区间最多约数

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N=1e7+5,M=3200;vector<int> prime;int pri[M]={0};int mx,ans;int pricnt;int li,ri;void f(){    pri[0]=pri[1]=1;    for(int i=2;i<M;i++)       if(!pri[i])       {            prime.push_back(i);            for(int j=i*i;j<M;j+=i)                pri[j]=1;       }}void dfs(int cur,int cnt,int pos,int L,int R){    if(cnt>mx&&pos>=li)        mx=cnt,ans=pos;    else if(cnt==mx&&ans>pos&&pos>=li)        ans=pos;    if(L==R&&L>pos)       dfs(cur,cnt<<1,pos*L,1,1);    for(int i=cur;i<pricnt;i++)    {        if(prime[i]>R) return;        int curp=prime[i],l=L-1,r=R,LL=L,num=pos,sum=cnt,m=1;        while(1)        {            m++;            sum+=cnt;            l/=curp;            r/=curp;            LL/=curp;            if(l==r) break;            num*=prime[i];            dfs(i+1,sum,num,LL,r);        }        if(cnt<mx/(1<<m))return;    }}int main(){    f();    pricnt=prime.size();//    cout<<pricnt<<endl;    while(~scanf("%d%d",&li,&ri))    {        if(li==1&&ri==1)        {            puts("[1,1] 1 1\n1");            continue;        }        mx=1;        dfs(0,1,1,li,ri);        printf("[%d,%d] %d %d\n",li,ri,ans,mx);        vector<int> out;        int tmpp=sqrt(ans+0.5);        for(int j=1;j<=tmpp;j++)            if(ans%j==0)            {                out.push_back(j);                out.push_back(ans/j);            }        if(tmpp*tmpp==ans)out.pop_back();        sort(out.begin(),out.end());        for(int i=0;i<out.size();i++)        {            if(i) printf(" ");            printf("%d",out[i]);        }        puts("");    }}