我这里用了个类lazytag的写法，关于二分我这里用的都是stl

hack数据都能过，但是最后一个点错了

数据

10 2
1 1 1 1 1 1 1 1 1 1
M 1 10 999
A 4 5 999

代码如下

#include<bits/stdc++.h>
#define int long long
using namespace std;
int n, arr[1001000], add[1001000], block, t, st[1000100], ed[1000100], pos[1001000], brr[1001000], q;
void build() {
	block = sqrt(n);
	t = n/block;
	if(n%block) ++t;
	for(int i=1; i<=t; ++i) {
		st[i] = (i-1)*block+1;
		ed[i] = i*block;
	}
	ed[t]=n;
	for(int i=1; i<=t; ++i) {
		sort(brr+st[i],brr+ed[i]+1);
	}
	for(int i=1; i<=n; ++i) {
		pos[i] = (i-1)/block+1;
	}
}
void Add(int L, int R, int W) {
	if(pos[L] == pos[R]) {
		int index = pos[L];
		for(int i=L; i<=R; ++i) {
			arr[i] = arr[i]+W;
		}
		for(int i=st[index]; i<=ed[index]; ++i) {
			brr[i] = arr[i];
		}
		sort(brr+st[index],brr+ed[index]+1);
		return;
	}
	else
	{
		int left = pos[L], right = pos[R];
		for(int i=L; i<=ed[left]; ++i) {
			arr[i] = arr[i]+W;
		}
		for(int i=st[left];i<=ed[left];++i) {
			brr[i] = arr[i];
		}
		sort(brr+st[left], brr+ed[left]+1);
		for(int i=left+1;i<right;++i) {
			add[i] += W;
		}
		for(int i=st[right]; i<=R; ++i) {
			arr[i] = arr[i]+W;
		}
		for(int i=st[right];i<=ed[right];++i) {
			brr[i] = arr[i];
		}
		sort(brr+st[right], brr+ed[right]+1);
	}
}
int que(int L, int R, int C) {
	if(pos[L] == pos[R]) {
		int sum = 0;
		for(int i=L; i<=R; ++i) {
			if(arr[i] >= C) {
				sum++;
			}
		}
		return sum;
	} else {
		int left = pos[L], right = pos[R], sum = 0, tempC;
		tempC = C-add[left];
		for(int i=L; i<=ed[left]; ++i) {
			if(arr[i] >= tempC) {
				sum++;
			}
		}
		for(int i=left+1; i<right; ++i) {
			tempC = C-add[i];
			int u = lower_bound(brr+st[i],brr+ed[i]+1,tempC)-brr;
			sum+=max(0ll, ed[i]-u+1);
		}
		tempC = C-add[right];
		for(int i=st[right]; i<=R; ++i) {
			if(arr[i] >= tempC) {
				sum++;
			}
		}
		return sum;
	}
}
signed main() {
	ios::sync_with_stdio(0);cin.tie(0);
	cin >> n >> q;
	for(int i=1; i<=n; ++i) {cin >> arr[i];brr[i]=arr[i];}
	build();
	for(int i=1; i<=q; ++i){
		char ch; cin >> ch;
		if(ch == 'M') {
			int L,R,W; cin >> L >> R >> W;
			Add(L,R,W);
		} else {
			int L,R,C; cin >> L >> R >> C;
			cout << que(L,R,C) << endl;
		}
	}
	return 0;
}