BZOJ 1211 树的计数(purfer序列)

首先考虑无解的情况， 根据purfer序列，当dee[i]=0并且n!=1的时候，必然无解。否则为1.

且sum(dee[i]-1)!=n-2也必然无解。

剩下的使用排列组合即可推出公式。需要注意的是题目虽然说最终答案不会超过1e17，但是中间过程可能超。

由于n<=150, 所以sum最多是148. 于是我们可以打出150*150的组合表。实现O(1)计算组合数。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...int dee[];
LL cc[][];void init()
{
    FOR(i,,) {
        cc[i][]=;
        FOR(j,,i) cc[i][j]=cc[i-][j-]+cc[i-][j];
    }
}
int main ()
{
    init();
    int n, sum=;
    LL ans=;
    scanf("%d",&n);
    if (n==) {
        scanf("%d",dee);
        puts(dee[]==?"":"");
        return ;
    }
    FOR(i,,n) {
        scanf("%d",dee+i), --dee[i], sum+=dee[i];
        if (dee[i]<) {puts(""); return ;}
    }
    if (sum!=n-) {puts(""); return ;}
    FOR(i,,n) {
        if (!dee[i]) continue;
        ans*=cc[sum][dee[i]];
        sum-=dee[i];
    }
    printf("%lld\n",ans);
    return ;
}