P2522 [HAOI2011]Problem b

$Problem\ b$

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$\mathcal{Description}$
对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

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$最初想法$
$Ans=\sum_{i=1}^{\lfloor \frac{b}{k} \rfloor}\sum_{j=1}^{\lfloor \frac{d}{k} \rfloor}[gcd(i,j)=k]- \sum_{i=1}^{\lfloor \frac{a-1}{k} \rfloor}\sum_{j=1}^{\lfloor \frac{d}{k} \rfloor} [ gcd(i,j)=k ] - \sum_{i=1}^{\lfloor \frac{b}{k} \rfloor}\sum_{j=1}^{\lfloor \frac{c-1}{k} \rfloor} [ gcd(i,j)=k ] + \sum_{i=1}^{\lfloor \frac{a-1}{k} \rfloor}\sum_{j=1}^{\lfloor \frac{c-1}{k} \rfloor} [ gcd(i,j)=k ]$

但是要使用 莫比乌斯反演, 于是回去学习…

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$正解部分$
$N$ 年过后… 莫比乌斯反演学会了.
click here
这里不再赘述 .

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$实现部分$

#include<cstdio>
#include<algorithm>
#define reg register

const int maxn = 50005;

int prime_num;
int mu[maxn];
int mu_sum[maxn];
int prime[maxn];

bool is_prime[maxn];

void sieve(){
        prime_num = 0, mu[1] = 1;
        for(reg int i = 2; i < maxn; i ++){
                if(!is_prime[i]) prime[++ prime_num] = i, mu[i] = -1;
                for(reg int j = 1; j <= prime_num && i*prime[j] < maxn; j ++){
                        int t = i*prime[j];
                        is_prime[t] = 1;
                        if(i%prime[j] == 0){ mu[t] = 0; break ; }
                        mu[t] = -mu[i];
                }
        }
        for(reg int i = 1; i < maxn; i ++) mu_sum[i] = mu_sum[i-1] + mu[i];
}

int Calc(int n, int m){
        int lim = std::min(n, m), s = 0;
        for(reg int l = 1, r; l <= lim; l = r+1){
                r = std::min(n/(n/l), m/(m/l));
                s += (mu_sum[r] - mu_sum[l-1]) * (n/l) * (m/l);
        }
        return s;
}

void Work(){
        int a, b, c, d, k;
        scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
        int Ans = Calc(b/k, d/k) - Calc((a-1)/k, d/k) - Calc(b/k, (c-1)/k) + Calc((a-1)/k, (c-1)/k);
        printf("%d\n", Ans);
}

int main(){
        sieve();
        int T;
        scanf("%d", &T);
        while(T --) Work();
        return 0;
}