【jzoj3623】【SDOI2014】【数表】【莫比乌斯反演】

题目大意

解题思路

ans=∑ni=1∑mj=1[f(gcd(i,j))<=a]其中f[x]=∑d|xd

经过一波莫比乌斯反演，转换主题之后，ans=∑nt=1⌊nt⌋⌊mt⌋g(t)其中g(t)=∑d|tf(d)∗μ(⌊td⌋)

可以用埃氏筛法求出f，用欧拉筛法求出mu，对f从小到大排序，对询问按a从小到大离线，每次询问之前把符合条件的f枚举另一个数组合成t，在树状数组中添加。询问时分快处理，在树状数组中查询即可。

code

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define LF double#define LL long long#define ULL unsigned int#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)#define fr(i,j) for(int i=begin[j];i;i=next[i])using namespace std;int const mn=1e5+9,mq=2*1e4+9,inf=1e9+7;int T,lim,b[mn],ss[mn],mu[mn],tag[mn];ULL f[mn],t[mn],anss[mn];struct rec{    int n,m,a,p;};rec a[mq];bool cmp(int i,int j){    return f[i]<f[j];}bool cm2(rec i,rec j){    return i.a<j.a;}void add(int x,int y){    while(x<=lim){        t[x]+=y;        x+=x&(-x);    }}ULL qury(int x){    LL ans=0;    while(x>0){        ans+=t[x];        x-=x&(-x);    }    return ans;}int main(){    //freopen("table.in","r",stdin);    //freopen("table.out","w",stdout);    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d",&T);    lim=1e5;    fo(i,1,lim)fo(j,1,lim/i)f[i*j]+=i;    mu[1]=1;    fo(i,2,lim){        if(!tag[i])ss[++ss[0]]=i,mu[i]=-1;        fo(j,1,ss[0]){            if(i*ss[j]>lim)break;            tag[i*ss[j]]=1;            mu[i*ss[j]]=-mu[i];            if(i%ss[j]==0){mu[i*ss[j]]=0;break;}        }    }    fo(i,1,lim)b[i]=i;    sort(b+1,b+lim,cmp);    fo(i,1,T)scanf("%d%d%d",&a[i].n,&a[i].m,&a[i].a),a[i].p=i;    sort(a+1,a+T+1,cm2);    int tmp=1;    fo(cas,1,T){        int tt=a[cas].a;        for(;(tmp<=lim)&&(f[b[tmp]]<=tt);tmp++){            int t1=lim/b[tmp];            fo(j,1,t1)add(b[tmp]*j,f[b[tmp]]*mu[j]);        }        int n=a[cas].n,m=a[cas].m,i=1;        if(n>m)swap(n,m);        ULL ans=0;        while(i<=n){            int t1=(n/i),t2(m/i),t3=n/t1,t4=m/t2,j=min(t3,t4);            ans+=t1*t2*(qury(j)-qury(i-1));            i=j+1;        }        anss[a[cas].p]=ans;    }    fo(i,1,T)printf("%llu\n",(unsigned long long)((anss[i]>2147483647)?anss[i]-2147483648:anss[i]));    return 0;}