HDU

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==0?a:gcd(b,a%b);}
ll qPow(ll a,ll b,ll mod){ ll ret=1ll;while(b){ if(b&1) ret=ret*a%mod;a=a*a%mod;b>>=1;} return ret; }

const int maxN=1e5+5;
bool check[maxN+10];
int prime[maxN+10];
int mu[maxN+10];
void init(){
    memset(check,false,sizeof(check));
    mu[1]=1;
    int tot=0;
    for(int i=2;i<=maxN;++i){
        if(!check[i]) { prime[tot++]=i; mu[i]=-1; }
        for(int j=0;j<tot;++j){
            long long k=i*prime[j];// may overflow , 
            if(k>maxN) break;
            check[k]=true;
            if(i%prime[j]==0){ mu[k]=0; break; }
            else mu[k]=-mu[i];
        }
    }
    /*
    for(int i=1;i<=100;++i)
        printf(" mobi %d : %d
",i,mu[i]);
    */
}

int main(){
    init();
    int T;
    scanf("%d",&T);
    for(int cntT=1;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("0");continue; }
        ll ans=0ll;
        ll rep=0ll;
        B/=K,D/=K;
        int bound=min(B,D);
        for(int i=1;i<=bound;++i) {
            ans+=1ll*mu[i]*(B/i)*(D/i);
            //printf("after %d : %lld
",i,ans);
        }
        for(int i=1;i<=bound;++i) rep+=1ll*mu[i]*(bound/i)*(bound/i);
        //printf(" 
%lld   %lld
",ans,rep);
        printf("%lld
",ans-rep/2);
    }
    return 0;
}