抄的题解，不算最优解

#include <cstdio>
#include <cstring>
namespace io
{
const int SIZE=(1<<21)+1;
char ibuf[SIZE],*iS,*iT,obuf[SIZE],*oS=obuf,oT=oS+SIZE-1,c,qu[55];
int f,qr;
// getchar
#define gc() (iSiT?(iT=(iS=ibuf)+fread(ibuf,1,SIZE,stdin),(iSiT?EOF:iS++)):iS++)
// print the remaining part
inline void flush(){fwrite(obuf,1,oS-obuf,stdout);oS=obuf;}
// putchar
inline void putc(char x){oS++=x;if(oSoT)flush();}
// input a signed integer
template <class I>
inline void read(I &x)
{
for(f=1,c=gc();c<'0'||c>'9';c=gc()) if(c'-') f=-1;
for(x=0;c<='9'&&c>='0';c=gc()) x=x10+(c&15);x=f;
}
// print a signed integer
template <class I>
inline void print(I &x)
{
if(!x)putc('0');if(x<0) putc('-'),x=-x;
while(x)qu[++qr]=x%10+'0',x/=10;
while(qr)putc(qu[qr--]);
}
//no need to call flush at the end manually
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using ioread;
using ioputc;
using io::print;
const int N=250010;
const int mod=998244353;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
#define mul(a,b) ((int)(1ll(a)(b)%mod))
struct matrix
{
int dx[4][4];
matrix(){memset(dx,0,sizeof dx);}
matrix friend operator +(matrix a,matrix b)
{
for(int i=0;i<=3;i++)
a.dx[0][i]=add(a.dx[0][i],b.dx[0][i]);
return a;
}
bool friend operator (matrix a,matrix b)
{
for(int i=0;i<=3;i++)
for(int j=0;j<=3;j++)
if(a.dx[i][j]!=b.dx[i][j])
return false;
return true;
}
}I,t,yuy[N<<2],tag[N<<2];
int n,m;
#define ls id<<1
#define rs id<<1|1
void build(int id,int l,int r)
{
tag[id]=I;
if(lr)
{
read(yuy[id].dx[0][0]),read(yuy[id].dx[0][1]),read(yuy[id].dx[0][2]);
yuy[id].dx[0][3]=1;
return;
}
int mid=l+r>>1;
build(ls,l,mid),build(rs,mid+1,r);
yuy[id]=yuy[ls]+yuy[rs];
}
void beecute(int id,matrix &a)
{
matrix bee=matrix();
for(int i=0;i<=3;i++)
for(int j=0;j<=3;j++)
bee.dx[0][i]=add(bee.dx[0][i],mul(yuy[id].dx[0][j],a.dx[j][i]));
yuy[id]=bee;
memset(bee.dx,0,sizeof bee.dx);
for(int i=0;i<=3;i++)
for(int j=0;j<=3;j++)
{
bee.dx[i][j]=add(bee.dx[i][j],mul(tag[id].dx[i][0],a.dx[0][j]));
bee.dx[i][j]=add(bee.dx[i][j],mul(tag[id].dx[i][1],a.dx[1][j]));
bee.dx[i][j]=add(bee.dx[i][j],mul(tag[id].dx[i][2],a.dx[2][j]));
bee.dx[i][j]=add(bee.dx[i][j],mul(tag[id].dx[i][3],a.dx[3][j]));
}
tag[id]=bee;
}
void pushdown(int id)
{
if(tag[id]I) return;
beecute(ls,tag[id]);
beecute(rs,tag[id]);
tag[id]=I;
}
void change(int id,int L,int R,int l,int r)
{
if(lL&&rR){beecute(id,t);return;}
pushdown(id);
int Mid=L+R>>1;
if(r<=Mid) change(ls,L,Mid,l,r);
else if(l>Mid) change(rs,Mid+1,R,l,r);
else change(ls,L,Mid,l,Mid),change(rs,Mid+1,R,Mid+1,r);
yuy[id]=yuy[ls]+yuy[rs];
}
matrix query(int id,int L,int R,int l,int r)
{
if(lL&&rR) return yuy[id];
pushdown(id);
int Mid=L+R>>1;
if(r<=Mid) return query(ls,L,Mid,l,r);
else if(l>Mid) return query(rs,Mid+1,R,l,r);
else return query(ls,L,Mid,l,Mid)+query(rs,Mid+1,R,Mid+1,r);
}
int main()
{
read(n);
for(int i=0;i<=3;i++) I.dx[i][i]=1;
build(1,1,n);
read(m);
for(int op,l,r,v,i=1;i<=m;i++)
{
read(op),read(l),read(r);
if(op7)
{
t=query(1,1,n,l,r);
print(t.dx[0][0]);
putc(' ');
print(t.dx[0][1]);
putc(' ');
print(t.dx[0][2]);
putc('\n');
continue;
}
t=I;
if(op1) t.dx[1][0]=1;
else if(op2) t.dx[2][1]=1;
else if(op3) t.dx[0][2]=1;
else if(op4) read(v),t.dx[3][0]=v;
else if(op==5) read(v),t.dx[1][1]=v;
else read(v),t.dx[3][2]=v,t.dx[2][2]=0;
change(1,1,n,l,r);
}
return 0;
}