BZOJ 2111: [ZJOI2010]Perm 排列计数（简单组合数学）

2111: [ZJOI2010]Perm 排列计数

组合计数

求1-n排列中对于任意2<=i<=N 有 Pi>P(i/2) 的排列个数 mod 质数 p 的余数;
问题等价于求1~n组成一棵满足小根堆性质的完全二叉树的方案数;
定义f[i]为当这棵完全二叉树有i个节点时的方案数;
则 f[i] = C(i-1,left) * f[left] * f[i-1-left];
f[0] = f[1] = 1;

下面来求left
深度 - 1 = floor(log2(i));
记 tot = 前深度 - 1 层总结点数 = pow(2,深度-1)-1;
则 left = (tot-1)/2 + min(i-tot,(tot+1)/2); 


下面来求C(n,k) mod p 
直接 预处理 n! mod p , 费马小定理求乘法逆元; 

 

#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;}