#莫比乌斯反演，整除分块，线性筛#洛谷 2522 bzoj 2301 problem b

题目

求$a\le x\le b,c\le y\le d$中$gcd(x,y)=k$的个数

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分析

其实只要做了洛谷 3455就会理解很多
蒟蒻的题解
然后这道题可以用前缀和表达出来
$TotalAns=Ans(b,d)-Ans(a-1,d)-Ans(b,c-1)+Ans(a-1,c-1)$
那么如何求一个呢
$$f(k)=\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)=k]$$
$$F(d)=\sum _{k|d}f(k)=\lfloor \frac{n}{d}\rfloor \lfloor \frac{m}{d}\rfloor$$
根据莫比乌斯反演可以得到
$$f(k)=\sum _{k|d}\mu (\lfloor \frac{d}{k}\rfloor )\times F(d)$$
$$Ans=\sum _{k|d}\mu (\lfloor \frac{d}{k}\rfloor )\times F(d)$$
枚举$\lfloor \frac{d}{k}\rfloor$，那么也就是
A n s = ∑ i = 1 m i n { n , m } μ ( i ) × ⌊ n i ∗ d ⌋ × ⌊ m i ∗ d ⌋ Ans=\sum_{i=1}^{min\{n,m\}}\mu(i)\times\lfloor\frac{n}{i*d}\rfloor\times\lfloor\frac{m}{i*d}\rfloor Ans=i=1∑min{n,m}​μ(i)×⌊i∗dn​⌋×⌊i∗dm​⌋
接着用整除分块一波推就好了

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代码

#include <cstdio>
#define rr register
#define min(a,b) ((a)<(b))?(a):(b)
#define N 50000
using namespace std;
typedef unsigned uit;
uit cnt,mu[N+1],v[N+1],prime[N+1];
inline signed iut(){
    rr uit ans=0; rr char c=getchar();
    while (c<48||c>57) c=getchar();
    while (c>47&&c<58) ans=(ans<<3)+(ans<<1)+c-48,c=getchar();
    return ans;
}
inline void iiiii(){//线性筛莫比乌斯函数
    mu[1]=1;
    for (rr uit i=2;i<=N;++i){
        if (!v[i]) v[i]=i,prime[++cnt]=i,mu[i]=-1;
        for (rr uit j=1;j<=cnt&&prime[j]*i<=N;++j){
            v[i*prime[j]]=prime[j];
            if (i%prime[j]==0) break;
            mu[i*prime[j]]=-mu[i];
        }
    }
    for (rr uit i=2;i<=N;++i) mu[i]+=mu[i-1];
}
inline void print(uit ans){
    if (ans>9) print(ans/10);
    putchar(ans%10+48);
}
inline signed answ(uit n,uit m,uit k){
    rr uit ans=0,t=min(n,m);
    for (rr uit l=1,r;l<=t;l=r+1){
        r=min(n/(n/l),m/(m/l));//确定区间
        ans+=(n/l/k)*(m/l/k)*(mu[r]-mu[l-1]);//计算答案
    }
    return ans;
}
signed main(){
    iiiii();
    for (rr uit t=iut();t;--t){
        rr uit a=iut(),b=iut(),c=iut(),d=iut(),k=iut();
        rr uit ans=answ(b,d,k)+answ(a-1,c-1,k)-answ(b,c-1,k)-answ(a-1,d,k);
        if (ans) print(ans); else putchar(48); putchar(10);
    }
    return 0;
}