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
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + ;
int n, m, tim, L, R, tot, sz, qsz;
int sum[maxn], cnt[maxn], ans[maxn], now[maxn];
int a[maxn], b[maxn];
struct node {
int l, r, id, t;
node() {}
node(int l, int r, int id, int t): l(l), r(r), id(id), t(t) {}
} qu[maxn];
struct node1 {
int pos, x, y;
node1() {}
node1(int pos, int x, int y): pos(pos), x(x), y(y) {}
} c[maxn];
int cmp(node a, node b) {
if (a.l / sz == b.l / sz) {
if (a.r / sz == b.r / sz) return a.t < b.t;
return a.r < b.r;
}
return a.l < b.l;
}
void add(int val) {
cnt[sum[val]]--;
sum[val]++;
cnt[sum[val]]++;
}
void del(int val) {
cnt[sum[val]]--;
sum[val]--;
cnt[sum[val]]++;
}
void change(int pos, int x) {
if (L <= pos && pos <= R) {
del(now[pos]);
add(x);
}
now[pos] = x;
}
int main() {
scanf("%d%d", &n, &m);
sz = (int)pow(n, 0.66666667);
for (int i = ; i <= n ; i++) {
scanf("%d", &a[i]);
b[++tot] = a[i];
now[i] = a[i];
}
qsz = tim = ;
for (int i = ; i <= m ; i++) {
int op;
scanf("%d", &op);
if (op == ) {
int l, r;
scanf("%d%d", &l, &r);
qu[++qsz] = node(l, r, qsz, tim);
} else {
int pos, x;
scanf("%d%d", &pos, &x);
b[++tot] = x;
c[++tim] = node1(pos, x, now[pos]);
now[pos] = x;
}
}
sort(qu + , qu + qsz + , cmp);
sort(b + , b + tot + );
tot = unique(b + , b + tot + ) - b;
for (int i = ; i <= n ; i++)
now[i] = lower_bound(b + , b + tot + , a[i]) - b;
for (int i = ; i <= tim ; i++) {
c[i].x = lower_bound(b + , b + tot + , c[i].x) - b;
c[i].y = lower_bound(b + , b + tot + , c[i].y) - b;
}
tim = ;
for (int i = ; i <= qsz ; i++) {
while(L > qu[i].l) add(now[--L]);
while(R < qu[i].r) add(now[++R]);
while(L < qu[i].l) del(now[L++]);
while(R > qu[i].r) del(now[R--]);
while(tim < qu[i].t) tim++, change(c[tim].pos, c[tim].x);
while(tim > qu[i].t) change(c[tim].pos, c[tim].y), tim--;
int mex = ;
while(cnt[mex] > ) mex++;
ans[qu[i].id] = mex;
}
for (int i = ; i <= qsz ; i++ )
printf("%d\n", ans[i]);
return ;
}