题目描述

输入输出格式

输入格式：

输入文件第一行是一个正整数n，表示树有n个结点。第二行有n个数，第i个数表示di，即树的第i个结点的度数。其中1<=n<=150，输入数据保证满足条件的树不超过10^17个。

输出格式：

输出满足条件的树有多少棵。

输入输出样例

输入样例#1： 复制

4                     
2 1 2 1

输出样例#1： 复制

2
首先不知道prufer序列的可以学一下；
https://blog.csdn.net/update7/article/details/77587329
知道以后，其实就是依据该序列来还原树；
prufer的长度为n-2，所以全排列为(n-2)!;
考虑重复排列；
那么:

然后分解质因数即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-11
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n;
ll d[200];
ll ct[200];

void sol(int x, int k) {
	int dv = 2;
	while (x > 1) {
		if (x%dv == 0) {
			ct[dv] += k; x /= dv;
		}
		else dv++;
	}
}

int main() {
//	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	n = rd(); int tot = 0;
	for (int i = 1; i <= n; i++) {
		rdllt(d[i]);
		if (d[i] > 1)tot += (d[i] - 1);
	}
	if (n == 1) {
		if (!d[1])cout << 1 << endl;
		else cout << 0 << endl;
		return 0;
	}
	if (tot != n - 2) { cout << 0 << endl; return 0; }
	for (int i = 2; i <= n - 2; i++) {
		sol(i, 1);
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 2; j < d[i]; j++) {
			sol(j, -1);
		}
	}
	ll ans = 1;
	for (ll i = 1; i <= n; i++) {
		for (ll j = 1; j <= ct[i]; j++)ans *= i;
	}
	printf("%lld\n", ans * 1ll);
	return 0;
}