题目描述

求(对 \(20101009\) 取模，\(n,m\le10^7\) )

\[\sum_{i=1}^n\sum_{j=1}^m\operatorname{lcm}(i,j)
\]

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大体思路

推式子:

\[\begin{aligned}\sum_{i=1}^n\sum_{j=1}^m\operatorname{lcm}(i,j)
&=\sum_{i=1}^n\sum_{j=1}^m\frac{i\times j}{\gcd(i,j)} \\
&=\sum_{i=1}^n\sum_{j=1}^m\sum_{d|i,d|j,\gcd(i/d,j/d)=1}\frac{i\times j}{d} \\
&=\sum_{d=1}^{\min(n,m)}\times d\times\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}[\gcd(i,j)=1]\times i\times j\end{aligned}\]

把式子后面那一大堆设为 \(sum(n,m)\) ：

\[sum(n,m)=\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=1]\times i\times j
\]

考虑化简一下 \(sum\) :

\[\begin{aligned}sum(n,m) &= \sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=1]\times i\times j\\
&=\sum_{i=1}^n\sum_{j=1}^m\sum_{d|\gcd(i,j)}\mu(d)\times i\times j \\
&=\sum_{d=1}^{\min(n,m)}\mu(d)\times d^2\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}i\times j\end{aligned}\]

可以发现 \(sum\) 后面那一大堆(设为 \(g(n,m)\) )可以 \(O(1)\) 求：

\[\begin{aligned}g(n,m)&=\sum_{i=1}^n\sum_{j=1}^m i\times j \\
&=\frac{n\times(n+1)}{2}\times \frac{m\times(m+1)}{2}\end{aligned}\]

那么 \(sum(n,m)\) 可以化为：

\[sum(n,m)=\sum_{d=1}^{\min(n,m)}\mu(d)\times d^2\times g(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)
\]

这个可以数论分块 \(\lfloor\frac{n}{\lfloor\frac{n}{d}\rfloor}\rfloor\) 求。

再回到定义 \(sum\) 的地方，那么：

\[Ans=\sum_{d=1}^{\min(n,m)}\times d\times sum(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)
\]

好像这个还是可以数论分块 \(QwQ\)

至此这道题就解决了。

---

细节注意事项

• \(long\ long\)一定要开呀。
• 不要写挂呀！！！

---

参考代码

/*--------------------------------
  Code name: crash.cpp
  Author: The Ace Bee
  This code is made by The Ace Bee
--------------------------------*/
#include <cstdio>
#define rg register
#define int long long
#define fileopen(x)                                \
    freopen(x".in", "r", stdin);                \
    freopen(x".out", "w", stdout);
#define fileclose                                \
    fclose(stdin);                              \
    fclose(stdout);
const int mod = 20101009;
const int MAXN = 10000010;
inline int min(int a, int b) { return a < b ? a : b; }
inline int read() {
    int s = 0; bool f = false; char c = getchar();
    while (c < '0' || c > '9') f |= (c == '-'), c = getchar();
    while (c >= '0' && c <= '9') s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
    return f ? -s : s;
}
int vis[MAXN], mu[MAXN];
int num, pri[MAXN], sum[MAXN];
inline void seive() {
    mu[1] = 1;
    for (rg int i = 2; i < MAXN; ++i) {
        if (!vis[i]) mu[i] = -1, pri[++num] = i;
        for (rg int j = 1; j <= num && i * pri[j] < MAXN; ++j) {
            vis[i * pri[j]] = 1;
            if (i % pri[j]) mu[i * pri[j]] = - mu[i];
            else { mu[i * pri[j]] = 0; break; }
        }
    }
    for (rg int i = 1; i < MAXN; ++i)
        sum[i] = (sum[i - 1] + 1ll * i * i % mod * (mu[i] + mod) % mod) % mod;
}
inline int g(int n, int m)
{ return 1ll * n * (n + 1) / 2 % mod * (m * (m + 1) / 2 % mod) % mod; }
inline int f(int n, int m) {
    int res = 0;
    for (rg int i = 1, j; i <= min(n, m); i = j + 1) {
        j = min(n / (n / i), m / (m / i));
        res = (res + 1ll * (sum[j] - sum[i - 1] + mod) * g(n / i, m / i) % mod) % mod;
    }
    return res;
}
inline int solve(int n, int m) {
    int res = 0;
    for (rg int i = 1, j; i <= min(n, m); i = j + 1) {
        j = min(n / (n / i), m / (m / i));
        res = (res + 1ll * (j - i + 1) * (i + j) / 2 % mod * f(n / i, m / i) % mod) % mod;
    }
    return res;
}
signed main() {
//    fileopen("crash");
    seive();
    int n = read(), m = read();
    printf("%lld\n", solve(n, m));
//    fileclose;
    return 0;
}

完结撒花\(qwq\)