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 8
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 <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 }