与 线段树 2 几乎一模一样，直接放出代码。
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//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<stack>
#include<list>
using namespace std;
#define int long long
const int N=1e5;
struct node{
//tag1:乘法标记,tag2:加法标记
int l,r,value,tag1,tag2;
}t[4*N+1];
int n,q,M,a[N+1];
inline void up(int p){
t[p].value=t[p*2].value+t[p*2+1].value;
t[p].value%=M;
}
void build(int p,int l,int r){
t[p].l=l,t[p].r=r,t[p].tag1=1;
if(l==r)t[p].value=a[l]%M;
else{
int mid=(l+r)/2;
build(p*2,l,mid);
build(p*2+1,mid+1,r);
up(p);
}
}
inline int size(int p){
return t[p].r-t[p].l+1;
}
inline void down(int p){
t[p*2].value=(t[p*2].value * t[p].tag1 + size(p*2)*t[p].tag2)%M;
t[p*2].tag1=t[p*2].tag1*t[p].tag1%M;
t[p*2].tag2=(t[p*2].tag2*t[p].tag1+t[p].tag2)%M;
t[p*2+1].value=(t[p*2+1].value * t[p].tag1 + size(p*2+1)*t[p].tag2)%M;
t[p*2+1].tag1=t[p*2+1].tag1*t[p].tag1%M;
t[p*2+1].tag2=(t[p*2+1].tag2*t[p].tag1+t[p].tag2)%M;
t[p].tag1=1;
t[p].tag2=0;
}
//乘法
void solve1(int p,int l,int r,int k){
if(l<=t[p].l&&t[p].r<=r){
t[p].value*=k;
t[p].value%=M;
t[p].tag1*=k;
t[p].tag1%=M;
t[p].tag2*=k;
t[p].tag2%=M;
}else{
down(p);
int mid=(t[p].l+t[p].r)/2;
if(l<=mid)solve1(p*2,l,r,k);
if(mid<r)solve1(p*2+1,l,r,k);
up(p);
}
}
//加法
void solve2(int p,int l,int r,int k){
if(l<=t[p].l&&t[p].r<=r){
t[p].value+=size(p)*k;
t[p].value%=M;
t[p].tag2+=k;
t[p].tag2%=M;
}else{
down(p);
int mid=(t[p].l+t[p].r)/2;
if(l<=mid)solve2(p*2,l,r,k);
if(mid<r)solve2(p*2+1,l,r,k);
up(p);
}
}
int query(int p,int l,int r){
if(l<=t[p].l&&t[p].r<=r)return t[p].value;
down(p);
int mid=(t[p].l+t[p].r)/2,ans=0;
if(l<=mid)ans+=query(p*2,l,r);
ans%=M;
if(mid<r)ans+=query(p*2+1,l,r);
ans%=M;
return ans;
}
main(){
// freopen("test.in","r",stdin);
// freopen("test.out","w",stdout);
scanf("%lld%lld",&n,&M);
for(int i=1;i<=n;i++)scanf("%lld",a+i);
build(1,1,n);
scanf("%lld",&q);
while(q--){
int op,x,y,k;
scanf("%lld %lld %lld",&op,&x,&y);
switch(op){
case 1:
scanf("%lld",&k);
solve1(1,x,y,k%M);
break;
case 2:
scanf("%lld",&k);
solve2(1,x,y,k%M);
break;
case 3:
printf("%lld\n",query(1,x,y));
break;
}
}
/*fclose(stdin);
fclose(stdout);*/
return 0;
}
