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:
- You are given two integers l and r. Let ci be the number of occurrences of i inal: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, ..., c109}
- You are given two integers p to x. Change ap 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, ..., an (1 ≤ ai ≤ 109).
Each of the next q lines describes a single query.
The first type of query is described by three integers ti = 1, li, ri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.
The second type of query is described by three integers ti = 2, pi, xi, where 1 ≤ pi ≤ n is the index of the element, which must be changed and 1 ≤ xi ≤ 109 is the new value.
Output
For each query of the first type output a single integer — the Mex of{c0, c1, ..., c109}.
Example
10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 82
3
2Note
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 <bits/stdc++.h>
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 }
来源：oschina
链接：https://my.oschina.net/u/4326830/blog/3893984