AtCoder Regular Contest 128 E K Different Values

洛谷传送门

AtCoder 传送门

考虑判断有无解。把序列分成 \(c = \left\lceil\frac{len}{k}\right\rceil\) 段，则 \(\forall a_i \le c\) 且 \(\sum\limits_{i=1}^n [a_i = c] \le ((len - 1) \bmod k) + 1\)。

必要性显然。充分性可以考虑直接构造，不难证明。

考虑如何构造字典序最小。贪心，如果当前不得不填 \(a_i = c\) 的数了，那么找到最小的并且能填的填；否则就直接找最小然后填即可。

感觉挺神奇的贪心题，怎么想到的！

code

// Problem: E - K Different Values
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2021（AtCoder Regular Contest 128）
// URL: https://atcoder.jp/contests/arc128/tasks/arc128_e
// 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 long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;

int n, m, len, a[maxn], b[maxn], c[maxn];

void solve() {
	scanf("%d%d", &n, &m);
	mems(c, -0x3f);
	int cnt = 0;
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
		len += a[i];
	}
	for (int i = 1; i <= n; ++i) {
		if (a[i] > (len + m - 1) / m) {
			puts("-1");
			return;
		}
		cnt += (a[i] == (len + m - 1) / m);
	}
	if (cnt > (len - 1) % m + 1) {
		puts("-1");
		return;
	}
	for (int i = 1; i <= len; ++i) {
		cnt = 0;
		for (int j = 1; j <= n; ++j) {
			cnt += (a[j] == (len + m - i) / m);
		}
		if (cnt == (len - i) % m + 1) {
			for (int j = 1; j <= n; ++j) {
				if (c[j] + m <= i && a[j] == (len + m - i) / m) {
					printf("%d ", j);
					--a[j];
					c[j] = i;
					break;
				}
			}
		} else {
			for (int j = 1; j <= n; ++j) {
				if (c[j] + m <= i && a[j]) {
					printf("%d ", j);
					--a[j];
					c[j] = i;
					break;
				}
			}
		}
	}
}

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