[BZOJ1005][HNOI2008]明明的烦恼
试题描述
自从明明学了树的结构,就对奇怪的树产生了兴趣......给出标号为1到N的点,以及某些点最终的度数,允许在
任意两点间连线,可产生多少棵度数满足要求的树?
输入
第一行为N(0 < N < = 1000),
接下来N行,第i+1行给出第i个节点的度数Di,如果对度数不要求,则输入-1
输出
一个整数,表示不同的满足要求的树的个数,无解输出0
输入示例
3 1 -1 -1
输出示例
2 数据规模及约定
见“输入”
题解
知道 prufer 序列这题就是删边题了。这题不仅要写高精度，还不能随便用除法，使劲压常数，组合数要分解质因数才能过！！
= =就当练高精度吧。
#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;
inline 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 * 10 + c - '0'; c = Getchar(); }return x * f;
}#define maxn 1010
int 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] / 10, A[len-1] %= 10, 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] / 10, ans.A[i] %= 10;int j = len + 1;for(; ans.A[j-1] > 9;) ans.A[j] += ans.A[j-1] / 10, ans.A[j-1] %= 10, 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] / 10, ans.A[i] %= 10;int j = ans.len + 1;for(; ans.A[j-1] > 9;) ans.A[j] += ans.A[j-1] / 10, ans.A[j-1] %= 10, 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] += 10, 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(int 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 tmp = 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();prime_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;
}
除法的定义其实没用，我第一次用了除法 T 飞了，现在懒得删了。
...)
