bzoj 2962 Sequence operation line segment tree

This question card is constant. .

First of all, c is relatively small, so dp transfer can be considered. For merged subsequences, just enumerate the cross-intervals directly

For the reversal operation, it should be noted that only the odd position will become the opposite number

For the increase operation, you can consider abstracting it into a combinatorial mathematical problem:

For (a1+b)*(a2+b)+(a1+b)*(a3+b)+(a2+b)*(a3+b), the disassembly means that each item will be related to two different items. Multiply,

Converting to the above picture is equivalent to selecting k left and ck right ones in each group, and then the combinatorial math can be calculated.

code:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define P 19940417
#define N 100005
#define zuo o<<1,l,mid
#define you o<<1|1,mid+1,r
int i,j,L[N<<2],R[N<<2],a,b,n,m;
long long lin3[N],lin2[N],lin[N],c,ci[25],ci2[25],jiabj[N<<2],fbj[N<<2],C[N][25],v[N<<2][25];
char op[5];
void up(int o)
{ 
    int ll=o<<1;
    int rr=o<<1|1;
    for(int i=1;i<=20;i++)
    { v[o][i]=0;
    for(int j=0;j<=i;j++)
    {
        v[o][i]+=(v[ll][j]*v[rr][i-j])%P;
        
    }v[o][i]%=P;
    }       
}
void jian(int o,int l,int r)
{
    L[o]=l;R[o]=r;
    v[o][0]=1;
    if(l==r)
    {
          v[o][1]=lin[l];
          return;       
    }
    int mid=(l+r)>>1; 
    jian(zuo);
    jian(you);
    up(o);  
}
void push(int o)
{int i,j;
    int ll=o<<1,rr=o<<1|1;
if(fbj[o])  
{
    fbj[o]=0;
    fbj[ll]^=1;
    fbj[rr]^=1;jiabj[ll]*=-1,jiabj[rr]*=-1;
    for(i=1;i<=20;i++)if(i&1)v[ll][i]*=-1,v[rr][i]*=-1;
}
    if(jiabj[o]!=0)
    {   
for(i=1;i<=20;i++)ci2[i]=(ci2[i-1]*jiabj[o])%P;
for(i=1;i<=20;i++)
for(j=0;j<=i;j++)if(R[ll]-L[ll]+1-j>=0)lin[i]=(lin[i]+v[ll][j]*C[R[ll]-L[ll]+1-j][i-j]%P*ci2[i-j]%P)%P;
    for(i=1;i<=20;i++)v[ll][i]=lin[i],lin[i]=0;
jiabj[ll]+=jiabj[o];
jiabj[ll]%=P;
for(i=1;i<=20;i++)
for(j=0;j<=i;j++)
if(R[rr]-L[rr]+1-j>=0)lin[i]=(lin[i]+v[rr][j]*C[R[rr]-L[rr]+1-j][i-j]%P*ci2[i-j]%P)%P;
    for(i=1;i<=20;i++)v[rr][i]=lin[i],lin[i]=0;
jiabj[rr]+=jiabj[o];
jiabj[rr]%=P;
jiabj[o]=0;     
    }
}
void gai(int o,int l,int r)
{
    if(a<=l&&r<=b)
    {int i,j;
        if(op[0]=='Q')
        {
            for(i=1;i<=c;i++)
            for(j=0;j<=i;j++)
            lin2[i]+=v[o][j]*lin3[i-j],lin2[i]%=P;
            for(i=1;i<=c;i++)
            lin3[i]=lin2[i],lin2[i]=0;
        }
        if(op[0]=='R')
        {   
            jiabj[o]*=-1;
            for(i=1;i<=20;i++)if(i&1)v[o][i]*=-1;
            fbj[o]^=1;
        }
        if(op[0]=='I')
        {
    for(i=1;i<=20;i++)
for(j=0;j<=i;j++)
if(R[o]-L[o]+1-j>=0)lin[i]=(lin[i]+v[o][j]*C[R[o]-L[o]+1-j][i-j]%P*ci[i-j]%P)%P;
 
    for(i=1;i<=20;i++)v[o][i]=lin[i],lin[i]=0;
jiabj[o]+=c;
        jiabj[o]%=P;    
        }   
    return ;    
    }
    push(o);
    int mid=(l+r)>>1;
    if(a<=mid)gai(zuo);
    if(b>mid)gai(you);
     
    up(o);  
}
long long  duru()
{
    long long x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int main()
{ci2[0]=1;
C[0][0]=1;
    scanf("%d%d",&n,&m);
    for(j=1;j<=n;j++)
    {C[j][0]=1;
    for(i=1;i<=20;i++)
    C[j][i]=(C[j-1][i-1]+C[j-1][i])%P;  
}
    for(i=1;i<=n;i++)
    {
    lin[i]=duru();  
    }
    jian(1,1,n);
    memset(lin,0,sizeof(lin));
    while(m--)
    {
        scanf("%s",&op);
      if(op[0]=='I')
      {
        a=duru();b=duru();c=duru();
        ci[0]=1;
        for(i=1;i<=20;i++)
        {   
            ci[i]=(ci[i-1]*c)%P;                 
        }
        gai(1,1,n);  for(i=1;i<=20;i++)lin[i]=0;     
      }
      if(op[0]=='R')
      { 
        a=duru();b=duru();
        gai(1,1,n);for(i=1;i<=20;i++)lin[i]=0;
      }
      if(op[0]=='Q')
      {
        a=duru();b=duru();c=duru();
      lin3[0]=lin[0]=1;
        gai(1,1,n);
        printf("%lld\n",(lin3[c]+P)%P);
            for(i=1;i<=20;i++)lin3[i]=lin[i]=0;
      }
    }
}