[BZOJ1005] [HNOI2008] Clearly the trouble

[BZOJ1005] [HNOI2008] Clearly the trouble

Question Description

Ever since I learned the structure of a tree, I have been interested in strange trees ... Gives the point labeled 1 to N, and the final degree of some points, allowing
Any two-point connection, how many trees can be produced to meet the requirements?

Input

The first behavior n (0 < n < = 1000),
Next n rows, the i+1 line gives the degree di of the I node, and if not required for the degree, enter-1

Output

An integer that represents the number of different trees that satisfy the requirement, no output 0

Input example

3 1-1-1

Output example

2

Data size and conventions

See " Input "

Exercises

The problem with knowing the prufer sequence is to delete the problem. This problem not only to write high-precision, but also can not use division, hard pressure constant, combined number to decompose factorization before!!

= = Just practice high precision.

#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack > #include <vector> #include <queue> #include <cstring> #include <string> #include <map > #include <set>using namespace std;const int buffersize = 1 << 16;char buffer[buffersize], *head, *TAIL;INL        ine Char Getchar () {if (Head = = Tail) {int L = fread (buffer, 1, buffersize, stdin);    Tail = (Head = buffer) + L; } return *head++;}    int read () {int x = 0, f = 1; char c = Getchar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = Getchar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = Getchar ();} return x * f;} #define MAXN 1010int N, Deg[maxn], cnt;struct bign {int len, a[5010];bign () {len = 1; memset (A, 0, sizeof (a));} Bign operator = (const int& t) {len = 1; A[0] = T;while (a[len-1] > 9) A[len] = a[len-1]/ten, a[len-1]%=, Len++;return *this;} void Clear () {for (;! A[len-1] && Len; len--); IF (!len) len = 1;return;} Bign operator * (const int& t) const {bign ans; ans.len = len;memcpy (ans. A, a, sizeof (a)), for (int i = 0; i < len; i++) ans. A[i] *= t;for (int i = 0; i < len; i++) ans. A[i+1] + = ans. A[i]/ten, ans. A[i]%= 10;int j = len + 1;for (; ans. A[J-1] > 9;) ans. A[j] + = ans. A[J-1]/ten, ans. A[j-1]%=, J++;ans.len = J;ans.clear (); return ans;} Bign operator *= (const int& t) {*this = *this * T;return *this;} Bign operator * (const bign& t) const {bign ans; ans.len = len + t.len-1;for (int i = 0; i < len; i++) for (int j = 0; J < T.len; J + +) ans. A[I+J] + = a[i] * t.a[j];for (int i = 0; i < Ans.len; i++) ans. A[i+1] + = ans. A[i]/ten, ans. A[i]%= 10;int j = ans.len + 1;for (; ans. A[J-1] > 9;) ans. A[j] + = ans. A[J-1]/ten, ans. A[j-1]%=, J++;ans.len = J;ans.clear (); return ans;} Bign operator *= (const bign& t) {*this = *this * T;return *this;} Bign operator-(const bign& t) const {bign ans; ans.len = len;memcpy (ans. A, a, sizeof (a)); for (int i = 0; i < Len; i++) {if (I < t.len) ans. A[i]-= t.a[i];if (ans. A[i] < 0) ans. A[i] + = ten, ans. a[i+1]--;} while (!ans. A[ans.len-1]) Ans.len--;return ans;} Bign operator-= (const bign &t) {*this = *this-t;return *this;} Bign operator/(const bign& t) const {bign ans, f; f = 0; Ans.len = -1;for (int i = len-1; I >= 0; i--) {f *= 10; F.a[0] = A[i];while (f >= t) {f-= T;ans. A[i]++;if (Ans.len = =-1) Ans.len = i + 1;}} return ans;}  BOOL operator >= (const bign& t) const {if (len! = T.len) return len > t.len;for (int i = len-1; I >= 0; i--) if (a[i]! = T.a[i]) return a[i] > T.a[i];return 1;} void print () {for (int i = len-1; I >= 0; i--) Putchar (a[i] + ' 0 '); return;}} ; int PRIME[MAXN], cp;bool vis[maxn];void prime_table () {for (int i = 2; I <= n; i++) {if (!vis[i]) PRIME[++CP] = I;for (in T j = 1; J <= CP && I * prime[j] <= N; J + +) {Vis[i*prime[j]] = 1;if (i% prime[j] = = 0) break;}} return;} Bign Pow (int a, int n) {bign sum, t; sum = 1; t = A;while(n) {if (n & 1) sum *= t;t *= t; n >>= 1;} return sum;} int cp[maxn];bign C (int n, int m) {for (int i = 1; I <= Cp; i++) Cp[i] = 0;for (int i = n; I >= n-m + 1; i--) {int T MP = i;for (int j = 1; J <= CP; J + +) if (tmp% prime[j] = = 0) while (tmp% prime[j] = 0) cp[j]++, tmp/= PRIME[J];} for (int i = m; i; i--) {int tmp = i;for (int j = 1; J <= CP; J + +) if (tmp% prime[j] = = 0) while (tmp% prime[j] = = 0) cp[j ]--, tmp/= PRIME[J];} Bign sum; sum = 1;for (int i = 1; I <= CP; i++) if (cp[i]) sum *= Pow (Prime[i], cp[i]); return sum;} int main () {n = read ();p rime_table (), int tot = 0;bool OK = 1;for (int i = 1; I <= n; i++) {int x = read (); if (x >= 0) DEG[++CNT] = x-1, tot + = (x-1); if (!x) ok = 0;} if (!ok | | tot > N-2) return puts ("0"), 0;tot = N-2;bign sum; sum = 1;for (int i = 1; I <= cnt; i++) {sum *= C (tot, deg[i]); tot-= Deg[i];} Sum *= Pow (n-cnt, tot); Sum.print (); Putchar (' \ n '); return 0;}

The definition of division in fact useless, I first used a division T flew, now lazy to delete.

[BZOJ1005] [HNOI2008] Clearly the trouble