Adnan and the Burned drivers

題目鏈接：Adnan and the Burned drivers

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顯然維護正反字符串的哈希值，那麼直接判斷即可。

由於有修改，所以用線段樹維護。

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AC代碼：

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+10;
const int mod=1e9+3,base=131;
int n,m,s1[N<<2],s2[N<<2],pw[N]={1},q1,q2; char str[N],ch[5];
#define mid (l+r>>1)
void change(int p,int l,int r,int x,int v){
	if(l==r){s1[p]=s2[p]=v; return ;}
	if(x<=mid)	change(p<<1,l,mid,x,v);
	else	change(p<<1|1,mid+1,r,x,v);
	s1[p]=(s1[p<<1]*pw[r-mid]+s1[p<<1|1])%mod;
	s2[p]=(s2[p<<1]+s2[p<<1|1]*pw[mid-l+1])%mod;
}
int ask(int p,int l,int r,int ql,int qr,int k){
	if(l==ql&&r==qr)	return k?s1[p]:s2[p];
	if(qr<=mid)	return ask(p<<1,l,mid,ql,qr,k);
	else if(ql>mid)	return ask(p<<1|1,mid+1,r,ql,qr,k);
	else if(k)	return (ask(p<<1,l,mid,ql,mid,k)*pw[qr-mid]+ask(p<<1|1,mid+1,r,mid+1,qr,k))%mod;
	else	return (ask(p<<1,l,mid,ql,mid,k)+ask(p<<1|1,mid+1,r,mid+1,qr,k)*pw[mid-ql+1])%mod;
}
inline void solve(){
	cin>>n>>m; scanf("%s",str+1);
	for(int i=1;i<=n;i++)	change(1,1,n,i,str[i]);
	for(int i=1,op,l,r;i<=m;i++){
		scanf("%lld %lld",&op,&l);
		if(op==1)	scanf("%s",ch),change(1,1,n,l,str[l]=ch[0]);
		else{
			scanf("%lld",&r);
			if(r==l){puts("Adnan Wins");	continue;}		
			else if((r-l+1)&1LL) q1=ask(1,1,n,l,mid-1,1),q2=ask(1,1,n,mid+1,r,0);
			else q1=ask(1,1,n,l,mid,1),q2=ask(1,1,n,mid+1,r,0);
			puts(q1==q2?"Adnan Wins":"ARCNCD!");
		}
	}
}
signed main(){
	for(int i=1;i<=1e5;i++)	pw[i]=pw[i-1]*base%mod;
	int T; cin>>T; while(T--) solve();
	return 0;
}