?n?≤?20000) — the initial number of sculptures. The second line contains a sequence of integers t1,?t2,?...,?tn, ti — the degree of the i-th sculpture‘s attractiveness (?-?1000?≤?ti?≤?1000). The numbers on the line are separated by spaces.

Output

Print the required maximum sum of the sculptures‘ attractiveness.

Sample test(s)
input
8
1 2 -3 4 -5 5 2 3
output
14

本题还是有一定难度的。

注意点：

1 要使用逆向思维，不是要计算melt掉多少sculptures， 而是要计算需要剩下多少个sculptures

2 所有情况都需要考虑到，最大间隔melt掉的sculptures是n/3个

3 要判断这样间隔melt掉sculptures之后，剩下的sculptures是否能组成多边形

要很仔细想清楚才能做出来。

It is never too careful to scrutinize a problem!

It is never too careful to design an algorithm!

#include <vector>
#include <iostream>
using namespace std;

void IceSculptures()
{
	unsigned n;
	cin>>n;
	vector<int> sculptures(n);
	for (unsigned i = 0; i < n; i++)
	{
		cin>>sculptures[i];
	}
	int curSum = 0, maxSum = (1<<31);
	for (unsigned span = 1; span <= n / 3; span++)
	{
		if (n % span) continue;
		for (unsigned b = 0; b < span; b++)
		{
			curSum = 0;
			for (unsigned i = b; i < n; i += span)
			{
				curSum += sculptures[i];
			}
			maxSum = curSum > maxSum? curSum : maxSum;
		}
	}
	cout<<maxSum;
}


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