luogu P1198 最大数

传送门

起手线段树肯定可以做

但是更优的做法也有 更好的利用了题面的性质

就是单调栈

单调栈功能没有单调队列那么强大 但是非常适合这道题

修改的时候按照单调栈的原则插入

查询的时候暴力二分

由于栈里是递减的 所以找一个最左边的满足要求的就行

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<iomanip>
#define itn int
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 100006;
int T,n;
ll mod,ans;
struct node {
    int idx;
    ll x;
}stk[N];

int top;
char cmd[5];

void insert(ll x) {
    while(top && stk[top].x < x) top--;
    stk[++top].x = x,stk[top].idx = ++n;
}

void query(int L) {
    int l = 1,r = top;
    while(l < r) {
        int m = l+r >> 1;
        if(stk[m].idx < n-L+1) l = m+1;
        else r = m;
    }
    printf("%lld
",ans = stk[r].x);
}

int main() {
    T = read(),mod = read();
    while(T--) {
        scanf("%s",cmd);
        if(cmd[0] == 'Q') query(read());
        if(cmd[0] == 'A') insert((read()+ans)%mod);
    }
}