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AtCoder Regular Contest 163 C Harmonic Mean

发布时间 2023-07-03 08:17:58作者: zltzlt

洛谷传送门

AtCoder 传送门

这题是不是想到裂项就会了但是我赛时没想到

下面复读 Ender32k 做法。

根据 \(\frac{1}{i(i + 1)} = \frac{1}{i} - \frac{1}{i + 1}\)，可以构造 \(\frac{1}{2}, \frac{1}{6}, ..., \frac{1}{(n - 1)n}, \frac{1}{n}\)。但是如果存在正整数 \(t\) 使得 \(t(t + 1) = n\) 那就寄了，因为会重。

于是我们把 \(\frac{1}{n}\) 和 \(\frac{1}{t(t + 1)}\) 合并变成 \(\frac{1}{\frac{t(t + 1)}{2}}\)，把 \(\frac{1}{(n - 1)n}\) 变成 \(\frac{1}{\frac{3(n - 1)n}{2}}, \frac{1}{3(n - 1)n}\) 即可。

但是 \(n = 12, 420\) 时 \(\frac{1}{\frac{t(t + 1)}{2}}\) 还可以跟前面的继续合并，因此要特判。

code 
// Problem: C - Harmonic Mean
// Contest: AtCoder - AtCoder Regular Contest 163
// URL: https://atcoder.jp/contests/arc163/tasks/arc163_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

void solve() {
	int n;
	scanf("%d", &n);
	if (n == 1) {
		puts("Yes\n1");
		return;
	}
	if (n == 2) {
		puts("No");
		return;
	}
	int t = -1;
	for (int i = 1; i <= n; ++i) {
		if (i * (i + 1) == n) {
			t = i;
			break;
		}
	}
	puts("Yes");
	if (t == -1) {
		for (int i = 1; i < n; ++i) {
			printf("%d ", i * (i + 1));
		}
		printf("%d\n", n);
		return;
	}
	if (n == 12) {
		printf("2 3 ");
		for (int i = 4; i <= n - 2; ++i) {
			printf("%d ", i * (i + 1));
		}
		puts("264 396 792");
		return;
	}
	if (n == 420) {
		for (int i = 1; i <= n - 2; ++i) {
			if (i * (i + 1) == 210) {
				continue;
			}
			if (i == t) {
				printf("105 ");
			} else {
				printf("%d ", i * (i + 1));
			}
		}
		int t = 175980;
		printf("%d %d %d\n", t * 2, t * 3, t * 6);
		return;
	}
	for (int i = 1; i <= n - 2; ++i) {
		if (i * (i + 1) == n) {
			printf("%d ", i * (i + 1) / 2);
			continue;
		}
		printf("%d ", i * (i + 1));
	}
	int k = (n - 1) * n;
	printf("%d %d\n", k * 3 / 2, k * 3);
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}