给定一个大小为$n\times n$的表格，初始时$i,j$位置上填的是$f(i,j)=i\times j$，有$m$个操作，每次操作给定$a,b,x,k$，把格子$a,b$上的值改成$x$，求$\sum _{i=1}^k\sum _{j=1}^{k}f(i,j)$。

我们定义，在任何时刻，表格里的值都满足$f(i,j)=f(j,i),j\times f(i,i+j)=(i+j)\times f(i,j)$。
$$\begin{gathered} b\times f(a,a+b)=(a+b)\times f(a,b) \\ \frac{f(a,a+b)}{a\times (a+b)}=\frac{f(a,b)}{a\times b} \\ \frac{f(a,b)}{a\times b}=\frac{f(b,a\%b)}{b\times a\%b} \\ \frac{f(a,b)}{a\times b}=\frac{f(d,d)}{d\times d},d=\gcd (a,b) \\ f(a,b)=\frac{a\times b}{d\times d}f(d,d) \end{gathered}$$

考虑统计答案：
$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{n}f(i,j) \\ \sum _{d=1}^{n}f(d,d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}ij[\gcd (i,j)=1] \\ \sum _{d=1}^{n}f(d,d)\left(\sum _{i=1}^{\frac{n}{d}}i\sum _{j=1}^{i}j[\gcd (i,j)-1]-1\right) \\ \sum _{d=1}^{n}f(d,d)\sum _{i=1}^{\frac{n}{d}}{i}^2\varphi (i) \\ g(n)=\sum _{i=1}^n{i}^2\times \varphi (i) \end{gathered}$$
容易发现$g(n)$可以线性筛得到，所以我们只要动态维护$f(d,d)$的前缀和即可，考虑用分块维护前缀和，满足$\sqrt{n}$修改，$O(1)$查询。

#include <bits/stdc++.h>using namespace std;const int N = 4e6 + 10, mod = 1e9 + 7;int prime[N], phi[N], g[N], f[N], cnt, n, m;int L[N], R[N], id[N], sum[N], lazy[N], block, blocks;bool st[N];inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}void init() {phi[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;phi[i] = i - 1;}for (int j = 1; j <= cnt && 1ll * 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++) {g[i] = add(g[i - 1], 1ll * i * i % mod * phi[i] % mod);f[i] = 1ll * i * i % mod, sum[i] = add(sum[i - 1], f[i]);}block = sqrt(n);for (int i = 1; i <= n; i += block) {L[++blocks] = i, R[blocks] = min(i + block - 1, n);for (int j = L[blocks]; j <= R[blocks]; j++) {id[j] = blocks;}}
}inline int query(int x) {return add(sum[x], lazy[id[x]]);
}inline void update(int x, int v) {int pos = id[x];for (int i = x; i <= R[pos]; i++) {sum[i] = add(sum[i], v);}pos++;while (pos <= blocks) {lazy[pos] = add(lazy[pos], v);pos++;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &m, &n);init();for (int cas = 1, a, b, k; cas <= m; cas++) {long long x;scanf("%d %d %lld %d", &a, &b, &x, &k);int p = __gcd(a, b);update(p, sub(0, f[p]));f[p] = (x / (a / p)) / (b / p) % mod;update(p, f[p]);int ans = 0;for (int l = 1, r; l <= k; l = r + 1) {r = k / (k / l);ans = add(ans, 1ll * g[k / l] * sub(query(r), query(l - 1)) % mod);}printf("%d\n", ans);}return 0;
}