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P3768 简单的数学题

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P3768 简单的数学题

先推式子：
$$
\begin {aligned}
& \sum _ {i = 1} ^n \sum _ {j = 1} ^ n i j (i, j) \\
= & \sum _ {d = 1} ^ n d \sum _ {i = 1} ^ n \sum _ {j = 1} ^ n i j [(i, j) = d] \\
= & \sum _ {d = 1} ^ n d \sum _ {i = 1} ^ {\left \lfloor \frac n d \right \rfloor} \sum _ {j = 1} ^ {\left \lfloor \frac n d \right \rfloor} id jd [(i, j) = 1] \\
= & \sum _ {d = 1} ^ n d ^ 3 \sum _ {i = 1} ^ {\left \lfloor \frac n d \right \rfloor} \sum _ {j = 1} ^ {\left \lfloor \frac n d \right \rfloor} i j \sum _ {x | i \wedge x | j} \mu (x)\\
= & \sum _ {d = 1} ^ n d ^ 3 \sum _ {x = 1} ^ {\left \lfloor \frac n d \right \rfloor} \mu (x) \sum _ {i = 1} ^ {\left \lfloor \frac n {xd} \right \rfloor} \sum _ {j = 1} ^ {\left \lfloor \frac n {xd} \right \rfloor} xi xj \\
= & \sum _ {d = 1} ^ n d ^ 3 \sum _ {x = 1} ^ {\left \lfloor \frac n d \right \rfloor} \mu (x) x ^ 2 F ^ 2(\left \lfloor \frac n {xd} \right \rfloor) \\
\end {aligned}
$$
其中 $F(x) = \displaystyle \sum _ {i = 1} ^ n i$ ，令 $T = xd$ ，枚举 $T$
$$
\begin {aligned}
& \sum _ {T = 1} ^ n F ^ 2(\left \lfloor \frac n T \right \rfloor) \sum _ {d | T} \mu (\frac T d) d ^ 3(\frac T d) ^ 2 \\
= & \sum _ {T = 1} ^ n F ^ 2(\left \lfloor \frac n T \right \rfloor) T ^ 2 \sum _ {d | T} \mu (\frac T d) d \\
\end {aligned}
$$

其中

$$
\sum _ {d | T} \mu (\frac T d) d = \mu * id = \varphi
$$
所以
$$
\begin {aligned}
\sum _ {T = 1} ^ n F ^ 2(\left \lfloor \frac n T \right \rfloor) T ^ 2 \varphi (T)\\
\end {aligned}
$$
计算时可以用数论分块，现在考虑如何计算 $T ^2 \varphi(T)$ 的前缀和。

记
$$
f(x) = x ^ 2 \varphi (x) \\
s(x) = \sum _ {i = 1} ^ n f(i) = \sum _ {i = 1} ^ n i ^ 2 \varphi (i) = \sum _ {i = 1} ^ n id ^ 2 \varphi (i)
$$
卷上 $id ^ 2$ ，
$$
s(x) = \sum _ {i = 1} ^ n ((id ^ 2 \varphi) * id ^ 2)(i) - \sum _ {i = 2} ^ n id ^ 2 (i) s(\left \lfloor \frac n i \right \rfloor)
$$
其中
$$
\begin {aligned}
& (id ^ 2 \varphi) * id ^ 2 (x) \\
= & \sum _ {d | x} id ^ 2 \varphi (d) id ^ 2(\frac x d) \\
= & \sum _ {d | x} d ^ 2 \varphi (d) ({\frac x d}) ^ 2 \\
= & \sum _ {d | x} x ^ 2 \varphi (d) \\
= & x ^ 2 (\varphi * 1) \\
= & id ^ 3 (x)
\end {aligned}
$$
因此，
$$
s(x) = \sum _ {i = 1} ^ n i ^ 3 - \sum _ {i = 2} ^ n i ^ 2 s(\left \lfloor \frac n i \right \rfloor)
$$

其中，
$$
\sum _ {i = 1} ^ n i ^ 3 = (\sum _ {i = 1} ^ n i) ^ 2 = F ^ 2(n)
$$

另
$$
\sum _ {i = 1} ^ n i ^ 2 = \frac {n (n + 1) (2n + 1)} 6
$$

查看代码 
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#include <cstdio>
#include <cmath>
#include <unordered_map>
using namespace std;
typedef long long LL;
const int N = 5e6 + 10;
LL n;
int m, mod, inv2, inv6;
bool vis[N];
int cnt, p[N], phi[N], s[N];
unordered_map<LL, int> _s;
int binpow(int b, int k)
{
    int res = 1;
    while (k)
    {
        if (k & 1)
            res = (LL)res * b % mod;
        b = (LL)b * b % mod;
        k >>= 1;
    }
    return res;
}
void init()
{
    vis[1] = true;
    phi[1] = 1;
    for (int i = 2; i <= m; i++)
    {
        if (!vis[i])
        {
            p[++cnt] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; j <= cnt && i * p[j] <= m; j++)
        {
            vis[i * p[j]] = true;
            if (i % p[j] == 0)
            {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * phi[p[j]];
        }
    }
    for (int i = 1; i <= m; i++)
        s[i] = ((LL)i * i % mod * phi[i] % mod + s[i - 1]) % mod;
}
int F_2(LL x)
{
    x %= mod;
    x = (LL)x * (x + 1) % mod * inv2 % mod;
    return (LL)x * x % mod;
}
int _2sum(LL x)
{
    x %= mod;
    return (LL)x * (x + 1) % mod * (x * 2 % mod + 1) % mod * inv6 % mod;
}
int S(LL x)
{
    if (x <= m)
        return s[x];
    if (_s.count(x))
        return _s[x];
    int res = F_2(x), pre = s[1], cur;
    for (LL l = 2, r; l <= x; l = r + 1)
    {
        r = x / (x / l);
        cur = _2sum(r);
        res = (res - ((LL)cur - pre) % mod * S(x / l) % mod) % mod;
        pre = cur;
    }
    return _s[x] = res;
}
int main()
{
    scanf("%d%lld", &mod, &n);
    inv2 = binpow(2, mod - 2), inv6 = binpow(6, mod - 2);
    m = pow(n, 0.666667);
    init();
    int res = 0, pre = 0, cur;
    for (LL l = 1, r; l <= n; l = r + 1)
    {
        r = n / (n / l);
        cur = S(r);
        res = (res + ((LL)cur - pre) % mod * F_2(n / l) % mod) % mod;
        pre = cur;
    }
    printf("%d", ((LL)res + mod) % mod);
    return 0;
}
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