P3700 [CQOI2017]小Q的表格
给定一个大小为 n × n n \times n n×n的表格，初始时 i , j i, j i,j位置上填的是 f ( i , j ) = i × j f(i, j) = i \times j f(i,j)=i×j，有 m m m个操作，每次操作给定 a , b , x , k a, b, x, k a,b,x,k，把格子 a , b a, b a,b上的值改成 x x x，求 ∑ i = 1 k ∑ j = 1 k f ( i , j ) \sum\limits_{i = 1} ^{k} \sum\limits_{j = 1} ^{k} f(i, j) i=1∑kj=1∑kf(i,j)。
我们定义，在任何时刻，表格里的值都满足
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f(i, j) = f(j, i), j \times f(i, i + j) = (i + j) \times f(i, j)
f(i,j)=f(j,i),j×f(i,i+j)=(i+j)×f(i,j)。
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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)\\
b×f(a,a+b)=(a+b)×f(a,b)a×(a+b)f(a,a+b)=a×bf(a,b)a×bf(a,b)=b×a % bf(b,a % b)a×bf(a,b)=d×df(d,d),d=gcd(a,b)f(a,b)=d×da×bf(d,d)
考虑统计答案：
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\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 \phi(i)\\ g(n) = \sum_{i = 1} ^{n} i ^ 2 \times \phi(i)\\
i=1∑nj=1∑nf(i,j)d=1∑nf(d,d)i=1∑dnj=1∑dnij[gcd(i,j)=1]d=1∑nf(d,d)⎝⎛i=1∑dnij=1∑ij[gcd(i,j)−1]−1⎠⎞d=1∑nf(d,d)i=1∑dni2ϕ(i)g(n)=i=1∑ni2×ϕ(i)
容易发现
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g(n)可以线性筛得到，所以我们只要动态维护
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f(d,d)的前缀和即可，考虑用分块维护前缀和，满足
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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;
}