P2522 [HAOI2011]Problem b

P2522 [HAOI2011]Problem b

$$
\begin {aligned}
f(n) = & \sum_{i = 1} ^ a \sum_{j = 1} ^ b [ gcd(i, j) = n ] \\
= & \sum_{i = 1} ^ {\frac a n} \sum_{j = 1} ^ {\frac b n} \sum_{d | gcd(i, j)} \mu(d) \\
= & \sum_{i = 1} ^ {\frac a n} \sum_{j = 1} ^ {\frac b n} \sum_{d | gcd(i, j)} \mu(d) \\
= & \sum_{d = 1} ^ {min({\frac a n}, {\frac b n})} \mu (d) \left \lfloor \frac a {dn} \right \rfloor \left \lfloor \frac b {dn} \right \rfloor \\
\end {aligned}
$$

容斥原理处理即可。

查看代码

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e6 + 10;
bool vis[N];
int cnt, primes[N], mu[N];
LL s[N];
void init()
{
    mu[1] = 1;
    for (int i = 2; i < N; i++)
    {
        if (!vis[i])
        {
            primes[++cnt] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= cnt && i * primes[j] < N; j++)
        {
            vis[primes[j] * i] = true;
            if (i % primes[j] == 0)
            {
                mu[primes[j] * i] = 0;
                break;
            }
            mu[primes[j] * i] = -mu[i];
        }
    }
    for (int i = 1; i < N; i++)
        s[i] = s[i - 1] + mu[i];
}
LL f(int n, int m, int k)
{
    n /= k, m /= k;
    LL res = 0;
    for (int l = 1, r; l <= min(n, m); l = r + 1)
    {
        r = min(n / (n / l), m / (m / l));
        res += (s[r] - s[l - 1]) * (LL)(n / l) * (LL)(m / l);
    }
    return res;
}
int main()
{
    init();
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int a, b, c, d, k;
        scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
        printf("%lld\n", f(b, d, k) - f(a - 1, d, k) - f(b, c - 1, k) + f(a - 1, c - 1, k));
    }
    return 0;
}