HDU 1695 GCD（莫比乌斯反演）

Description
求1<=x<=b，1<=y<=d且gcd(x,y)=k的对数，(x,y)和(y,x)视为一组
Input
第一行一整数T表示用例组数，每组用例五个整数a,b,c,d,k
(T<=3000,a=c=1,b,d<=100000)
Output
对于每组用例，输出1<=x<=b，1<=y<=d且gcd(x,y)=k的对数
Sample Input
2
1 3 1 5 1
1 11014 1 14409 9
Sample Output
Case 1: 9
Case 2: 736427
Solution
设f(x,y,k)为1<=i<=x,1<=j<=y且gcd(i,j)=k的(i,j)对数，则有

假设n<=m，则ans=f(n,m,k)-f(n,n,k)/2
Code

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define maxn 1111111 typedef long long ll;bool check[maxn];int prime[maxn],mu[maxn];void Moblus(int n){    memset(check,0,sizeof(check));    mu[1]=1;    int tot=0;    for(int i=2;i<=n;i++)    {        if(!check[i])        {            prime[tot++]=i;            mu[i]=-1;        }        for(int j=0;j<tot;j++)        {            if(i*prime[j]>n)break;            check[i*prime[j]]=1;            if(i%prime[j]==0)            {                mu[i*prime[j]]=0;                break;            }            else mu[i*prime[j]]=-mu[i];        }    }}int T,a,b,c,d,k,res=1;int main(){    Moblus(100000);    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        if(k==0)        {            printf("Case %d: 0\n",res++);            continue;        }        int n=b/k,m=d/k,t=min(n,m);        ll ans=0,plus=0;        for(int i=1;i<=n&&i<=m;i++)        {            ans+=1ll*mu[i]*(n/i)*(m/i);            plus+=1ll*mu[i]*(t/i)*(t/i);        }        printf("Case %d: %I64d\n",res++,ans-plus/2);    }    return 0;}