POI2008PER-Permutation

POI #Year2008 #数学 #康拓展开 #逆元

如果是一个排列，根据康拓展开，答案为

\[\sum\limits_{i=1}^nsum_{i}\times (n-i)! \]

其中

\[sum_{i}=\sum\limits_{j=i+1}^n[a_i>a_j] \]

那么再加入了重复的数字之后，答案变为

\[ \sum\limits_{i=1}^n \frac{sum_i\times (n-i)!}{\prod\limits_{j\in s_i} cnt_j} \]

其中 \(s_i=\{a_j|j\in [i,n]\}\)

考虑怎么再非质数模数下计算这个东西，可以先对模数质因子分解，然后对于每个质因子统计个数，对于非质因子，即存在逆元可以直接算

// Author: xiaruize
const int N = 3e5 + 10;

int n, m;
int a[N];
int sum[N];
int pr[N], tot, cnt[N];
int ph;
int inv[N];

void init(int x)
{
	rep(i, 2, x)
	{
		if (i * i > x)
			break;
		if (x % i == 0)
		{
			pr[++tot] = i;
			ph = ph / i * (i - 1);
			while (x % i == 0)
				x /= i;
		}
	}
	if (x != 1)
	{
		pr[++tot] = x;
		ph = ph / x * (x - 1);
	}
}

int qpow(int a, int b)
{
	int res = 1;
	while (b)
	{
		if (b & 1)
			res = res * a % m;
		a = a * a % m;
		b >>= 1;
	}
	return res;
}

int cur = 1, res = 0;

void mul(int x, int &t)
{
	rep(i, 1, tot)
	{
		while (x % pr[i] == 0)
		{
			x /= pr[i];
			cnt[i]++;
		}
	}
	t = t * x % m;
}

void divi(int x, int &t)
{
	rep(i, 1, tot)
	{
		while (x % pr[i] == 0)
		{
			x /= pr[i];
			cnt[i]--;
		}
	}
	t = t * qpow(x, ph - 1) % m;
}

struct BIT
{
	int tr[N];

	void add(int x, int v)
	{
		while (x <= 3e5)
		{
			tr[x] += v;
			x += x & -x;
		}
	}

	int sum(int x)
	{
		int res = 0;
		while (x)
		{
			res += tr[x];
			x -= x & -x;
		}
		return res;
	}
} tr;

int get()
{
	int res = cur % m;
	rep(i, 1, tot) res = res * qpow(pr[i], cnt[i]) % m;
	return res;
}

void solve()
{
	cin >> n >> m;
	ph = m;
	init(m);
	rep(i, 1, n) cin >> a[i];
	sum[a[n]]++;
	tr.add(a[n], 1);
	per(i, n - 1, 1)
	{
		int w = tr.sum(a[i] - 1);
		debug(i);
		mul(n - i, cur);
		debug(i);
		tr.add(a[i], 1);
		sum[a[i]]++;
		divi(sum[a[i]], cur);
		if (!w)
			continue;
		mul(w, cur);
		res = (res + get()) % m;
		divi(w, cur);
	}
	cout << (res + 1) % m << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed
main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:"
		 << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB"
		 << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms"
		 << endl;
#endif
	return 0;
}