洛谷 p3455 [POI2007]ZAP-Queries

题意：

给定$a, b, d$,求满足$1 \leq x \leq a ,  1 \leq y \leq b$且$\gcd(x, y)=d$的二元组$(x, y)$的数量

 

思路：

$\sum_{x=1}^{n}\sum_{y=1}^{b}[\gcd(x, y)=d]$

简化式子：

$\sum_{x=1}^{\lfloor\frac{a}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{b}{d}\rfloor}[\gcd(x, y)=1]$

将$[\gcd(x, y) = 1]$替换为$\varepsilon(\gcd(x, y))$:

$\sum_{x=1}^{\lfloor\frac{a}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{b}{d}\rfloor}\varepsilon(\gcd(x, y))$

展开$\varepsilon$函数得到：

$\sum_{x=1}^{\lfloor\frac{a}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{b}{d}\rfloor}\sum_{k|\gcd(x,y)} \mu(k)$

变换求和顺序，先枚举$k|\gcd(x, y)$得：

$\sum_{k=1}\mu(k)\sum_{x=1}^{\lfloor\frac{a}{d}\rfloor}[k|x]\sum_{y=1}^{\lfloor\frac{b}{d}\rfloor}[k|y]$

已知$1~\lfloor\frac{a}{d}\rfloor$中$k$的倍数有$\lfloor\frac{a}{kd}\rfloor$个：

$\sum_{k=1}\mu(k)\lfloor\frac{a}{kd}\rfloor\lfloor\frac{b}{kd}\rfloor$

Code:

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
// #include <array>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
// #include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 1e5 + 20;
const int M = 1e5 + 20;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int cnt, primes[N], mu[N];
bool st[N];

void get(int n)
{
    mu[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!st[i])
        {
            primes[cnt++] = i;
            mu[i] = -1;
        }
        for (int j = 0; primes[j] <= n / i; j++)
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0)
            {
                mu[i * primes[j]] = 0;
                break;
            }
            mu[i * primes[j]] = -mu[i];
        }
    }
    for(int i = 2; i <= n; i ++) mu[i] += mu[i - 1];
}

int a, b, d;

void solve()
{
    sdd(a, b);
    sd(d);


    ll ans = 0;
    a /= d, b /= d;
    for(int l = 1, r; l <= min(a, b); l = r + 1){
        r = min(a / (a / l), b / (b / l));
        ans = ans + (ll)(mu[r] - mu[l - 1]) * (a / l) * (b / l);
    }

    printf("%lld\n", ans);
}


int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
    freopen("/home/jungu/code/practice/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    sd(T);
    get(100000);
    // init();
    // int cas = 1;
    while (T--)
    {
        // printf("Case #%d:", cas++);
        solve();
    }

    return 0;
}