P3312 [SDOI2014]数表

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^m\sigma (gcd(i,j)) \\ \sum _{d=1}^n\sigma (d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}[gcd(i,,j)==1] \\ \sum _{d=1}^n\sigma (d)\sum _{d=1}^{\frac{n}{d}}\mu (k)\frac{n}{kd}\frac{m}{kd} \\ t=kd \\ \sum _{t=1}^n\frac{n}{t}\frac{m}{t}\sum _{d\mid t}\sigma (d)\mu (\frac{t}{d}) \end{gathered}$$

这里就是这道题目的最简表达式了，所以我们只要离线处理，先读入所有的询问，然后按照$a$从大到小排序，也就是$\sigma (d)$，不断地插入树状数组中就行了。

接下来的操作就是基本的数论分块。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e5 + 10;int prime[N], mu[N], ans[N], tree[N], cnt;bool st[N];struct ask {int n, m, a, id;bool operator < (ask t) {return a < t.a;}
}q[N];struct node {int sigma, id;bool operator < (node t) {return sigma < t.sigma;}
}a[N];void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i *prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {a[i].id = i;for(int j = i; j < N; j += i)a[j].sigma += i;}
}int lowbit(int x) {return x & (-x);
}void update(int x, int value) {while(x < N) {tree[x] += value;x += lowbit(x);}
}int query(int x) {int ans = 0;while(x) {ans += tree[x];x -= lowbit(x);}return ans;
}int calc(int n, int m) {if(n > m) swap(n, m);int ans= 0;for(int l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans += (n / l) * (m / l) * (query(r) - query(l - 1));}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T = read();for(int i = 1; i <= T; i++) {q[i].n = read(), q[i].m = read(), q[i].a = read(), q[i].id = i;}sort(q + 1, q + T + 1);sort(a + 1, a + N);int pos=0;for(int i = 1; i <= T; i++){while(pos + 1 < N && a[pos + 1].sigma <= q[i].a){pos++;for(int j = 1; j * a[pos].id < N; j++) update(j * a[pos].id, mu[j] * a[pos].sigma);}ans[q[i].id] = calc(q[i].n, q[i].m);}for(int i = 1; i <= T; i++) printf("%d\n", ans[i] & ((1ll << 31) - 1));return 0;
}