51nod1238 最小公倍数之和 V3(莫比乌斯反演)

题意

题目链接

sol

不想打公式了，最后就是求一个

\(\sum_{i=1}^n ig(\frac{n}{i})\)

\(g(i) = \sum_{i=1}^n \phi(i) i^2\)

拉个\(id2\)卷一下

#include<bits/stdc++.h> 
#define int long long 
#define ll long long 
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10, inv2 = 500000004, inv6 = 166666668, b = 1e6;
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
void print(int x) {
    if(!x) return ;
    print(x / 10);
    putchar(x % 10 + '0');
}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = 1ll * x * 10 + c - '0', c = getchar();
    return x * f;
}
int g[maxn], phi[maxn], mu[maxn], vis[maxn], prime[maxn], tot;
map<int, int> mp;
int sum(int n) {return mul(mul(n % mod, add(n, 1)), inv2);}
int sum2(int n) {return mul(mul(n % mod, mul(add(n, 1), mul(2, n) + 1)), inv6);}
int sum3(int n) {return mul(sum(n), sum(n));}
void sieve(int n) {
    vis[1] = phi[1] = mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1, phi[i] = i - 1;
        for(int j = 1; j <= tot && i * prime[j] <= n; j++) {
            vis[i * prime[j]] = 1;
            if(i % prime[j]) phi[i * prime[j]] = phi[i] * phi[prime[j]], mu[i * prime[j]] = -mu[i];
            else {mu[i * prime[j]] = 0; phi[i * prime[j]] = phi[i] * prime[j]; break;};
        }
    }
    for(int i = 1; i <= n; i++) g[i] = add(g[i - 1], mul(phi[i], mul(i, i)));
}
ll dsieve(int n) {
    if(n <= b) return g[n];
    else if(mp[n]) return mp[n];
    ll t = sum3(n);
    for(int i = 2, nxt; i <= n; i = nxt + 1) {
        nxt = n / (n / i); 
        add2(t, -mul(add(sum2(nxt), -sum2(i - 1)), dsieve(n / i)));
    }
    return mp[n] = t;
}
signed main() {
    sieve(b);
    int n = read(), ans = 0;
    for(int i = 1, nxt; i <= n; i = nxt + 1) {
        nxt = n / (n / i);
        add2(ans, mul(add(sum(nxt), -sum(i - 1)), dsieve(n / i)));
    }
    print(ans);
    return 0;
}