D - Different Sums

最新推荐文章于 2024-06-04 07:30:23 发布

原创最新推荐文章于 2024-06-04 07:30:23 发布 · 285 阅读

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本文探讨了如何构造一个整数数组，该数组包含至少k种不同的整数，并且所有非空子数组的和尽可能少地重复。文章提供了一个简单的算法实现思路及代码示例。

Alex is a very serious mathematician and he likes to solve serious problems. For example, this problem.

You are to construct an array of n integers in which the amount of different integers is not less than k. Among all such arrays you have to construct the one with the minimal amount of different sums on non-empty subarrays. In other words, lets compute the sums of every non-empty subarray and remove repeating sums. You have to minimize the number of remaining sums.

Input

In the only line of input there are two integers n, k (1 ≤ k ≤ n ≤ 500), separated by a space.

Output

Print n integers separated by spaces — the answer for the problem. All the numbers must not be greater than 10 ⁶ by absolute value. It is guaranteed that there exists an optimal solution with numbers up to 10 ⁵ by absolute value. If there are multiple possible answers, you may print any of them.

Example

input	output
1 1	-987654
3 2	0 7 0

Notes

Let’s take a closer look on the second sample. We will denote the sum on the segment [ l, r] by sum( l, r) (elements are numbered starting with 1). sum(1, 1) = sum(3, 3) = 0, sum(1, 2) = sum(1, 3) = sum(2, 2) = sum(2, 3) = 7, so there are only two different sums.

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n, k, i, qwq = 1, j = 0;
    scanf("%d %d", &n, &k);
    for(i = 0; i < n - k; i++)cout << "0 ";
    for(i = n - k; i < n; i++)
    {
        if(qwq)
        {
            cout << j++ << " ";
            qwq = 0;
        }
        else
        {
            cout << -j << " ";
            qwq = 1;
        }
    }
    return 0;
}