Machine Learning CodeForces - 940F (带修改的莫队)

You come home and fell some unpleasant smell. Where is it coming from?

You are given an array a. You have to answer the following queries:

1. You are given two integers l and r. Let c_i be the number of occurrences of i ina_l: r, where a_l: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c₀, c₁, ..., c_10⁹}
2. You are given two integers p to x. Change a_p to x.

The Mex of a multiset of numbers is the smallest non-negative integer not in the set.

Note that in this problem all elements of a are positive, which means that c₀ = 0 and 0 is never the answer for the query of the second type.

Input

The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.

The second line of input contains n integers — a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Each of the next q lines describes a single query.

The first type of query is described by three integers t_i = 1, l_i, r_i, where 1 ≤ l_i ≤ r_i ≤ n — the bounds of the subarray.

The second type of query is described by three integers t_i = 2, p_i, x_i, where 1 ≤ p_i ≤ n is the index of the element, which must be changed and 1 ≤ x_i ≤ 10⁹ is the new value.

Output

For each query of the first type output a single integer  — the Mex of{c₀, c₁, ..., c_10⁹}.

Example

Input

10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 8

Output

2
3
2

Note

The subarray of the first query consists of the single element — 1.

The subarray of the second query consists of four 2s, one 3 and two 1s.

The subarray of the fourth query consists of three 1s, three 2s and one 3.

 

注意离散化 ，更换点也要加入数组里面进行离散化

 

 1 #include 
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 2e5 + 10;
 5 int n, m, tim, L, R, tot, sz, qsz;
 6 int sum[maxn], cnt[maxn], ans[maxn], now[maxn];
 7 int a[maxn], b[maxn];
 8 struct node {
 9     int l, r, id, t;
10     node() {}
11     node(int l, int r, int id, int t): l(l), r(r), id(id), t(t) {}
12 } qu[maxn];
13 struct node1 {
14     int pos, x, y;
15     node1() {}
16     node1(int pos, int x, int y): pos(pos), x(x), y(y) {}
17 } c[maxn];
18 int cmp(node a, node b) {
19     if (a.l / sz == b.l / sz) {
20         if (a.r / sz == b.r / sz) return a.t < b.t;
21         return a.r < b.r;
22     }
23     return a.l < b.l;
24 }
25 void add(int val) {
26     cnt[sum[val]]--;
27     sum[val]++;
28     cnt[sum[val]]++;
29 }
30 void del(int val) {
31     cnt[sum[val]]--;
32     sum[val]--;
33     cnt[sum[val]]++;
34 }
35 void change(int pos, int x) {
36     if (L <= pos && pos <= R) {
37         del(now[pos]);
38         add(x);
39     }
40     now[pos] = x;
41 }
42 int main() {
43     scanf("%d%d", &n, &m);
44     sz = (int)pow(n, 0.66666667);
45     for (int i = 1 ; i <= n ; i++) {
46         scanf("%d", &a[i]);
47         b[++tot] = a[i];
48         now[i] = a[i];
49     }
50     qsz = tim = 0;
51     for (int i = 1 ; i <= m ; i++) {
52         int op;
53         scanf("%d", &op);
54         if (op == 1) {
55             int l, r;
56             scanf("%d%d", &l, &r);
57             qu[++qsz] = node(l, r, qsz, tim);
58         } else {
59             int pos, x;
60             scanf("%d%d", &pos, &x);
61             b[++tot] = x;
62             c[++tim] = node1(pos, x, now[pos]);
63             now[pos] = x;
64         }
65     }
66     sort(qu + 1, qu + qsz + 1, cmp);
67     sort(b + 1, b + tot + 1);
68     tot = unique(b + 1, b + tot + 1) - b;
69     for (int i = 1 ; i <= n ; i++)
70         now[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
71     for (int i = 1 ; i <= tim ; i++) {
72         c[i].x = lower_bound(b + 1, b + tot + 1, c[i].x) - b;
73         c[i].y = lower_bound(b + 1, b + tot + 1, c[i].y) - b;
74     }
75     tim = 0;
76     for (int i = 1 ; i <= qsz ; i++) {
77         while(L > qu[i].l) add(now[--L]);
78         while(R < qu[i].r) add(now[++R]);
79         while(L < qu[i].l) del(now[L++]);
80         while(R > qu[i].r) del(now[R--]);
81         while(tim < qu[i].t) tim++, change(c[tim].pos, c[tim].x);
82         while(tim > qu[i].t) change(c[tim].pos, c[tim].y), tim--;
83         int mex = 1;
84         while(cnt[mex] > 0) mex++;
85         ans[qu[i].id] = mex;
86     }
87     for (int i = 1 ; i <= qsz ; i++ )
88         printf("%d\n", ans[i]);
89     return 0;
90 }

 

转载于:https://www.cnblogs.com/qldabiaoge/p/9359174.html