显然题目就是要我们求 ∑ i = 1 n ∑ j = 1 m 2 g c d ( i , j ) − 1 \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} 2gcd(i, j) - 1 ∑i=1n∑j=1m2gcd(i,j)−1
= 2 ∑ i = 1 n ∑ j = 1 m g c d ( i , j ) − n m = 2\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} gcd(i, j) - nm =2i=1∑nj=1∑mgcd(i,j)−nm
转化为我们要求 ∑ i = 1 n ∑ j = 1 m g c d ( i , j ) \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} gcd(i, j) ∑i=1n∑j=1mgcd(i,j)
= ∑ d = 1 n d ∑ i = 1 n d ∑ j = 1 m d g c d ( i , j ) = = 1 = \sum_{d = 1} ^{n}d\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} gcd(i, j) == 1 =d=1∑ndi=1∑dnj=1∑dmgcd(i,j)==1
套上 m o b i u s mobius mobius
= ∑ d = 1 n d ∑ i = 1 n d ∑ j = 1 m d ∑ k ∣ g c d ( i , j ) μ ( k ) = \sum_{d = 1} ^{n}d\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} \sum_{k \mid gcd(i, j)} \mu(k) =d=1∑ndi=1∑dnj=1∑dmk∣gcd(i,j)∑μ(k)
= ∑ d = 1 n d ∑ i = 1 n d ∑ j = 1 m d ∑ k ∣ g c d ( i , j ) μ ( k ) = \sum_{d = 1} ^{n}d\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} \sum_{k \mid gcd(i, j)} \mu(k) =d=1∑ndi=1∑dnj=1∑dmk∣gcd(i,j)∑μ(k)
= ∑ d = 1 n d ∑ k = 1 n d μ ( k ) ⌊ n d k ⌋ ⌊ m d k ⌋ = \sum_{d = 1} ^{n} d\sum_{k = 1} ^{\frac{n}{d}}\mu(k) \lfloor\frac{n}{dk}\rfloor \lfloor\frac{m}{dk}\rfloor =d=1∑ndk=1∑dnμ(k)⌊dkn⌋⌊dkm⌋
另 t = d k t = dk t=dk
= ∑ t = 1 n ⌊ n t ⌋ ⌊ m t ⌋ ∑ d ∣ t d μ ( t d ) = \sum_{t = 1} ^{n} \lfloor\frac{n}{t}\rfloor \lfloor\frac{m}{t}\rfloor \sum_{d \mid t}d\mu(\frac{t}{d}) =t=1∑n⌊tn⌋⌊tm⌋d∣t∑dμ(dt)
m o b i u s mobius mobius反演有 ∑ d ∣ n μ ( d ) d = ϕ ( n ) n \sum_{d\mid n}\frac{\mu(d)}{d} = \frac{\phi(n)}{n} ∑d∣ndμ(d)=nϕ(n)
= ∑ t = 1 n ⌊ n t ⌋ ⌊ m t ⌋ ϕ ( t ) = \sum_{t = 1} ^{n} \lfloor\frac{n}{t}\rfloor \lfloor\frac{m}{t}\rfloor \phi(t) =t=1∑n⌊tn⌋⌊tm⌋ϕ(t)
/*
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 = 1e7 + 10;
bool st[N];
vector<int> prime;
int n, m;
ll phi[N];
void mobius() {
st[0] = st[1] = phi[1] = 1;
for(int i = 2; i < N; i++) {
if(!st[i]) {
prime.pb(i);
phi[i] = i - 1;
}
for(int j = 0; j < prime.size() && 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] += phi[i - 1];
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
mobius();
ll n = read(), m = read();
if(n > m) swap(n, m);
ll ans = 0;
for(ll l = 1, r; l <= n; l = r + 1) {
r = min(n / (n / l), m / (m / l));
ans += (n / l) * (m / l) * (phi[r] - phi[l - 1]);
}
printf("%lld\n", 2 * ans - n * m);
return 0;
}