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  • agc38C LCMs

    https://atcoder.jp/contests/agc038/tasks/agc038_c
    题意:给(a_i),求(sum_{i=1}^nsum_{j=i+1}^nlcm(a_i,a_j))
    题解:设(sum_{d|i}c_d=frac{1}{i}),(O(nlogn))求出(c_i)
    (sum_{i=1}^nsum_{j=i+1}^nlcm(a_i,a_j))
    (=sum_{i=1}^nsum_{j=i+1}^nfrac{a_icdot a_j}{gcd(a_i,a_j)})
    (=sum_{i=1}^nsum_{j=i+1}^na_icdot a_j cdot sum_{d|a_i,d|a_j}c_d)
    (=sum_{i=1}^{ma}c_isum_{d|a_i,d|a_j,i<j}a_icdot a_j)
    总复杂度(O(nlogn))

    //#pragma GCC optimize(2)
    //#pragma GCC optimize(3)
    //#pragma GCC optimize(4)
    //#pragma GCC optimize("unroll-loops")
    //#pragma comment(linker, "/stack:200000000")
    //#pragma GCC optimize("Ofast,no-stack-protector")
    //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    #include<bits/stdc++.h>
    //#include <bits/extc++.h>
    #define fi first
    #define se second
    #define db double
    #define mp make_pair
    #define pb push_back
    #define mt make_tuple
    //#define pi acos(-1.0)
    #define ll long long
    #define vi vector<int>
    #define mod 998244353
    #define ld long double
    //#define C 0.5772156649
    #define ls l,m,rt<<1
    #define rs m+1,r,rt<<1|1
    #define pll pair<ll,ll>
    #define pil pair<int,ll>
    #define pli pair<ll,int>
    #define pii pair<int,int>
    #define ull unsigned long long
    #define bpc __builtin_popcount
    #define base 1000000000000000000ll
    #define fin freopen("a.txt","r",stdin)
    #define fout freopen("a.txt","w",stdout)
    #define fio ios::sync_with_stdio(false);cin.tie(0)
    #define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
    inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
    inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
    template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
    template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
    inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
    inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
    inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c),b>>=1;}return ans;}
    
    using namespace std;
    //using namespace __gnu_pbds;
    
    const ld pi = acos(-1);
    const ull ba=233;
    const db eps=1e-5;
    const ll INF=0x3f3f3f3f3f3f3f3f;
    const int N=1000000+10,maxn=2000000+10,inf=0x3f3f3f3f;
    
    ll a[N],b[N],c[N];
    int main()
    {
        int n,ma=0;scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]),b[a[i]]++,ma=max(ma,(int)a[i]);
        ll ans=0;
        for(int i=1;i<=ma;i++)c[i]=qp(i,mod-2);
        for(int i=1;i<=ma;i++)
            for(int j=2*i;j<=ma;j+=i)
                sub(c[j],c[i]);
        for(int i=1;i<=ma;i++)
        {
            ll te=0,p=0;
            for(int j=i;j<=ma;j+=i)
                add(te,b[j]*j%mod),add(p,b[j]*j%mod*j%mod);
            te=((te*te-p)%mod+mod)%mod;
            te=te*qp(2,mod-2)%mod;
            add(ans,te*c[i]%mod);
        }
        printf("%lld
    ",ans);
        return 0;
    }
    /********************
    
    ********************/
    
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/11585237.html
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