hdu1695莫比乌斯反演模板题

hdu1695

求1<=i<=n&&1<=j<=m,gcd(i,j)=k的（i，j）的对数

最后的结果f(k)=Σ(1<=x<=n/k)mu[x]*(n/(x*k))*(m/(x*k))

遍历的复杂度是O(n/k)，按理来说是会t的，但是这题过了，更好的办法是用分块降低到O(sqrt(n/k))

详细介绍请看：链接

这题要（i，j）和（j，i）算重复的，所以要减去

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

//#pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define mod 1000000007

#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pil pair<int,ll>

#define pii pair<int,int>

#define ull unsigned long long

#define base 1000000000000000000

#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

int mu[N],prime[N],sum[N];

bool mark[N];

void init()

{

    mu[]=;

    int cnt=;

    for(int i=;i<N;i++)

    {

        if(!mark[i])prime[++cnt]=i,mu[i]=-;

        for(int j=;j<=cnt;j++)

        {

            int t=i*prime[j];

            if(t>N)break;

            mark[t]=;

            if(i%prime[j]==){mu[t]=;break;}

            else mu[t]=-mu[i];

        }

    }

    for(int i=;i<N;i++)sum[i]=sum[i-]+mu[i];

}

int main()

{

    init();

    int t,cnt=;

    scanf("%d",&t);

    while(t--)

    {

        ll a,b,c,d,k;

        scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k);

        if(!k)

        {

            printf("Case %d: 0\n",++cnt);

            continue;

        }

        if(b>d)swap(b,d);

        b/=k,d/=k;

        ll ans=,ans1=;

        for(ll i=,last=;i<=b;i=last+)

        {

            last=min(b/(b/i),d/(d/i));

            ans+=(ll)(sum[last]-sum[i-])*(b/i)*(d/i);

        }

        for(ll i=,last=;i<=b;i=last+)

        {

            last=b/(b/i);

            ans1+=(ll)(sum[last]-sum[i-])*(b/i)*(b/i);

        }

        printf("Case %d: %lld\n",++cnt,ans-ans1/);

    }

    return ;

}

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