为什么我N开到2e6就过了，开1e6会RE两个点

//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
#include<vector>
#define ll long long
#define lx (x<<1)
#define rx (x<<1|1)
#define db double
#define pb push_back
#define mp make_pair
#define pa pair<int,int>
#define mid ((l+r)>>1)
#define mem(a) memset(a,0,sizeof(a))
#define memf(a) memset(a,127,sizeof(a))//浮点，int数最大 
#define memm(a) memset(a,0x3f,sizeof(a))//1e9
#define memn(a) memset(a,0xcf,sizeof(a))//int最小 
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ct(a) (__builtin_popcount(a))
#define lowbit(i) (i&(-i))
using namespace std;
const int N=2e6+7,M=1e6+10;
const double eps=1e-9,pi=acos(-1.0);
int tt;
int n,m,k,cnt=0;
int a[N];
vector<int>v[20105];
int lazy[20150];
void gai(int l,int r,int x,int w){
	for(int i=l;i<=r;i++){
		a[i]+=w;
	}
	vector<int>().swap(v[x]);
	for(int i=(x-1)*k;i<x*k;i++){
		v[x].pb(a[i]);
	}
	sort(v[x].begin(),v[x].end());
}
int qu(int l,int r,int x,int c){
	if(lazy[x]!=0){
		for(int i=(x-1)*k;i<x*k;i++){
			a[i]+=lazy[x];
		}
		for(int i=0;i<v[x].size();i++){
			v[x][i]+=lazy[x];
		}
		lazy[x]=0;
	}
	int sum=0;
	for(int i=l;i<=r;i++){
		if(a[i]>=c)sum++;
	}
	return sum;
}
int check(int x,int c){
	int l=0,r=v[x].size()-1;
	while(l<=r){
		if(v[x][mid]+lazy[x]>=c)r=mid-1;
		else l=mid+1;
	}
	return v[x].size()-r-1;
}
void solve(){
	scanf("%d%d",&n,&m);
	k=sqrt(n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		v[i/k+1].pb(a[i]);
	}
	int mx=n/k+1;
	for(int i=1;i<=mx;i++){
		sort(v[i].begin(),v[i].end());
	}
	char op[2];
	while(m--){
		scanf("%s",op);
		if(op[0]=='M'){
			int l,r,w;
			scanf("%d%d%d",&l,&r,&w);
			int sl=l/k+1;
			int sr=r/k+1;
			if(sl==sr){
				gai(l,r,sl,w);
				continue;
			}
			gai(l,sl*k-1,sl,w);
			gai((sr-1)*k,r,sr,w);
			for(int i=sl+1;i<=sr-1;i++){
				lazy[i]+=w;
			}
		}
		if(op[0]=='A'){
			int l,r,c;
			scanf("%d%d%d",&l,&r,&c);
			int sl=l/k+1;
			int sr=r/k+1;
			int ans=0;
			if(sl==sr){
				ans+=qu(l,r,sl,c);
				printf("%d\n",ans);
				continue;
			}
			ans+=qu(l,sl*k-1,sl,c);
			ans+=qu((sr-1)*k,r,sr,c);
			for(int i=sl+1;i<=sr-1;i++){
				ans+=check(i,c);
			}
			printf("%d\n",ans);
		}
	}
}
/*
5 3
1 2 3 4 5
A 1 5 4
M 3 5 1
A 1 5 4
*/
signed main(){
//	scanf("%d",&tt);
	tt=1;
	while(tt--){
		solve();
	} 
	return 0;
}
/*
1
1 2 3 4 5
1 2 2 3 3
*/