【LOJ#573】【LNR#2】单枪匹马（线段树）

题面

LOJ

题解

考虑拿线段树维护这个值，现在的问题就是左右怎么合并，那么就假设最右侧进来的那个分数是\(\frac{x}{y}\)的形式，那么就可以维护一下每一个值的系数，就可以直接合并了。
我代码又臭又长，还写得贼复杂

#include<iostream>
#include<cstdio>
using namespace std;
#define MOD 998244353
#define MAX 1000500
#define lson (now<<1)
#define rson (now<<1|1)
inline int read()
{int x=0;bool t=false;char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')t=true,ch=getchar();while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();return t?-x:x;
}
int n,m,type,a[MAX],N,MX;
struct Num{int a,b,c;};
Num operator+(Num a,Num b){return (Num){(a.a+b.a)%MOD,(a.b+b.b)%MOD,(a.c+b.c)%MOD};}
Num operator*(Num a,int b){return (Num){1ll*a.a*b%MOD,1ll*a.b*b%MOD,1ll*a.c*b%MOD};}
struct Fact{Num a,b;};
Fact operator+(Fact a,int b){return (Fact){a.b*b+a.a,a.b};}
Fact Rev(Fact a){return (Fact){a.b,a.a};}
Fact Value(Fact a,int x,int y)
{int u=(1ll*a.a.a*x+1ll*a.a.b*y+a.a.c)%MOD;int v=(1ll*a.b.a*x+1ll*a.b.b*y+a.b.c)%MOD;return (Fact){(Num){0,0,u},(Num){0,0,v}};
}
struct Node{Fact s,v;}t[MAX<<2];
Node Calc(Node l,Node r)
{Node ret;swap(r.v.a,r.v.b);swap(r.s.a,r.s.b);ret.s=Value(l.v,r.s.a.c,r.s.b.c);Num a=(Num){(1ll*l.v.a.a*r.v.a.a+1ll*l.v.a.b*r.v.b.a)%MOD,(1ll*l.v.a.a*r.v.a.b+1ll*l.v.a.b*r.v.b.b)%MOD,(1ll*l.v.a.a*r.v.a.c+1ll*l.v.a.b*r.v.b.c+l.v.a.c)%MOD};Num b=(Num){(1ll*l.v.b.a*r.v.a.a+1ll*l.v.b.b*r.v.b.a)%MOD,(1ll*l.v.b.a*r.v.a.b+1ll*l.v.b.b*r.v.b.b)%MOD,(1ll*l.v.b.a*r.v.a.c+1ll*l.v.b.b*r.v.b.c+l.v.b.c)%MOD};ret.v=(Fact){a,b};return ret;
}
void Modify(int now,int l,int r,int p)
{if(l==r){t[now].s=(Fact){(Num){0,0,a[l]},(Num){0,0,1}};t[now].v=(Fact){(Num){1,0,0},(Num){0,1,0}}+a[l];return;}int mid=(l+r)>>1;if(p<=mid)Modify(lson,l,mid,p);else Modify(rson,mid+1,r,p);t[now]=Calc(t[lson],t[rson]);
}
Node Query(int now,int l,int r,int L,int R)
{if(L==l&&r==R)return t[now];int mid=(l+r)>>1;if(R<=mid)return Query(lson,l,mid,L,R);if(L>mid)return Query(rson,mid+1,r,L,R);return Calc(Query(lson,l,mid,L,mid),Query(rson,mid+1,r,mid+1,R));
}
int main()
{n=read();m=read();type=read();N=n+m;MX=n;for(int i=1;i<=n;++i)a[i]=read(),Modify(1,1,N,i);for(int i=1,lans=0;i<=m;++i){int opt=read();if(opt==1){int x=read();if(type==1)x^=lans;a[++MX]=x;Modify(1,1,N,MX);}else{int l=read(),r=read();if(type==1)l^=lans,r^=lans;Node u=Query(1,1,N,l,r);printf("%d %d\n",u.s.a.c,u.s.b.c);lans=u.s.a.c^u.s.b.c;}}return 0;
}