解题：UOJ #46 玄学

题面

二进制分组，修改把区间拆开丢在后面，合并的时候归并最后两块；查询在对应节点上二分答案

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=,Max=1e5;
 struct a{int p,k,b;}ope[N*];
 int seq[*N],lef[*N],rig[*N];
 int n,T,t1,t2,t3,t4,op,ans,cnt,tot,mod,oni;
 void Change(int nde,int l,int r,int ll,int rr,int k,int b)
 {
     if(l==r)
     {
         lef[nde]=cnt+;
         if(ll>) ope[++cnt]=(a){ll-,,};
         ope[++cnt]={rr,k,b};
         if(rr<n) ope[++cnt]=(a){n,,};
         rig[nde]=cnt;
     }
     else
     {
         int mid=(l+r)>>,ls=*nde,rs=*nde+,l1,l2,r1,r2;
         if(tot<=mid) Change(ls,l,mid,ll,rr,k,b);
         else Change(rs,mid+,r,ll,rr,k,b);
         if(tot>=r)
         {
               lef[nde]=cnt+,l1=lef[ls],l2=lef[rs],r1=rig[ls],r2=rig[rs];
               while(l1<=r1&&l2<=r2)
               {
                   ope[++cnt]=(a){min(ope[l1].p,ope[l2].p),
                                  1ll*ope[l1].k*ope[l2].k%mod,
                                (1ll*ope[l2].k*ope[l1].b+ope[l2].b)%mod};
                 if(ope[l1].p==ope[l2].p) l1++,l2++;
                 else ope[l1].p<ope[l2].p?l1++:l2++;
             }
             rig[nde]=cnt;
         }
     }
 }
 void Calc(int nde,int tsk)
 {
     int l=lef[nde],r=rig[nde],p=;
     while(l<=r)
     {
         int mid=(l+r)>>;
         if(ope[mid].p>=tsk) p=mid,r=mid-;
         else l=mid+;
     }
     ans=(1ll*ans*ope[p].k+ope[p].b)%mod;
 }
 void Query(int nde,int l,int r,int ll,int rr,int tsk)
 {
     if(l>rr||r<ll)
         return ;
     else if(l>=ll&&r<=rr)
         Calc(nde,tsk);
     else
     {
         int mid=(l+r)>>,ls=*nde,rs=*nde+;
         Query(ls,l,mid,ll,rr,tsk),Query(rs,mid+,r,ll,rr,tsk);
     }
 }
 int main()
 {
     scanf("%d",&oni),oni&=;
     scanf("%d%d",&n,&mod);
     for(int i=;i<=n;i++)
         scanf("%d",&seq[i]);
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&op);
         if(op==)
         {
             scanf("%d%d%d%d",&t1,&t2,&t3,&t4);
             t1^=ans*oni,t2^=ans*oni,tot++;
             Change(,,Max,t1,t2,t3,t4);
         }
         else
         {
             scanf("%d%d%d",&t1,&t2,&t3);
             t1^=ans*oni,t2^=ans*oni,t3^=ans*oni;
             ans=seq[t3],Query(,,Max,t1,t2,t3);
             printf("%d\n",ans);
         }
     }
     return ;
 }