题解 P3455 【[POI2007]ZAP-Queries】

题目链接

Solution [POI2007]ZAP-Queries

题目大意：多组数据，每次给定\(a,b,d\)，询问\(\sum_{i=1}^{a}\sum_{j=1}^{b}[gcd(i,j)=d]\)

莫比乌斯反演

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解析：

\[\begin{aligned}ans &= \sum_{i=1}^{a}\sum_{j=1}^{b}[gcd(i,j)=d] \\ &=\sum_{i=1}^{\lfloor \frac{a}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{b}{d} \rfloor}[gcd(i,j)=1]\end{aligned}\]

方便起见，令\(n=\lfloor \frac{a}{d} \rfloor\)，\(m=\lfloor \frac{b}{d} \rfloor\)
\[\begin{aligned}ans &= \sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=1] \\ &= \sum_{i=1}^{n}\sum_{j=1}^{m}\epsilon(gcd(i,j))\end{aligned}\]

因为\(\mu * 1=\epsilon\)，且\(d \mid gcd(i,j) \Longleftrightarrow d\mid i,d \mid j\)

\[\begin{aligned}ans &= \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d\mid i,d \mid j}\mu(d) \\ &= \sum_{d=1}^{min(n,m)}\mu(d)\sum_{i=1}^{n}\sum_{j=1}^{m}[d \mid i][d \mid j] \\ &= \sum_{d=1}^{min(n,m)}\mu(d)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor \end{aligned}\]

筛出\(\mu\)，整除分块求解即可

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int maxn = 1e5;
int vis[maxn],mu[maxn];
inline int sum(int a,int b){return mu[b] - mu[a - 1];}
vector<int> pri;
inline void sieve(){
    mu[1] = 1;
    for(int i = 2;i < maxn;i++){
        if(!vis[i]){
            pri.push_back(i);
            mu[i] = -1;
        }
        for(int x : pri){
            if(1ll * x * i >= maxn)break;
            vis[i * x] = 1;
            if(i % x){
                mu[i * x] = mu[i] * mu[x];
            }else{
                mu[i * x] = 0;
                break;
            }
        }
    }
    for(int i = 1;i < maxn;i++)mu[i] += mu[i - 1];
}
int t,a,b,n,m,d;
inline void solve(){
    cin >> a >> b >> d;
    n = a / d,m = b / d;
    int ans = 0;
    for(int l = 1,r;l <= min(n,m);l = r + 1){
        r = min(n / (n / l),m / (m / l));
        ans += sum(l,r) * (n / l) * (m / l);
    }
    cout << ans << '\n';
}
int main(){ 
    sieve();
    cin >> t;
    while(t--)solve();
    return 0;
}