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字尾平衡樹模板題
用平衡樹維護每一個字尾的排名
關鍵在於查詢兩個字尾的大小
可以用二分加hash，複雜度 \(log^2n\) 插入
或者：
每次前面插入一個字元，先比較兩個字尾第一個字元的大小
而後面的大小我們已經在平衡樹上維護好了
像這樣分配權值

給樹上每個子樹一個實數權值區間 \([l,r]\)，這個點權值為 \(mid=\frac{l+r}{2}\)
左子樹 \([l,mid]\) 右子樹 \([mid,r]\)

那麼可以做到 \(O(1)\) 比較
只需要選擇一個樹高 \(log\) 的樹(treap/替罪羊樹)使得滿足精度要求即可
最後用線段樹維護一下每個幻影的最小的字尾

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(1e6 + 5);
const double alpha(0.75);

int ls[maxn], rs[maxn], rt, tot, size[maxn], que[maxn], cnt;
double val[maxn];
int mn[maxn << 1], id[maxn], n, m, p[maxn], pos[maxn];
char s[maxn];

void Dfs(int u) {
    if (!u) return;
    Dfs(ls[u]), que[++cnt] = u, Dfs(rs[u]);
}

int Build(int l, int r, double vl, double vr) {
    if (l > r) return 0;
    double midv;
    int mid, o;
    mid = (l + r) >> 1, o = que[mid], midv = (vl + vr) * 0.5;
    ls[o] = rs[o] = 0, val[o] = midv;
    ls[o] = Build(l, mid - 1, vl, midv);
    rs[o] = Build(mid + 1, r, midv, vr);
    size[o] = size[ls[o]] + size[rs[o]] + 1;
    return o;
}

int Rebuild(int x, double vl, double vr) {
    cnt = 0, Dfs(x);
    return Build(1, cnt, vl, vr);
}

inline int Compare(int x, int y) {
    return (s[x] ^ s[y]) ? s[x] < s[y] : val[id[x - 1]] < val[id[y - 1]];
}

int Insert(int &x, double vl, double vr, int ps) {
    double midv;
    int ret;
    midv = (vl + vr) * 0.5;
    if (!x) {
        x = ++tot, val[x] = midv, pos[x] = ps, size[x] = 1;
        return x;
    }
    if (alpha * size[x] < max(size[ls[x]], size[rs[x]])) x = Rebuild(x, vl, vr);
    if (Compare(ps, pos[x])) ret = Insert(ls[x], vl, midv, ps);
    else ret = Insert(rs[x], midv, vr, ps);
    size[x] = size[ls[x]] + size[rs[x]] + 1;
    return ret;
}

void Modify(int x, int l, int r, int ps) {
    int mid;
    if (l == r) mn[x] = l;
    else {
        mid = (l + r) >> 1;
        ps <= mid ? Modify(x << 1, l, mid, ps) : Modify(x << 1 | 1, mid + 1, r, ps);
        mn[x] = val[id[p[mn[x << 1]]]] <= val[id[p[mn[x << 1 | 1]]]] ? mn[x << 1] : mn[x << 1 | 1];
    }
}

int Query(int x, int l, int r, int ql, int qr) {
    int mid, ret, v;
    if (ql <= l && qr >= r) return mn[x];
    mid = (l + r) >> 1, ret = -1, v;
    if (ql <= mid) ret = Query(x << 1, l, mid, ql, qr);
    if (qr > mid) {
        v = Query(x << 1 | 1, mid + 1, r, ql, qr);
        ret = (ret == -1 || val[id[p[v]]] < val[id[p[ret]]]) ? v : ret;
    }
    return ret;
}

int main() {
    int i, l, r, k, ans, type, len;
    char op;
    scanf("%d%d%d%d %s", &n, &m, &len, &type, s + 1);
    reverse(s + 1, s + len + 1), val[0] = -1;
    for (i = 1; i <= len; ++i) id[i] = Insert(rt, 0, 1, i);
    for (i = 1; i <= n; ++i) scanf("%d", &p[i]), Modify(1, 1, n, i);
    ans = 0;
    for (i = 1; i <= m; ++i) {
        scanf(" %c", &op);
        if (op == 'I') {
            scanf(" %d", &k);
            if (type) k = (k ^ ans);
            s[++len] = k + 'a';
            id[len] = Insert(rt, 0, 1, len);
        }
        else if (op == 'C') {
            scanf("%d%d", &l, &r);
            p[l] = r, Modify(1, 1, n, l);
        }
        else scanf("%d%d", &l, &r), printf("%d\n", ans = Query(1, 1, n, l, r));
    }
    return 0;
}