Problem Five:[双亲数/POI Zap（多组）]

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=21352

∑i=1a∑j=1bgcd(i,j)==d=∑i=1a/d∑j=1b/dgcd(i,j)==1=∑i=1a/d∑j=1b/d∑d|i,d|jmin(i,j)μ(d)=∑d1=1min(a/d,b/d)μ(d1)∗(a/d/d1)∗(b/d/d1)

这题还要去一下重

#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int N=1e5+100;typedef __int64 LL;int pri[N],pn,mu[N],p[N];void init(){    mu[1]=1;    memset(p,0,sizeof(p));    int n=100000;    pn=0;    for(int i=2;i<=n;i++){        if(!p[i]){            pri[pn++]=i;            mu[i]=-1;        }        for(int j=0;j<pn;j++){            if((LL)i*pri[j]>n)break;            p[i*pri[j]]=1;            if(i%pri[j]==0){                mu[i*pri[j]]=0;break;            }            else {                mu[i*pri[j]]=-mu[i];            }        }    }//    for(int i=1;i<=10;i++){//        printf("%d %d %d\n",i,p[i],mu[i]);//    }    for(int i=1;i<=n;i++){        mu[i]+=mu[i-1];    }}LL f(int k1,int k2){    if(k1>k2)swap(k1,k2);    return (LL)(k2-k1)*k1+(LL)(k1-1)*k1/2+k1;}int main(){    #ifdef DouBi    freopen("in.cpp","r",stdin);    #endif // DouBi    int a,b,c,d,k;    init();    int cas=0;    int T;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;        }        b/=k;d/=k;        //printf("%d %d\n",b,d);        int r=min(b,d);        LL ans=0;        while(r>0){            //printf("%d ",r);            int k1=b/r;            int k2=d/r;            int l1=b/(k1+1)+1;            int l2=d/(k2+1)+1;            int l=max(l1,l2);            ans+=(LL)f(k1,k2)*(mu[r]-mu[l-1]);            //printf("%d %d %I64d\n",l,r,ans);            r=l-1;        }        printf("Case %d: %I64d\n",++cas,ans);    }    return 0;}