题目已经没有了
思路：
莫队+分块
首先有一个结论：所有的答案都在0到n之间，用反正法就能证明，所以所有大于n的数都可以看成n
离线，对询问区间进行莫队，再对答案的范围0到n进行分块
复杂度（n+2*m)√n
代码：
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head const int N = 2e5 + ;
int a[N], cnt[N], bl[N], ans[N], l[], r[], block[], blo, n;
struct node {
int l, r, bl, id;
bool operator < (const node & t) const {
if(bl == t.bl) return r < t.r;
else bl < t.bl;
}
}Q[N];
void add(int x) {
if(!cnt[x]) block[bl[x]]++;
cnt[x] ++;
}
void del(int x) {
cnt[x]--;
if(!cnt[x]) block[bl[x]]--;
}
int query() {
int i;
for (i = bl[]; i <= bl[n]; i++) if(block[i] != r[i] - l[i] + ) break;
for (int j = l[i]; j <= r[i]; j++) if(!cnt[j]) return j;
}
int main() {
int m;
while(~scanf("%d%d", &n, &m)) {
for (int i = ; i <= n; i++) {
scanf("%d", &a[i]);
if(a[i] >= n) a[i] = n;
}
blo = sqrt(n+);
for (int i = ; i <= n; i++) bl[i] = i/blo + ;
for (int i = bl[]; i <= bl[n]; i++) {
l[i] = (i-)*blo;
r[i] = min(i*blo-, n);
}
blo = sqrt(n);
for (int i = ; i <= m; i++) {
scanf("%d%d", &Q[i].l, &Q[i].r);
Q[i].bl = (Q[i].l - )/blo + ;
Q[i].id = i;
}
sort(Q+, Q++m);
mem(cnt, );
mem(block, );
int l = , r = ;
for (int i = ; i <= m; i++) {
while(r < Q[i].r) r++, add(a[r]);
while(r > Q[i].r) del(a[r]), r--;
while(l < Q[i].l) del(a[l]), l++;
while(l > Q[i].l) l--, add(a[l]);
ans[Q[i].id] = query();
}
for (int i = ; i <= m; i++) printf("%d\n", ans[i]);
}
return ;
}