[ZJOI2006]书架（树状数组水过）

这道题显然平衡树，splay，treap什么的随便切
然而我不想打，决定水过这道题
把空间开3倍，树状数组维护它前面的树的个数，开个id数组记录位置
找一个数排名直接二分加求前缀和，log^2的搞一搞
把一个数放在顶/低 直接丢在当前顶/低的前后就可以了不然开3倍数组干嘛
c常数小堪比log的平衡树居然还快一些

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(3e5 + 10), INF(2e9), PF(1e5);

IL ll Read(){
    char c = '%'; ll x = 0, z = 1;
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

int n, m, t[_], id[_], a[_];
char opt[10];

IL void Add(RG int x, RG int v){  for(; x <= PF * 3; x += x & -x) t[x] += v;  }

IL int Query(RG int x){  RG int cnt = 0; for(; x; x -= x & -x) cnt += t[x]; return cnt;  }

IL int Find(RG int x){
    RG int l = 1, r = PF * 3;
    while(l < r){
        RG int mid = (l + r) >> 1;
        if(Query(mid) >= x) r = mid;
        else l = mid + 1;
    }
    return l;
}

int main(RG int argc, RG char *argv[]){
    n = Read(); m = Read();
    for(RG int i = 1; i <= n; ++i) a[PF + i] = Read(), Add(PF + i, 1), id[a[PF + i]] = PF + i;
    while(m--){
        scanf(" %s", opt); RG int x = Read(), y, p, q;
        if(opt[0] == 'T'){
            p = Find(1);
            a[id[x]] = 0; Add(id[x], -1);
            id[x] = p - 1;
            a[id[x]] = x; Add(id[x], 1);
        }
        else if(opt[0] == 'B'){
            p = Find(n);
            a[id[x]] = 0; Add(id[x], -1);
            id[x] = p + 1;
            a[id[x]] = x; Add(id[x], 1);
        }
        else if(opt[0] == 'I'){
            y = Read(); q = Query(id[x]);
            p = Find(q + y); RG int t = a[p];
            swap(a[p], a[id[x]]); swap(id[x], id[t]);
        }
        else if(opt[0] == 'A') printf("%d
", Query(id[x]) - 1);
        else printf("%d
", a[Find(x)]);
    }
    return 0;
}