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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