bzoj2301 [HAOI2011]Problem b （莫比乌斯函数）

首先可以想到分为四个前缀区间进行加加减减，考虑[1,a]，[1,b]这组：

∑i=1a∑j=1b[gcd(i,j)==k]

=∑i=1a/k∑j=1b/k[gcd(i,j)==1]

=∑i=1a/k∑j=1b/k∑t|i,t|jμ(t)

令n=a/k,m=b/k,则原式可改写为

∑t=1min(n,m)μ(t)[nt][mt]

这部分可以在O(max{b,d}−−−−−−−−√)分段求

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<set>#include<map>#include<queue>#include<stack>#include<vector>#include<bitset>using namespace std;typedef long long LL;int miu[51000];int prim[51000],primm;bool valid[51000];void mobius(int N){    miu[1]=1;    for (int i=2;i<=N;i++)    {        if (!valid[i]) prim[++primm]=i,miu[i]=-1;        for (int j=1;j<=primm&&i*prim[j]<=N;j++)        {            valid[i*prim[j]]=1;            if (i%prim[j]==0)            {                miu[i*prim[j]]=0;                break;            }            else miu[i*prim[j]]=-miu[i];        }    }}int a,b,c,d,n,k;LL sum[51000];LL F(int n,int m){    n/=k,m/=k;    LL res=0;    for (int i=1;i<=min(n,m);i++)    {        int pos=i;        i=min(n/(n/i),m/(m/i));        LL a=n/i,b=m/i;        res+=(sum[i]-sum[pos-1])*a*b;    }    return res;}int main(){    mobius(50000);    for (int i=1;i<=50000;i++)        sum[i]=sum[i-1]+(LL)miu[i];    scanf("%d",&n);    while(n--)    {        scanf("%d %d %d %d %d",&a,&b,&c,&d,&k);        LL ans=F(b,d)-F(a-1,d)-F(b,c-1)+F(a-1,c-1);        printf("%lld\n",ans);    }    return 0;}