清橙A1339. JZPLCM(顾昱洲)

http://www.tsinsen.com/ViewGProblem.page?gpid=A1339

题解：https://blog.csdn.net/LOI_DQS/article/details/51251737

对着题解A掉了。。。然而并不知道为什么要这么转化问题。。。

复杂度nlog^2n级别吧

 #include<cstdio>
 #include<algorithm>
 #include<map>
 #include<cmath>
 #define fi first
 #define se second
 #define md 1000000007
 using namespace std;
 typedef long long LL;
 typedef pair<int,int> P;
 int a[],d[],nxt[],pos[],len;
 bool vis[];
 int prime[];
 bool fir[];
 void dprime(int x,map<int,int> &ma)
 {
     ma.clear();
     if(x<=)    return;
     int i,end=floor(sqrt(x+0.5));
     for(i=;prime[i]<=end;i++)
         while(x!=prime[i])
         {
             if(x%prime[i]==)
             {
                 ma[prime[i]]++;
                 x/=prime[i];
             }
             else
                 break;
         }
     ma[x]++;
 }
 int n,qq;
 namespace S
 {
 #define mid (l+((r-l)>>1))
 #define lc (num<<1)
 #define rc (num<<1|1)
 int dat[];
 void build(int l,int r,int num)
 {
     if(l==r){dat[num]=fir[l]?d[l]:;return;}
     build(l,mid,lc);build(mid+,r,rc);
     dat[num]=LL(dat[lc])*dat[rc]%md;
 }
 int L,R,x;
 void upd(int l,int r,int num)
 {
     if(l==r)    {dat[num]=x;return;}
     if(L<=mid)    upd(l,mid,lc);
     else    upd(mid+,r,rc);
     dat[num]=LL(dat[lc])*dat[rc]%md;
 }
 int que(int l,int r,int num)
 {
     if(L<=l&&r<=R)    return dat[num];
     int ans=;
     if(L<=mid)    ans=LL(ans)*que(l,mid,lc)%md;
     if(mid<R)    ans=LL(ans)*que(mid+,r,rc)%md;
     return ans;
 }
 #undef mid
 #undef lc
 #undef rc
 }
 int poww(int a,int b)
 {
     LL ans=,base=a;
     for(;b;base=base*base%md,b>>=)
         if(b&)
             ans=ans*base%md;
     return ans;
 }
 struct Q
 {
     int l,r,num,ans;
 }q[];
 bool operator<(const Q &a,const Q &b)    {return a.l<b.l||(a.l==b.l&&a.r<b.r);}
 bool cmp(const Q &a,const Q &b)    {return a.num<b.num;}
 int main()
 {
     int i,j,t;map<int,int> tans;
     map<int,int>::iterator it;
     for(i=;i<=;i++)
     {
         if(!vis[i]) prime[++prime[]]=i;
         for(j=;j<=prime[]&&i*prime[j]<=;j++)
         {
             vis[i*prime[j]]=;
             if(i%prime[j]==)    break;
         }
     }
     scanf("%d%d",&n,&qq);
     for(i=;i<=n;i++)
     {
         scanf("%d",&t);dprime(t,tans);
         for(it=tans.begin();it!=tans.end();it++)
             for(j=;j<=it->second;j++)
                 ++len,a[len]=poww(it->first,j),d[len]=it->first;
         pos[i]=len;
     }
     tans.clear();
     for(i=;i<=len;i++)
     {
         if(!tans[a[i]])    fir[i]=;
         else    nxt[tans[a[i]]]=i;
         tans[a[i]]=i;
     }
     S::build(,len,);
     for(i=;i<=qq;i++)    scanf("%d%d",&q[i].l,&q[i].r),q[i].num=i,q[i].l=pos[q[i].l-]+,q[i].r=pos[q[i].r];
     sort(q+,q+qq+);
     for(i=,j=;i<=qq;i++)
     {
         for(;j<q[i].l;)
         {
             if(nxt[j])
             {
                 S::L=nxt[j];S::x=d[nxt[j]];S::upd(,len,);
             }
             ++j;
         }
         S::L=q[i].l;S::R=q[i].r;
         q[i].ans=S::que(,len,);
     }
     sort(q+,q+qq+,cmp);
     for(i=;i<=qq;i++)    printf("%d\n",q[i].ans);
     return ;
 }