还没学莫队，感觉纯分块也能做，但只有40分。

#include<bits/stdc++.h>
#define fo(x,y,z) for(int (x)=(y);(x)<=(z);(x)++)
#define fu(x,y,z) for(int (x)=(y);(x)>=(z);(x)--)
typedef long long ll;
using namespace std;
inline ll qr()
{
	char ch=getchar();ll x=0,f=1;
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=(x<<3)+(x<<1)+(ch^48);
	return x*f;
}
#define qr qr()
const int Ratio=0;
const int N=2000005;
const int maxi=INT_MAX;
ll n,m,cnt;
ll a[N],sv[N],st[N],ed[N],pre[N],las[N],bl[N];
void mdf(ll p,ll v)
{
	ll cox=a[p],no=las[cox];
	a[p]=v;
	if(no==p)
		las[cox]=pre[p];
	else if(no>p)
	{
		while(pre[no]!=p)
			no=pre[no];
		pre[no]=pre[p];
		fo(i,st[bl[no]],ed[bl[no]])
			sv[i]=pre[i];
		sort(sv+st[bl[no]],sv+ed[bl[no]]+1);
	}
	no=las[v];
	if(no<p)
	{
		las[v]=p;
		pre[p]=no;
		fo(i,st[bl[p]],ed[bl[p]])
			sv[i]=pre[i];
		sort(sv+st[bl[p]],sv+ed[bl[p]]+1);
	}
	else
	{
		while(pre[no]>p)
			no=pre[no];
		pre[p]=pre[no];
		pre[no]=p;
		fo(i,st[bl[p]],ed[bl[p]])
			sv[i]=pre[i];
		sort(sv+st[bl[p]],sv+ed[bl[p]]+1);
		fo(i,st[bl[no]],ed[bl[p]])
			sv[i]=pre[i];
		sort(sv+st[bl[no]],sv+ed[bl[no]]+1);
	}
}
ll Acheron(ll l,ll r)
{
	ll ans=0;
	if(bl[l]==bl[r])
	{
		fo(i,l,r)
			if(pre[i]<l)
				ans++;
		return ans;
	}
	else
	{
		fo(i,l,ed[bl[l]])
			if(pre[i]<l)
				ans++;
		fo(i,st[bl[r]],r)
			if(pre[i]<l)
				ans++;
		fo(i,bl[l]+1,bl[r]-1)
		{
			ll ls=st[i],rs=ed[i],re=0,mid;
			while(ls<=rs)
			{
				mid=(ls+rs)>>1;
				if(sv[mid]<l)
					ls=mid+1,re=mid-st[i]+1;
				else
					rs=mid-1;
			}
			ans+=re;
		}
		return ans;
	}
}
signed main()
{
	n=qr,m=qr;ll sq=sqrt(n);
	fo(i,1,n)
	{
		a[i]=qr;
		pre[i]=las[a[i]];
		las[a[i]]=i;
	}
	fo(i,1,sq)
		st[i]=n/sq*(i-1)+1,ed[i]=n/sq*i;
	ed[sq]=n;
	fo(i,1,sq)
		fo(j,st[i],ed[i])
			bl[j]=i;
	fo(i,1,n)
		sv[i]=pre[i];
	fo(i,1,sq)
		sort(sv+st[i],sv+ed[i]+1);
	while(m--)
	{
		string s;
		cin>>s;
		ll a=qr,b=qr;
		if(s=="R")
			mdf(a,b);
		else
			printf("%lld\n",Acheron(a,b));
	} 
	return Ratio;
}