zoukankan      html  css  js  c++  java
  • 【BZOJ 3560】DZY Loves Math V(欧拉函数)

    传送门

    直接考虑欧拉函数的性质
    考虑每个质因数的贡献乘起来
    ppaia_i中出现bib_i
    那么pp的贡献为
    i1=0b1....in=0bn(pjij1)p1p+1ϕ(1)sum_{i_1=0}^{b_1}....sum_{i_n=0}^{b_n}(p^{sum_ji_j}-1)*frac {p-1}{p}+1(特殊处理phi(1))

    =[i(jpj)1]p1p+1=[prod_i(sum_jp^j)-1]*frac {p-1}p+1

    然后就完了

    #include<bits/stdc++.h>
    using namespace std;
    #define cs const
    #define re register
    #define pb push_back
    #define pii pair<int,int>
    #define ll long long
    #define fi first
    #define se second
    #define bg begin
    cs int RLEN=1<<20|1;
    inline char gc(){
        static char ibuf[RLEN],*ib,*ob;
        (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
        return (ib==ob)?EOF:*ib++;
    }
    inline int read(){
        char ch=gc();
        int res=0;bool f=1;
        while(!isdigit(ch))f^=ch=='-',ch=gc();
        while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
        return f?res:-res;
    }
    template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
    template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
    cs int mod=1e9+7;
    inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
    inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
    inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
    inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
    inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
    inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
    inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
    inline int Inv(int x){return ksm(x,mod-2);}
    cs int N=10000006,M=1000006;
    int pr[M],tot,mnp[N];
    vector<int> num[N];
    bitset<N> vis;
    inline void init(int len=N-6){
        for(int i=2;i<=len;i++){
            if(!vis[i])pr[++tot]=i,mnp[i]=i;
            for(int j=1,p;j<=tot&&i*pr[j]<=len;j++){
                p=i*pr[j],vis[p]=1,mnp[p]=pr[j];
                if(i%pr[j]==0)break;
            }
        }
    }
    int n,pri[100],cnt;
    int main(){
        #ifdef Stargazer
        freopen("lx.in","r",stdin);
        #endif
        init();
        n=read();
        for(int i=1;i<=n;i++){
            int x=read();cnt=0;
            while(x>1){pri[++cnt]=mnp[x],x/=mnp[x];}
            sort(pri+1,pri+cnt+1);
            for(int j=1;j<=cnt;j++){
                int c=1;
                while(pri[j+1]==pri[j]&&j<cnt)j++,c++;
                num[pri[j]].pb(c);
            }
        }
        int ans=1;
        for(int i=1;i<=tot;i++)if(num[pr[i]].size()){
            int ret=1,p=pr[i];
            for(int j=0,iv=Inv(p-1);j<num[p].size();j++){
                int b=num[p][j];
                Mul(ret,mul(dec(ksm(p,b+1),1),iv));
            }
            Dec(ret,1);
            Mul(ans,add(1,mul(ret,mul(p-1,Inv(p)))));
        }
        cout<<ans<<'
    ';
    }
    
  • 相关阅读:
    缓存Cache
    RDD的行动操作
    redis数据库的配置
    requests的封装(user-agent,proxies)
    phantjs
    python多线程
    etree-xpath
    Flask
    Flask
    Flask
  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328327.html
Copyright © 2011-2022 走看看