BZOJ3529: [Sdoi2014]数表(莫比乌斯反演 树状数组)

题意

题目链接

sol

首先不考虑\(a\)的限制

我们要求的是

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

用常规的套路可以化到这个形式

\[\sum_{d = 1}^n \sigma (d) \sum_{k = 1}^{\frac{n}{d}} \mu(k) \frac{n}{kd} \frac{m}{kd}\]

设\(kd = t\)

那么

\(\sum_{t = 1}^n \left\lfloor \frac{n}{t} \right\rfloor \left\lfloor \frac{m}{t} \right\rfloor \sum_{d \ | t} \sigma(d) \mu(\frac{t}{d})\)

然后按照套路筛出后面的卷积。

但是现在有\(a\)的限制了。。

那么可以直接离线之后树状数组维护一下。

时间复杂度：\(o(t\sqrt{n}logn)\)

约数个数和用埃氏筛两行就筛出来了

#include<bits/stdc++.h>
#define chmax(a, b) (a = (a > b ? a : b))
#define chmin(a, b) (a = (a < b ? a : b))
//#define int long long 
using namespace std;
const int maxn = 3e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int t, mx, mu[maxn], phi[maxn], prime[maxn], tot, vis[maxn], si[maxn], out[maxn], p[maxn];
struct query{
    int n, m, a, id;
    bool operator < (const query &rhs) const {
        return a < rhs.a;
    }
}q[maxn];
int comp(const query &a, const query &b) {
    return a.id < b.id;
}
int comp2(const int &a, const int &b) {
    return si[a] < si[b];
}
void get(int n) {
    for(int i = 1; i <= n; i++) 
        for(int j = i; j <= n; j += i) si[j] += i;
    vis[1] = 1; mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1;
        for(int j = 1; j <= tot && i * prime[j] <= n; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) {mu[i * prime[j]] = 0; break;}
            else mu[i * prime[j]] = -mu[i];
        }
    }
}
#define lb(x) (x & (-x))
int tr[maxn];
void add(int x, int val) {
    while(x <= mx) tr[x] += val, x += lb(x);
}
int sum(int x) {
    int ans = 0;
    while(x) ans += tr[x], x -= lb(x);
    return ans;
}
int solve(int n, int m) {
    int rt = 0, las = 0, now;
    for(int i = 1, nxt; i <= n; i = nxt + 1) {
        nxt = min(m / (m / i), n / (n / i)); 
        now = sum(nxt);
        rt += (n / i) * (m / i) * (now - las);
        las = now;
    }
    return rt;
}
signed main() {
    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;
        if(q[i].n > q[i].m) swap(q[i].n, q[i].m);;
        chmax(mx, q[i].n);
    }
    get(mx);
    for(int i = 1; i <= mx; i++) p[i] = i;
    sort(p + 1, p + mx + 1, comp2);
    sort(q + 1, q + t + 1);
    
    for(int t = 1, d = 1; t <= t; t++) {
        for(; d <= mx && si[p[d]] <= q[t].a; d++) {
            for(int k = p[d]; k <= mx; k += p[d]) {
                if(!mu[k / p[d]]) continue;
                add(k, si[p[d]] * mu[k / p[d]]);
            }
        }
        out[q[t].id] = solve(q[t].n, q[t].m);
    }
    for(int i = 1; i <= t; i++) 
        printf("%d\n", out[i] < 0 ? out[i] + 2147483648 : out[i]);
        //printf("%d\n", out[i] <);
    return 0;
}