Description

一个有n个结点的树，设它的结点分别为v1, v2, …, vn，已知第i个结点vi的度数为di，问满足这样的条件的不同的树有多少棵。给定n，d1, d2, …, dn，编程需要输出满足d(vi)=di的树的个数。

Solution

答案为$\prod \binom{n}{d_1-1}\binom{n-d_1+1}{d_2}\dots$
由于答案比较大但n只有150，我们可以把$151^7$作为模数，用欧拉定理处理逆元即可。

Code

/************************************************
 * Au: Hany01
 * Date: Jul 10th, 2018
 * Prob: BZOJ1211 树的计数
 * Email: [email protected]
 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 155;

int n, pi, res;
LL Mod, xt, fac[maxn], ifac[maxn], Ans = 1;

inline LL mult(LL a, LL b) {
    LL Ans = 0;
    for ( ; b; b >>= 1, (a = a + a) %= Mod) if (b & 1) (Ans += a) %= Mod;
    return Ans;
}

inline LL Pow(LL a, LL b) {
    LL Ans = 1;
    for ( ; b; b >>= 1, a = mult(a, a)) if (b & 1) Ans = mult(Ans, a);
    return Ans;
}

inline LL C(int n, int m) { if (n < m) return 0; return mult(fac[n], mult(ifac[m], ifac[n - m])); }

int main()
{
#ifdef hany01
    File("bzoj1211");
#endif

    n = read();

    if (n == 1) {
        if (!read()) puts("1"); else puts("0");
        return 0;
    }

    Mod = 151ll * 151ll * 151ll * 151ll * 151ll * 151ll * 151ll;
    fac[0] = 1;
    For(i, 1, n) fac[i] = mult(fac[i - 1], i);
    xt = Pow(151, 6) * 150 - 1, ifac[n] = Pow(fac[n], xt);
    Fordown(i, n, 1) ifac[i - 1] = mult(ifac[i], i);

    res = n - 2;
    For(i, 1, n) Ans = mult(Ans, C(res, pi = (read() - 1))), res -= pi;
    if (res != 0) puts("0");
    else printf("%lld\n", Ans);

    return 0;
}
//多情只有春庭月，犹为离人照落花。
//    -- 张泌《寄人》