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AtCoder Regular Contest 104 D Multiset Mean

发布时间 2023-04-15 13:16:39作者: zltzlt

洛谷传送门

AtCoder 传送门

很平凡的一道计数啊。

考虑将所有数都减去 \(x\)，那么就要求选的数和为 \(0\)。

正负分开考虑，\(0\) 可以任意选。需要多重背包求 \(f_{i,j}\) 表示选 \(1 \sim i\) 的数和为 \(j\) 的方案数。前缀和优化是平凡的。

code 
// Problem: D - Multiset Mean
// Contest: AtCoder - AtCoder Regular Contest 104
// URL: https://atcoder.jp/contests/arc104/tasks/arc104_d
// Memory Limit: 1024 MB
// Time Limit: 4000 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 = 110;
const int maxm = 1000100;

ll n, m, mod, f[maxn][maxm], g[maxn];

void solve() {
	scanf("%lld%lld%lld", &n, &m, &mod);
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		mems(g, 0);
		for (int j = 0; j <= i * (i + 1) / 2 * m; ++j) {
			g[j % i] = (g[j % i] + f[i - 1][j]) % mod;
			if (j - i * (m + 1) >= 0) {
				g[j % i] = (g[j % i] - f[i - 1][j - i * (m + 1)] + mod) % mod;
			}
			f[i][j] = g[j % i];
		}
	}
	for (int i = 1; i <= n; ++i) {
		ll ans = mod - 1;
		for (int j = 0; j <= n * (n + 1) / 2 * m; ++j) {
			ans = (ans + f[i - 1][j] * f[n - i][j] % mod * (m + 1) % mod) % mod;
		}
		printf("%lld\n", ans);
	}
}

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