2020 牛客多校 第六场 E-Easy Construction（思维）

题目链接: E-Easy Construction

Description

题意：给出n, k, 求是否存在连续任意个数的和模n为k，并输出序列

Input

• The first line contains two integers n, k(1 <= n <= 5000, 0 <= k < n).

Output

Print n integers, the answer permutation in one line if such permutation exists, or print “-1” in one line if no solution exists.

Sample Input

2 1

Sample Output

1 2

Method

• 显然要满足n个连续的数之和对n取模为K，则要保证 (n*(n+1)/2)%n == k， 而当满足这个条件之后，对于连续1~n，是一定存在 连续i个数的和模n为k的（可以动手试验一下）；
• 得到这个规律之后，再对n的奇偶进行分析即可；
  • 对于偶数： { n/2, n, 1, n-1, 2, n-2, … }
  • 对于奇数： { n, 1, n-1, 2, n-2, … }

Code

详见注释

#include 

using namespace std;
#pragma GCC optimize(2)
#define ios std::ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);
#define rtxt freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#define ll long long
const int Max = 1e6+3;
const int mod = 1e9+7;

int main()
{
	ios
	
	int n, k;
	cin >> n >> k;
	if(k != (n*(n+1)/2)%n) cout << -1 << endl;
	else {
		if(n&1) {
			for(int i=1; i<=n/2; i++)
				cout << i << " " << n-i << " ";
			cout << n << endl;
		}
		else {
			for(int i=1; i<n/2; i++)
				cout << i << " " << n-i << " ";
				cout << n/2 << " " << n << endl;
		}
	}
	return 0;
}

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蒟蒻一只，欢迎指正