hdu 5321 Beautiful Set（推式子+ 反演+組合）*

大牛博客：https://blog.csdn.net/FirstLucker/article/details/47128347

下面是看了大牛的博客後，自己依着思路推的
zstu的推式子：

hdu的話，最後注意有$2^n=\sum_{i=0}^nC_n^k$

#include <bits/stdc++.h>

using namespace std;
#define ll long long 
const int mod = 258280327;


const int N = 1e5+10;
int prime[N];
ll h[N];
ll bin[N];

int tmp_val[N];
int cnt[N];
ll jc[N];
ll inv[N];//i的逆元
ll fac_inv[N]; //i!的逆元


void get_prime()
{
    bin[0]=1;
    bin[1]=2;
    jc[0]=1;
    jc[1]=1;
    
    h[1]=1;
    inv[0]=1;
    inv[1]=1;
    fac_inv[0]=1;
    fac_inv[1]=1;
    for(int i=2;i<N;i++)
    {
        bin[i]=bin[i-1]*2%mod;
        jc[i]=jc[i-1]*i%mod;
        inv[i]=(mod-mod/i)*inv[mod%i]%mod;
        fac_inv[i]=fac_inv[i-1]*inv[i]%mod;
        if(!prime[i])
        {
            prime[++prime[0]]=i;
            h[i]=i-1;
        }
        for(int j=1;j<=prime[0]&&i*prime[j]<N;j++)
        {
            prime[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                h[i*prime[j]]=prime[j]*h[i]%mod;
                break;
            }
            h[i*prime[j]]=(prime[j]-1)*h[i]%mod;
        }
    }
    
}



void solve(int n,int max_val)
{
    for(int i=1;i<=max_val;i++)
    {
        cnt[i]=0;
        for(int j=i;j<=max_val;j+=i)
        {
            cnt[i]+=tmp_val[j];
        }
    }
    ll zstu=0,hdu=0;
    for(int d=1;d<=max_val;d++) if(cnt[d])
    {
        zstu=(zstu+h[d]*jc[cnt[d]]%mod*jc[n+1]%mod*fac_inv[cnt[d]-1]%mod*inv[n-cnt[d]+2]%mod)%mod;
        hdu=(hdu+h[d]*cnt[d]%mod*bin[(cnt[d]-1)]%mod)%mod;
    }
    // cout<<zstu<<" "<<hdu<<endl;
    if(zstu==hdu)
        printf("Equal %lld\n",hdu);
    else if(zstu<hdu)
        printf("Mr. Hdu %lld\n",hdu);
    else 
        printf("Mr. Zstu %lld\n",zstu);
    
}



int main()
{
    get_prime();
    int n,x;
    while(~scanf("%d",&n))
    {
        memset(tmp_val,0,sizeof(tmp_val));
        int max_val=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            tmp_val[x]++;
            max_val=max(max_val,x);
        }
        solve(n,max_val);
    }
    
    return 0;
}