「ARC106F」Figures

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给定一个 $n$ 个点的无向图，第 $i$ 个点有 $d_i$ 个接口，每条边连接两个不同的点的接口，每个接口只能被一条边连接。问这张图的生成树个数，对 $998244353$ 取模。

$2\le n\le 2\times10^5,1\le d_i<998244353$。

gugugu…

$code$

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#include <iostream>
#include <cstdlib>
#include <cstdio>
#define ll long long
using namespace std;
inline int read()
{
    int f = 1, x = 0;
    char ch;

    do{
        ch = getchar();
        if (ch == '-')
            f = -1;
    }while(ch < '0' || ch > '9');
    do{
        x = x * 10 + ch - '0';
        ch = getchar();
    }while(ch >= '0' && ch <= '9');
    return f * x;
}
const int N = 2e5;
const int mod = 998244353;

int n;
int ans = 1;
ll ssum;

inline int pow(int x, int y)
{
    int sum = 1;

    while (y) {
        if (y & 1)
            sum = 1ll * sum * x % mod;
        x = 1ll * x * x % mod;
        y >>= 1;
    }
    return sum;
}
inline int C(int x, ll y)
{
    if (x > y)
        return 0;
    int sum = 1, summ = 1;

    for (ll i = y - x + 1; i <= y; i++)
        sum = 1ll * (i % mod) * sum % mod;
    for (int i = 1; i <= x; i++)
        summ = 1ll * i * summ % mod;
    return 1ll * sum * pow(summ, mod - 2) % mod;
}
int main()
{
    n = read();
    for (int i = 1; i <= n; i++) {
        int x = read();
        if (i <= n - 2)
            ans = 1ll * ans * i % mod * x % mod;
        else
            ans = 1ll * ans * x % mod;
        ssum += x - 1;
    }

    ans = 1ll * ans * C(n - 2, ssum) % mod;

    printf("%d\n", ans);
    return 0;
}