nikkei2019_2_qual_e Non-triangular Triplets

nikkei2019_2_qual_e Non-triangular Triplets

https://atcoder.jp/contests/nikkei2019-2-qual/tasks/nikkei2019_2_qual_e

给出 \(N\) 和 \(K\) .

判断对于这 \(3N\) 个数 \(K,K+1,\cdots,K+3N-1\) .是否存在一种将它们分为 \(N\) 个三元组 \((a_i,b_i,c_i)\) 的方案,且满足每个数恰好出现一次,且

\[a_i + b_i \le c_i \]

判断是否存在方案,若不存在,输出 \(-1\) .否则输出一组方案

\(1 \le N \le 10^5\)

\(1 \le K \le 10^9\)

Tutorial

假设存在方案,那么满足

\[\sum (a_i+b_i) \le \sum c_i \\ \sum(a_i+b_i+c_i) \le 2\sum c_i \\ \sum_{i=K}^{K+3N-1} i \le 2\sum c_i \le 2\sum_{i=K+2N}^{K+3N-1} i \\ (K+K+3N-1)(3N) \le (K+2N+K+3N-1)(2N) \\ N \ge 2K-1 \]

那么当 \(N<2K-1\) 时则无解,否则考虑构造方案.

当 \(N\) 为奇数时,设 \(N=2L-1\) 则有 \(K \le L\) ,那么

\[(a_1,a_2,\cdots,a_N)=(K,K+2,K+4,\cdots,K+2L-2,K+1,K+3,\cdots,K+2L-3) \\ (b_1,b_2,\cdots,b_N)=(K+3L-2,K+3L-3,\cdots,K+2L-1,K+4L-3,K+4L-4,\cdots,K+3L-1) \\ (c_1,c_2,\cdots,c_N)=(K+4L-2,\cdots,K+6L-4) \]

主要思想为将 \((a,b)\) 以 \((x+y),(x+2,y-1),\cdots\) 的顺序安排以达到每次和增加 \(1\) 的效果.

\(N\) 为偶数时类似的构造即可.

Code

#include <cstdio>
#include <iostream>
#define debug(...) fprintf(stderr, __VA_ARGS)
using namespace std;
const int maxn = 1e5 + 50;
int N, K;
int a[maxn], b[maxn], c[maxn];
bool sol()
{
	if (N < (K << 1) - 1) return 0;
	if (N & 1)
	{
		int L = (N + 1) >> 1;
		int cnt = 0;
		for (int i = K; i <= K + 2 * L - 2; i += 2) 
			a[++cnt] = i;
		for (int i = K + 1; i <= K + 2 * L - 3; i += 2)
			a[++cnt] = i;
		cnt = 0;
		for (int i = K + 3 * L - 2; i >= K + 2 * L - 1; --i)
			b[++cnt] = i;
		for (int i = K + 4 * L - 3; i >= K + 3 * L - 1; --i)
			b[++cnt] = i;
		cnt = 0;
		for (int i = K + 4 * L - 2; i <= K + 6 * L - 4; ++i)
			c[++cnt] = i;
	}
	else
	{
		int L = N >> 1;
		int cnt = 0;
		for (int i = K; i <= K + 2 * L - 2; i += 2)
			a[++cnt] = i;
		for (int i = K + 1; i <= K + 2 * L - 1; i += 2)
			a[++cnt] = i;
		cnt = 0;
		for (int i = K + 3 * L - 1; i >= K + 2 * L; --i)
			b[++cnt] = i;
		for (int i = K + 4 * L - 1; i >= K + 3 * L; --i)
			b[++cnt] = i;
		cnt = 0;
		for (int i = K + 4 * L; i <= K + 6 * L; ++i)
			c[++cnt] = i;
	}
	for (int i = 1; i <= N; ++i)
		printf("%d %d %d\n", a[i], b[i], c[i]);
	return 1;
}
int main()
{
	scanf("%d%d", &N, &K);
	if (!sol())
		puts("-1");
	return 0;
}