HDU

题目：给你a,b,c,d,k，让你求gcd(x,y)=k的个数，a<=x<=b,c<=y<=d，a=1，b=1

gcd(x,y)和gcd(y,x)只算一次

思路：莫比乌斯就行了

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fconst int maxn=1e5+50;bool check[maxn];int prime[maxn];int mu[maxn];void Moblus(int n){    mm(check,false);    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]]=true;            if(i%prime[j]==0){                mu[i*prime[j]]=0;                break;            }            else{                mu[i*prime[j]]=-mu[i];            }        }    }}int main(){    Moblus(maxn-1);    int T,cas=0;    int a,b,c,d,k;    scanf("%d",&T);    while(T--){        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        if(k==0){            printf("Case %d: 0\n",++cas);            continue;        }        if(b>d)            swap(b,d);        b/=k;        d/=k;        LL ans1=0,ans2=0;        for(int i=1;i<=b;i++)            ans1+=(LL)mu[i]*(b/i)*(d/i);        for(int i=1;i<=b;i++)            ans2+=(LL)mu[i]*(b/i)*(b/i);        ans1-=ans2/2;        printf("Case %d: %lld\n",++cas,ans1);    }    return 0;}