Description
给你一个序列,初始为空。资瓷下列操作:
- 在序列末尾加上一个数;
- 查询后 \(L\) 个数中的最大值。
操作总数为 \(m\) , \(1\leq m\leq 200000\)
Solution
单调栈的板子。
Code
//It is made by Awson on 2018.2.7
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 200000;
void read(int &x) {
char ch; bool flag = 0;
for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }
int n, d, t, x, S[N+5], top, a[N+5], cnt;
char ch[5];
int divide(int x) {
int l = 1, r = top, ans;
while (l <= r) {
int mid = (l+r)>>1;
if (S[mid] >= x) r = mid-1, ans = mid;
else l = mid+1;
}
return ans;
}
void work() {
scanf("%d%d", &n, &d);
for (int i = 1; i <= n; i++) {
scanf("%s%d", ch, &x);
if (ch[0] == 'A') {
a[++cnt] = x = (1ll*x+1ll*t)%d;
while (top >= 1 && a[S[top]] <= x) --top;
S[++top] = cnt;
}else writeln(t = (x == 0 ? 0 : a[S[divide(cnt-x+1)]]));
}
}
int main() {
work(); return 0;
}