洛谷$P$2522 $Problem\ b\ [HAOI2011]$ 莫比乌斯反演

正解:莫比乌斯反演

解题报告:

传送门!

首先看到这个显然就想到莫比乌斯反演$QwQ$?

就先瞎搞下呗$QwQ$

$gcd(x,y)=k$,即$gcd(\left \lfloor \frac{x}{k} \right \rfloor,\left \lfloor \frac{y}{k} \right \rfloor)=1$

然后这个,虽然以前推过几次辣,,,但还是重新推下,,,太久没碰这些东西辣/$kel\ kel\ kel$

设$F[k]$表示$gcd(x,y)$为$k$的倍数的数量,显然有$F[k]=\left \lfloor \frac{a}{k} \right \rfloor\cdot \left \lfloor \frac{b}{k} \right \rfloor$

然后再设$f[x]$表示$gcd(x,y)=k$的数量,则显然有$F[k]=\sum_{k|d} f[d]$

然后就直接上莫比乌斯反演就欧克辣,,,$f[k]=\sum_{k|d}\mu (\frac{k}{d})\cdot F[d]$.

然后对于询问$a\leq x\leq b,c\leq y\leq d$,显然求下$x\leq b,y\leq d$ & $x\leq b,y\leq c$ & $x\leq a,y\leq d$ & $x\leq a,y\leq c$,然后瞎容斥下就做完辣,,,$QwQ$

然后看这个复杂度的样子估计还要个数论分块,,,?不知道反正加个数论分块显然不亏嘻嘻,那就加呗$QwQ$

$over$,等下放代码$QAQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)

const int N=+;
int miu[N],pr[N],pr_cnt,sum[N];
bool is_prim[N];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void pre()
{
    sum[]=miu[]=;
    rp(i,,N-)
    {
        if(!is_prim[i])miu[i]=-,pr[++pr_cnt]=i;sum[i]=sum[i-]+miu[i];
        rp(j,,pr_cnt){if(pr[j]*i>N-)break;is_prim[pr[j]*i]=;if(!(i%pr[j])){miu[i*pr[j]]=;break;}else miu[i*pr[j]]=-miu[i];}
    }
}
il int cal(ri x,ri y,ri z)
{
    x/=z;y/=z;int ret=;
    for(ri i=,j;i<=min(x,y);i=j+){j=min(x/(x/i),y/(y/i));ret+=1ll*(sum[j]-sum[i-])*(x/i)*(y/i);}
    return ret;
}

int main()
{
//    freopen("2522.in","r",stdin);freopen("2522.out","w",stdout);
    pre();ri T=read();
    while(T--){ri a=read()-,b=read(),c=read()-,d=read(),k=read();printf("%d\n",cal(a,c,k)+cal(b,d,k)-cal(b,c,k)-cal(a,d,k));}
    return ;
}