agc38C LCMs

https://atcoder.jp/contests/agc038/tasks/agc038_c
题意:给\(a_i\),求\(\sum_{i=1}^n\sum_{j=i+1}^nlcm(a_i,a_j)\)
题解:设\(\sum_{d|i}c_d=\frac{1}{i}\),\(O(nlogn)\)求出\(c_i\)
\(\sum_{i=1}^n\sum_{j=i+1}^nlcm(a_i,a_j)\)
\(=\sum_{i=1}^n\sum_{j=i+1}^n\frac{a_i\cdot a_j}{gcd(a_i,a_j)}\)
\(=\sum_{i=1}^n\sum_{j=i+1}^na_i\cdot a_j \cdot \sum_{d|a_i,d|a_j}c_d\)
\(=\sum_{i=1}^{ma}c_i\sum_{d|a_i,d|a_j,i
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
//#include 
#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
#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
#define pil pair
#define pli pair
#define pii pair
#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;}
templateinline T const& MAX(T const &a,T const &b){return a>b?a:b;}
templateinline T const& MIN(T const &a,T const &b){return a>=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\n",ans);
    return 0;
}
/********************

********************/

转载于:https://www.cnblogs.com/acjiumeng/p/11585237.html