fjwc2019 D6T1 堆（组合数+打表）

#193. 「2019冬令营提高组」堆

但是每个点都遍历一遍，有些点的子树完全相同却重复算了

忽然记起完全二叉树的性质之一：每个非叶节点的子树中至少有一个是满二叉树

那么我们预处理满二叉树的那一块，剩下的dfs就可以辣

求阶乘.......分块打表

设打表分成$k$段，则复杂度$O(logn+n/k)$

#include<cstdio>
const int mod=1e9+7,W=1e7;
const int tab[100]={1,924724006,582347126,500419162,881147799,693776109,435873621,279027658,727951124,398578768,678364145,204828554,345795998,116118093,359401113,236930793,856493327,207383191,617606889,933753281,26701748,329394893,360779992,416008308,187501984,165706817,328891607,16385287,117411011,404196042,765064133,239669664,761588352,566114869,673499119,840260100,352356536,53839501,178657924,373444237,227300165,207172723,444208499,367531373,297449176,605324209,729265513,567907756,125889461,250743107,666666670,598576559,632705086,295855233,185718228,414607857,737215408,863388390,182290465,707552496,881713600,417895708,490627919,364521407,775935292,972492338,473340273,920880265,530581,696910290,64037482,649527920,756691728,283805222,711255329,825205499,263679166,341083474,914727729,919247968,465317279,960145703,274813468,393588827,65909169,521964827,794328994,484551338,521297378,54488990,591837535,255746228,25827429,177799409,92011129,469664591,35708489,197025781,288851931,254032854};
int n,f[31],ans;
inline int Pow(int x,int y){
    int re=1;
    for(;y;y>>=1,x=1ll*x*x%mod)
        if(y&1) re=1ll*re*x%mod;
    return re;
}
int dfs(int x){
    int i=0;
    while((1<<i)-1<x) ++i;
    if((1<<i)-1==x) return f[i];
    if(x+(1<<(i-2))<=(1<<i)-1) return 1ll*Pow(x,mod-2)*f[i-2]%mod*dfs(x-(1<<(i-2)))%mod;
    else return 1ll*Pow(x,mod-2)*f[i-1]%mod*dfs(x-(1<<(i-1)))%mod;
}
int main(){
    freopen("heap.in","r",stdin);
    freopen("heap.out","w",stdout);
    scanf("%d",&n); f[0]=1;
    for(int i=1;i<=30;++i) f[i]=1ll*f[i-1]*f[i-1]%mod*Pow((1<<i)-1,mod-2)%mod;
    ans=tab[(n-1)/W];
    for(int i=(n-1)/W*W+2;i<=n;++i) ans=1ll*ans*i%mod;
    ans=1ll*ans*dfs(n)%mod;
    printf("%d",ans);
    return 0;
}