BZOJ3585 mex

 1 /**************************************************************
 2     Problem: 3585
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:9756 ms
 7     Memory:7068 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13  
14 using namespace std;
15 const int N = 200005;
16 const int B = 505;
17  
18 int n, Q;
19 int sz, cnt_block;
20 int a[N], b[N], cnt[N], st[B];
21 int l, r, block_cnt[N], ans[N];
22  
23 struct query{
24     int l, r, w;
25  
26     inline bool operator < (const query x) const {
27         return b[l] == b[x.l] ? r < x.r : b[l] < b[x.l];
28     }
29 } q[N];
30  
31 inline int read() {
32     int x = 0;
33     char ch = getchar();
34     while (ch < ‘0‘ || ‘9‘ < ch)
35         ch = getchar();
36     while (‘0‘ <= ch && ch <= ‘9‘) {
37         x = x * 10 + ch - ‘0‘;
38         ch = getchar();
39     }
40     return x;
41 }
42  
43 void update(int x, int del) {
44     if (x > n) return;
45     if (cnt[x] == 0) ++block_cnt[b[x]];
46     cnt[x] += del;
47     if (cnt[x] == 0) --block_cnt[b[x]];
48 }
49  
50 int Query() {
51     int i;
52     if (!cnt[0]) return 0;
53     for (i = 1; i <= cnt_block; ++i)
54         if (block_cnt[i] != st[i + 1] - st[i]) break;
55     for (i = st[i]; i < n; ++i)
56     if (!cnt[i]) return i;
57     return n;
58 }
59  
60 int main() {
61     int i;
62     n = read(), Q = read();
63     sz = (int) sqrt(n);
64     if (!sz) sz = 1;
65     for (i = 1; i <= n; ++i) {
66         if (i % sz == 1 || sz == 1) 
67         st[++cnt_block] = i;
68         b[i] = cnt_block;
69     }
70     st[cnt_block + 1] = n + 1;
71  
72     for (i = 1; i <= n; ++i) a[i] = read();
73     for (i = 1; i <= Q; ++i)
74         q[i].l = read(), q[i].r = read(), q[i].w = i;
75     sort(q + 1, q + Q + 1);
76     for (i = l = 1, r = 0; i <= Q; ++i) {
77         for (; l < q[i].l; ) update(a[l++], -1);
78         for (; l > q[i].l; ) update(a[--l], 1);
79         for (; r < q[i].r; ) update(a[++r], 1);
80         for (; r > q[i].r; ) update(a[r--], -1);
81         ans[q[i].w] = Query();
82     }
83     for (i = 1; i <= Q; ++i)
84         printf("%d\n", ans[i]);
85     return 0;
86 }