莫比乌斯反演（HDU5663）

题意：

思路：

代码：

const int maxn = 10000000 + 10;
ll n, m, a;
ll miu[maxn], v[maxn];
ll sum[maxn];
void Eratosthenes(int n) {
    for (int i = 1; i <= n; i++)miu[i] = 1, v[i] = 0;
    for (int i = 2; i <= n; ++i) {
        if (v[i])continue;
        miu[i] = -1;
        for (int j = 2 * i; j <= n; j += i) {
            v[j] = 1;
            if ((j / i) % i == 0)miu[j] = 0;
            else miu[j] *= -1;
        }
    }
    for (int i = 1; i * i <= n; ++i) {
        for (int j = i * i; j <= n; j += i * i) {
            sum[j] += miu[j / (i * i)];
        }
    }
    for (int i = 1; i <= n; ++i)sum[i] += sum[i - 1];
}
ll cal(int n, int m) {
    ll ans = 0, pos;
    for (int i = 1; i <= min(n, m); i = pos + 1) {
        pos = min(n / (n / i), m / (m / i));
        ans += 1ll * (sum[pos] - sum[i - 1]) * (n / i) * (m / i);
    }
    return ans;
}

int main() {
    Eratosthenes(10000000);
    int T = read();
    while (T--) {
        n = read(), m = read();
        printf("%lld
", n * m - cal(n, m));
    }
}