BZOJ1798: [Ahoi2009]Seq 维护序列seq

传送门

写这道题是为了get一个同时传送乘法下标和加法下标的小技巧，线段树模板题。不多说。

标记名字打错无限智力--

 //BZOJ 1798
 //by Cydiater
 //2016.9.13
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <queue>
 #include <map>
 #include <iomanip>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 using namespace std;
 #define ll unsigned long long
 #define up(i,j,n)        for(int i=j;i<=n;i++)
 #define down(i,j,n)        for(int i=j;i>=n;i--)
 #define FILE "seqb"
 ;
 const ll oo=0x3f3f3f3f;
 inline ll read(){
     ,f=;
     ;ch=getchar();}
     +ch-';ch=getchar();}
     return x*f;
 }
 struct SegTree{
     ll v,delta1,delta2,leftt,rightt;
 }t[MAXN<<];
 ll N,mod,M,op,x,y,Delta1,Delta2,a[MAXN];
 namespace solution{
     inline void downit(int node){
         if(t[node].leftt==t[node].rightt)return;
         ll delta1=t[node].delta1,delta2=t[node].delta2,leftt,rightt;
         t[node].delta1=;t[node].delta2=;
         ){
             t[node<<].delta1*=delta2;t[node<<].delta1%=mod;
             t[node<<].delta2*=delta2;t[node<<].delta2%=mod;
             t[node<<].v*=delta2;t[node<<].delta2%=mod;
             t[node<<|].delta1*=delta2;t[node<<|].delta1%=mod;
             t[node<<|].delta2*=delta2;t[node<<|].delta2%=mod;
             t[node<<|].v*=delta2;t[node<<|].delta2%=mod;
         }
         leftt=t[node<<].leftt;rightt=t[node<<].rightt;
         t[node<<].delta1+=delta1;t[node<<].delta1%=mod;
         t[node<<].v+=(rightt-leftt+)*delta1;t[node<<].v%=mod;
         leftt=t[node<<|].leftt;rightt=t[node<<|].rightt;
         t[node<<|].delta1+=delta1;t[node<<|].delta1%=mod;
         t[node<<|].v+=(rightt-leftt+)*delta1;t[node<<|].v%=mod;
     }
     inline ].v+t[node<<|].v)%mod;}
     inline void updata(int leftt,int rightt,int root){
         downit(root);
         if(leftt>y||rightt<x)                return;
         if(leftt>=x&&rightt<=y){
             if(Delta1){
                 t[root].delta1+=Delta1;
                 t[root].v+=(rightt-leftt+)*Delta1;
                 t[root].delta1%=mod;
                 t[root].v%=mod;
             }
             ){
                 t[root].delta1*=Delta2;t[root].delta1%=mod;
                 t[root].delta2*=Delta2;t[root].delta2%=mod;
                 t[root].v*=Delta2;t[root].v%=mod;
             }
             return;
         }
         ;
         updata(leftt,mid,root<<);
         updata(mid+,rightt,root<<|);
         reload(root);
     }
     inline ll find(int leftt,int rightt,int root){
         downit(root);
         ;
         if(leftt>=x&&rightt<=y)        return t[root].v;
         ;
         )+find(mid+,rightt,root<<|))%mod;
     }
     inline void build(int leftt,int rightt,int root){
         t[root].delta2=;
         if(leftt==rightt){
             t[root].leftt=t[root].rightt=leftt;
             t[root].v=a[leftt];t[root].delta1=;
             return;
         }
         t[root].leftt=leftt;t[root].rightt=rightt;t[root].delta1=;
         ;
         build(leftt,mid,root<<);
         build(mid+,rightt,root<<|);
         reload(root);
     }
     void slove(){
         N=read();mod=read();
         up(i,,N)a[i]=read();
         build(,N,);
         M=read();
         while(M--){
             op=read();x=read();y=read();
             ){
                 Delta1=;Delta2=read();
                 updata(,N,);
             }
             ){
                 Delta1=read();Delta2=;
                 updata(,N,);
             }
             ){
                 printf(,N,));
                 //cout<<find(1,N,1)<<endl;
             }
         }
     }
 }
 int main(){
     //freopen(FILE".in","r",stdin);
     //freopen("input.in","r",stdin);
     //freopen(FILE".out","w",stdout);
     using namespace solution;
     slove();
     ;
 }