51 NOD 1227 平均最小公倍数（杜教筛）

1227 平均最小公倍数

推式子

$$\begin{gathered} S(n)=\sum _{i=1}^n\sum _{j=1}^i\frac{lcm(i,j)}{i} \\ =\sum _{i=1}^n\sum _{j=1}^i\frac{ij}{igcd(i,j)} \\ =\sum _{i=1}^n\sum _{j=1}^i\frac{j}{gcd(i,j)} \\ =\sum _{i=1}^n\sum _{d=1}^i\sum _{j=1}^i\frac{j}{d}(gcd(i,j)==d) \\ =\sum _{i=1}^n\sum _{d=1}^i\sum _{j=1}^{\frac{i}{d}}j(gcd(j,\frac{i}{d})==1) \\ =\sum _{i=1}^n\sum _{d=1}^i\frac{\frac{i}{d}\varphi (\frac{i}{d})+(\frac{i}{d}==1)}{2} \\ =\sum _{d=1}^n\sum _{i=1}^{\frac{n}{d}}\frac{i\varphi (i)+(i==1)}{2} \\ 接下来就是杜教筛求\sum _{i=1}^{n}i\varphi (i)了，g(1)S(n)=\sum _{i=1}^n(f\ast g)(i)-\sum _{i=2}^{n}f(i)S(\frac{n}{i}) \\ 另f(i)=i，即可求得S(n)=\sum _{i=1}^n{i}^2-\sum _{i=2}^{n}iS(\frac{n}{d}) \end{gathered}$$

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include 

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 1e6 + 10, mod = 1e9 + 7;

int prime[N], cnt;

ll phi[N], inv2, inv6;

bool st[N];

ll quick_pow(ll a, ll n, ll mod) {
    ll ans = 1;
    while(n) {
        if(n & 1) ans = ans * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ans;
}

void init() {
    phi[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime[cnt++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; j < cnt && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
    for(int i = 1; i < N; i++) {
        phi[i] = (1ll * phi[i] * i + phi[i - 1]) % mod;
    }
    inv2 = quick_pow(2, mod - 2, mod), inv6 = quick_pow(6, mod - 2, mod);
}

unordered_map<ll, ll> ans_s;

ll S(ll n) {
    if(n < N) return phi[n];
    if(ans_s.count(n)) return ans_s[n];
    ll ans = n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
    for(ll l = 2, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans - (l + r) * (r - l + 1) / 2 % mod * S(n / l) % mod + mod) % mod;
    }
    return ans_s[n] = ans;
}

ll solve(ll n) {
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans + 1ll * (r - l + 1) * S(n / l) % mod) % mod;
    }
    return (ans + n) % mod;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    ll l = read(), r = read();
    printf("%lld\n", ((solve(r) - solve(l - 1)) % mod * inv2 % mod + mod) % mod);
	return 0;
}