D - Different Sums URAL - 2065 （规律）

最新推荐文章于 2021-05-28 21:35:32 发布

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最新推荐文章于 2021-05-28 21:35:32 发布

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博客围绕URAL - 2065题目展开，要求构造一个含n个整数的数组，不同整数数量不少于k，且非空子数组不同和的数量最少。给出了输入输出要求及示例，还提到解题思路是递增的正负正负可相互抵消。

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D - Different Sums

URAL - 2065

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.

思路:递增的正负正负就能相互抵消了

#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <string>
#include <functional>
#include <cstdio>
#include <cmath>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
using namespace std;
#define ll long long
#define F first
#define S second
#define p_b push_back
#define m_p make_pair
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int M = 1e7 + 7;
const int N = 1e6 + 7;
int main() {
	
	int n, k;
	scanf ("%d %d", &n, &k);
	printf ("0");
	int j = 1;
	for (int i = 1; i < k; ++i) {
		if (i&1) printf (" %d", j);
		else printf (" %d", -j);
		if (i%2 == 0) ++j;
	}
	for (int i = 1; i < (n - (k - 1)); ++i) {
		printf (" 0\n");
	}
	puts("");
	return 0;
}