这题好像有比较巧妙的单调做法，但它都强制在线了直接线段树尊重一下，按题意模拟即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#define int long long
#define lc (x<<1)
#define rc ((x<<1)|1)
#define mid ((l+r)>>1)
using namespace std;
const int N=2e5+17;
int n,m,sum[4*N],k;
void change(int x,int l,int r,int from,int to,int v){
	if(from<=l&&r<=to){
		sum[x]=v;return;
	}
	if(from<=mid) change(lc,l,mid,from,to,v);
	if(to>mid) change(rc,mid+1,r,from,to,v);
	sum[x]=max(sum[lc],sum[rc]);
}
int query(int x,int l,int r,int from,int to){
	if(from<=l&&r<=to){
		return sum[x];
	}int res=0;
	if(from<=mid) res=max(res,query(lc,l,mid,from,to));
	if(to>mid) res=max(res,query(rc,mid+1,r,from,to));
	return res;
}int top=0,ans=0;
signed main(){
	//freopen("C.in","r",stdin);
	//freopen("c.out","w",stdout);
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n >> m;
	for(int i=1;i<=n;i++){
		char op;int x;
		cin >> op >> x;
		if(op=='A'){
			top++;//记录当前的最后位置 
			change(1,1,n,top,top,(x+ans)%m);//鄙人不会写单点修改说是 
		}else{
			int res=query(1,1,n,top-x+1,top);
			ans=res;cout << res << '\n';//处理强制在线 
		}
	}
	return 0;
}