G. GCD Festival 推式子 COMPFEST 13

G. GCD Festival 推式子 COMPFEST 13 - Finals Online Mirror

题意

给定数组(a)，求

[sum_{i=1}^nsum_{j=1}^ngcd(a_i,a_j) cdot gcd(i,j) ]

[2 leq n leq 10^5\ 1 leq a_i leq 10^5 ]

分析

(n = sum_{d|n} varphi(d))

(gcd(i,j) = sum_{d|gcd(i,j)} varphi(d) = sum_{d|i,d|j} varphi(d))

[sum_{i=1}^nsum_{j=1}^n gcd(i,j)cdot gcd(a_i,a_j)\ =sum_{i=1}^nsum_{j=1}^n gcd(a_i,a_j) sum_{d|i,d|j} varphi(d)\ = sum_{d=1}^nvarphi(d) sum_{i=1}^{lfloorfrac{n}{d} floor} sum_{j=1}^{lfloor frac{n}{d} floor} gcd(a_{id},a_{jd})\ =sum_{d=1}^nvarphi(d) sum_{i=1}^{lfloorfrac{n}{d} floor} sum_{j=1}^{lfloor frac{n}{d} floor} sum_{k|a_{id},k|a_{jd}} varphi(k)\ =sum_{d=1}^nvarphi(d) sum_{k=1}^n varphi(k)(sum_{i=1}^{lfloorfrac{n}{d} floor}[a_{id} equiv 0 (mod k)])^2 ]

于是可以枚举(d)，显然对于(k)只需要考虑所有(a_{id})的因子，因此复杂度变为调和级数乘(d(a_i))

代码

    const int maxn = 1e5 + 5;
     
    int phi[maxn];
    int v[maxn];
    vector<int> vv[maxn];
     
    inline void phi_table(int n){
        phi[1] = 1;
        for(int i = 2;i <= n;i++)
            if(!phi[i]) {
                for(int j = i;j <= n;j += i) {
                    if(!phi[j]) phi[j] = j;
                    phi[j] = phi[j] / i * (i - 1);
                }
            }
    }   
     
    int main(){
        unordered_map<int,int> mp;
        phi_table(maxn - 3);
        int n = rd();
        for(int i = 1;i <= maxn - 3;i++){
            for(int j = i;j <= maxn - 3;j += i)
                vv[j].push_back(i);
        }
        for(int i = 1;i <= n;i++){
            v[i] = rd();
        }
        int ans = 0;
        for(int d = 1;d <= n;d++){
            int tot = 0;
            mp.clear();
            for(int j = 1;j <= n / d;j++){
                for(auto it:vv[v[j * d]]) 
                    mp[it]++;
            }
            for(auto it:mp){
                add(tot,mul(phi[it.fi],(ll)it.se * it.se % MOD));
            }
            add(ans,mul(tot,phi[d]));
        }
        printf("%d",ans);
    }