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2 years ago
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 200010;
int m; // m个操作
int p; // mod p
// 线段树求最大值模板
#define ls (u << 1)
#define rs (u << 1 | 1)
#define mid ((l + r) >> 1)
struct Node {
int l, r, len;
int v;
} tr[N << 2];
void pushup(int u) {
tr[u].v = max(tr[ls].v, tr[rs].v);
}
void build(int u, int l, int r) {
tr[u].l = l, tr[u].r = r;
if (l == r) return;
build(ls, l, mid), build(rs, mid + 1, r);
}
int query(int u, int L, int R) {
int l = tr[u].l, r = tr[u].r;
if (l >= L && r <= R) return tr[u].v;
if (l > R || r < L) return 0;
return max(query(ls, L, R), query(rs, L, R));
}
void modify(int u, int x, int v) {
int l = tr[u].l, r = tr[u].r;
if (l == r) {
tr[u].v = v;
return;
}
if (x <= mid)
modify(u << 1, x, v);
else
modify(u << 1 | 1, x, v);
pushup(u);
}
int n, last;
int main() {
// 加快读入
ios::sync_with_stdio(false), cin.tie(0);
cin >> m >> p;
// 单点修改,求区间最大值
// ① 初始化线段树,最多m次操作最多m个数区间范围[1, m]
build(1, 1, m);
int x;
char op;
while (m--) {
cin >> op >> x;
if (op == 'A') {
// 对于节点++n进行修改,值=(最后一次的last查询值 + x )%p
modify(1, ++n, ((LL)last + x) % p);
} else {
// u = 1:从根节点开始查询
// 查询序列中最后x个数的最大数: 查询[n - x + 1, n]内的最大值
last = query(1, n - x + 1, n);
printf("%d\n", last);
}
}
return 0;
}