HDU - 1695 GDU

莫比乌斯反演基础。

用rep 去掉重复的对数，rep一定是奇数( 因为有(1,1 ) )

 #include <bits/stdc++.h>
 using namespace std;
 #define fst first
 #define scd second
 #define pb(x) push_back((x))
 #define mkp(x,y) make_pair((x),(y))
 #define ist(x) insert((x))
 typedef long long ll;
 typedef pair<int ,int > pii;
 typedef pair<ll ,ll > pll;
 typedef vector< int > vi;
 ll gcd(ll a,ll b){ return b==?a:gcd(b,a%b);}
 ll qPow(ll a,ll b,ll mod){ ll ret=1ll;while(b){ if(b&) ret=ret*a%mod;a=a*a%mod;b>>=;} return ret; }

 const int maxN=1e5+;
 bool check[maxN+];
 int prime[maxN+];
 int mu[maxN+];
 void init(){
     memset(check,false,sizeof(check));
     mu[]=;
     int tot=;
     for(int i=;i<=maxN;++i){
         if(!check[i]) { prime[tot++]=i; mu[i]=-; }
         for(int j=;j<tot;++j){
             long long k=i*prime[j];// may overflow ,
             if(k>maxN) break;
             check[k]=true;
             if(i%prime[j]==){ mu[k]=; break; }
             else mu[k]=-mu[i];
         }
     }
     /*
     for(int i=1;i<=100;++i)
         printf(" mobi %d : %d\n",i,mu[i]);
     */
 }

 int main(){
     init();
     int T;
     scanf("%d",&T);
     for(int cntT=;cntT<=T;++cntT){
         printf("Case %d: ",cntT);
         int A,B,C,D,K;
         scanf("%d%d%d%d%d",&A,&B,&C,&D,&K);
         if(!K) { puts("");continue; }
         ll ans=0ll;
         ll rep=0ll;
         B/=K,D/=K;
         int bound=min(B,D);
         for(int i=;i<=bound;++i) {
             ans+=1ll*mu[i]*(B/i)*(D/i);
             //printf("after %d : %lld\n",i,ans);
         }
         for(int i=;i<=bound;++i) rep+=1ll*mu[i]*(bound/i)*(bound/i);
         //printf(" \n%lld   %lld\n",ans,rep);
         printf("%lld\n",ans-rep/);
     }
     return ;
 }