2111: [ZJOI2010]Perm 排列計數

組合計數

求1-n排列中對於任意2<=i<=N 有 Pi>P(i/2) 的排列個數 mod 質數 p 的餘數;
問題等價於求1~n組成一棵滿足小根堆性質的完全二叉樹的方案數;
定義f[i]為當這棵完全二叉樹有i個節點時的方案數;
則 f[i] = C(i-1,left) * f[left] * f[i-1-left];
f[0] = f[1] = 1;

下面來求left
深度 - 1 = floor(log2(i));
記 tot = 前深度 - 1 層總結點數 = pow(2,深度-1)-1;
則 left = (tot-1)/2 + min(i-tot,(tot+1)/2); 


下面來求C(n,k) mod p 
直接 預處理 n! mod p , 費馬小定理求乘法逆元; 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
using namespace std;
int fac[1000010],f[1000010];
int n,p;
inline int pow(int x,int y,int p)
{
	int res(1);
	x %= p;
	while(y)
	{
		if(y&1) res = (long long)res * x % p;
		x = (long long)x * x % p;
		y >>= 1;
	}
	return res;
}
inline int C(int n,int m,int p)
{
	return (long long)fac[n]*pow((long long)fac[m]*fac[n-m]%p,p-2,p) % p;
}
inline int solve(int x)
{	
	if(f[x]) return f[x];
	int depth = log2(x);
	int tot = pow(2,depth,10000000)-1;
	int left = (tot-1>>1) + min(x - tot,tot+1>>1);	
	return f[x] = (long long)C(x-1,left,p)*solve(left)%p*solve(x-left-1)%p;
}
int main()
{
	scanf("%d%d",&n,&p);
	fac[0] = 1;
	for(int i = 1; i <= n; i++)
		fac[i] = (long long)fac[i-1] * i % p;
	f[0] = f[1] = 1;
	printf("%d\n",solve(n));
	return 0;
}