Link

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Solution

假如一对值 $(A,B)=(a_0,b_0)$经过区间 $K$中指令的变换得到了 $(A,B)=(k_0a_0+k_1b_0,k_2a_0+k_3b_0)$

假如一对值 $(A,B)=(a_0,b_0)$经过区间 $P$中指令的变换得到了 $(A,B)=(p_0a_0+p_1b_0,p_2a_0+p_3b_0)$

那么，如果我把 $K$和 $P$相连，再给定一组 $(A,B)=(a_0,b_0)$。经过区间 $K+P$中指令的变换之后，得到的应该是
$A = (p_0(k_0a_0+k_1b_0)+p_1(p_2a_0+p_3b_0))\\ B = (p_2(k_0a_0+k_1b_0)+p_3(p_2a_0+p_3b_0))$

整理化简一下，既可以作为线段树更新的式子了，具体请看代码中的merge()

反转操作，我直接把 $A,B$的初值交换就行了，具体请看代码中的maketag()

Code

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
#define mod 1000000007ll
struct SegmentTree
{
    ll L[maxn<<2], R[maxn<<2], rev[maxn<<2];
    vector<ll> k[maxn<<2];
    void maketag(ll o)
    {
        swap(k[o][0],k[o][3]);
        swap(k[o][1],k[o][2]);
        rev[o]^=1;
    }
    void pushdown(ll o)
    {
        if(L[o]==R[o])return;
        if(rev[o])
        {
            maketag(o<<1);
            maketag(o<<1|1);
            rev[o]=0;
        }
    }
    vector<ll> merge(vector<ll> K, vector<ll> P)
    {
        vector<ll> ret(4);
        ret[0] = P[0]*K[0] + P[1]*K[2];
        ret[1] = P[0]*K[1] + P[1]*K[3];
        ret[2] = P[2]*K[0] + P[3]*K[2];
        ret[3] = P[2]*K[1] + P[3]*K[3];
        for(auto& x:ret)x%=mod;
        return ret;
    }
    void pushup(ll o)
    {
        auto v = merge(k[o<<1], k[o<<1|1]);
        ll i; rep(i,0,3)k[o][i]=v[i];
    }
    void build(ll o, ll l, ll r, char* s)
    {
        ll mid(l+r>>1);
        L[o]=l, R[o]=r;
        rev[o]=0;
        k[o].resize(4);
        if(l==r)
        {
            if(s[l]=='A')
            {
                k[o][0]=1;
                k[o][1]=1;
                k[o][2]=0;
                k[o][3]=1;
            }
            else
            {
                k[o][0]=1;
                k[o][1]=0;
                k[o][2]=1;
                k[o][3]=1;
            }
            return;
        }
        build(o<<1,l,mid,s);
        build(o<<1|1,mid+1,r,s);
        pushup(o);
    }
    void Rev(ll o, ll l, ll r)
    {
        ll mid(L[o]+R[o]>>1);
        if(l<=L[o] and r>=R[o]){maketag(o);return;}
        pushdown(o);
        if(l<=mid)Rev(o<<1,l,r);
        if(r>mid)Rev(o<<1|1,l,r);
        pushup(o);
    }
    vector<ll> q(ll o, ll l, ll r)
    {
        pushdown(o);
        ll mid(L[o]+R[o]>>1);
        vector<ll> ret({1,0,0,1});
        if(l<=L[o] and r>=R[o])return k[o];
        if(l<=mid)ret=merge(ret,q(o<<1,l,r));
        if(r>mid)ret=merge(ret,q(o<<1|1,l,r));
        return ret;
    }
}segtree;
char s[maxn];
int main()
{
    ll n=read(), q=read();
    scanf("%s",s+1);
    segtree.build(1,1,n,s);
    while(q--)
    {
        ll type=read(), L=read(), R=read();
        if(type==1)
        {
            segtree.Rev(1,L,R);
        }
        else
        {
            ll a0=read(), b0=read();
            auto k=segtree.q(1,L,R);
            ll A = k[0]*a0 + k[1]*b0, B = k[2]*a0 + k[3]*b0;
            printf("%lld %lld\n",A%mod,B%mod);
        }
    }
    return 0;
}