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BZOJ 1211: [HNOI2004]树的计数

2017-09-23 20:54 302 查看
二次联通门 : BZOJ 1211: [HNOI2004]树的计数

/*
BZOJ 1211: [HNOI2004]树的计数

prufer数列 + Cayley公式 + 组合数学
其实就是明明的烦恼弱化版
(n-2)!/((d1 - 1)! * (d2 - 1)! * .... * (dn - 1))
要分解质因数和特判几种情况
*/
#include <cstdio>
#include <iostream>
#define rg register
typedef long long LL;
#define Max 200
inline void read (LL &n)
{
rg char c = getchar ();
for (n = 0; !isdigit (c); c = getchar ());
for (; isdigit (c); n =    n * 10 + c - '0', c = getchar ());
}
LL d[Max], J[Max];
int main (int argc, char *argv[])
{
LL Answer = 1, N; rg int i, j; LL s = 0; read (N);
for (i = 1; i <= N; ++ i)
{
read (d[i]), s += d[i] - 1;
if (d[i] == 0 && N != 1) return printf ("0"), 0;
for (j = 2, J[i] = 1; j <= d[i] - 1; ++ j) J[i] *= j;
}
if (s != N - 2) return printf ("0"), 0;
for (i = j = 1; i <= N - 2; ++ i)
{
Answer *= i; if (j > N) continue;
if (Answer % J[j] == 0) Answer /= J[j ++];
}
printf ("%lld", Answer); return 0;
}
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