1227 平均最小公倍数
推式子
【51 NOD 1227 平均最小公倍数（杜教筛）】 S ( n ) = ∑ i = 1 n ∑ j = 1 i l c m ( i , j ) i = ∑ i = 1 n ∑ j = 1 i i j i g c d ( i , j ) = ∑ i = 1 n ∑ j = 1 i j g c d ( i , j ) = ∑ i = 1 n ∑ d = 1 i ∑ j = 1 i j d ( g c d ( i , j ) = = d ) = ∑ i = 1 n ∑ d = 1 i ∑ j = 1 i d j ( g c d ( j , i d ) = = 1 ) = ∑ i = 1 n ∑ d = 1 i i d ? ( i d ) + ( i d = = 1 ) 2 = ∑ d = 1 n ∑ i = 1 n d i ? ( i ) + ( i = = 1 ) 2 接 下 来 就 是 杜 教 筛 求 ∑ i = 1 n i ? ( i ) 了 ， g ( 1 ) S ( n ) = ∑ i = 1 n ( f ? g ) ( i ) ? ∑ i = 2 n f ( i ) S ( n i ) 另 f ( i ) = i ， 即 可 求 得 S ( n ) = ∑ i = 1 n i 2 ? ∑ i = 2 n i S ( n d ) 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} \phi(\frac{i}{d}) + (\frac{i}{d} == 1)}{2}\\ = \sum_{d = 1} ^{n} \sum_{i = 1} ^{\frac{n}{d}} \frac{i \phi(i) + (i == 1)}{2} \\ 接下来就是杜教筛求\sum_{i = 1} ^{n}i \phi(i) 了，g(1)S(n) = \sum_{i = 1} ^{n} (f * 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})\\ S(n)=i=1∑n?j=1∑i?ilcm(i,j)?=i=1∑n?j=1∑i?igcd(i,j)ij?=i=1∑n?j=1∑i?gcd(i,j)j?=i=1∑n?d=1∑i?j=1∑i?dj?(gcd(i,j)==d)=i=1∑n?d=1∑i?j=1∑di??j(gcd(j,di?)==1)=i=1∑n?d=1∑i?2di??(di?)+(di?==1)?=d=1∑n?i=1∑dn??2i?(i)+(i==1)?接下来就是杜教筛求i=1∑n?i?(i)了，g(1)S(n)=i=1∑n?(f?g)(i)?i=2∑n?f(i)S(in?)另f(i)=i，即可求得S(n)=i=1∑n?i2?i=2∑n?iS(dn?)
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 | 1using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair 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 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;
}
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