HDU 5869 Different GCD subarray Query (GCD type preprocessing + Tree array maintenance)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5869

Ask you about the GCD kind of continuous sequence between l~r.

First fixed right end, pretreatment gcd different as far as possible on the right position (at this time gcd species not more than loga[i] species).

Preprocessing GCD The following code, feel really a bit ingenious ...

 
 
1         for(int i = 1; i <= n; ++i) {
 2             int x = a[i], y = i;
 3             for(int j = 0; j < ans[i - 1].size(); ++j) {
 4                 int gcd = GCD(x, ans[i - 1][j].first);
 5                 if(gcd != x) {
 6                     ans[i].push_back(make_pair(x, y));
 7                     x = gcd, y = ans[i - 1][j].second;
 8                 }
 9             }
10             ans[i].push_back(make_pair(x, y));
11         }

A tree array is then used to maintain the fixed gcd position of the right endpoint.

1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long LL;
15 typedef pair <int, int> P;
16 const int N = 1e6 + 5;
17 int a[N/10 + 5], bit[N], _pos[N]; //_pos stores the position of the last occurrence of gcd
18 vector <P> ans[N/10 + 5]; //Save gcd with i as the right endpoint
19 struct query {
20 int l, r, pos;
21 bool operator <(const query& cmp) const {
22 return r <cmp.r;
twenty three     }
24 }q[N/10 + 5];
25 int res[N/10 + 5]; //Answer
26
27 void init(int n) {
28 memset(_pos, 0, sizeof(_pos));
29 memset(bit, 0, sizeof(bit));
30 for(int i = 1; i <= n; ++i) {
31 ans[i].clear();
32}
33}
34
35 int GCD(int a, int b) {
36 return b? GCD(b, a% b): a;
37}
38
39 void add(int i, int x) {
40 for(; i <= N; i += (i&-i))
41 bit[i] += x;
42}
43
44 int sum(int i) {
45 int s = 0;
46 for(; i >= 1; i -= (i&-i))
47 s += bit[i];
48 return s;
49}
50
51 int main()
52 {
53 int n, m;
54 while(scanf("%d %d", &n, &m) != EOF) {
55 init(n);
56 for(int i = 1; i <= n; ++i) {
57 scanf("%d", a + i);
58}
59 for(int i = 1; i <= m; ++i) {
60 scanf("%d %d", &q[i].l, &q[i].r);
61 q[i].pos = i;
62}
63 for(int i = 1; i <= n; ++i) {
64 int x = a[i], y = i;
65 for(int j = 0; j <ans[i-1].size(); ++j) {
66 int gcd = GCD(x, ans[i-1][j].first);
67 if(gcd != x) {
68 ans[i].push_back(make_pair(x, y));
69 x = gcd, y = ans[i-1][j].second;
70}
71}
72 ans[i].push_back(make_pair(x, y));
73}
74 sort(q + 1, q + m + 1);
75 int p = 1;
76 for(int i = 1; i <= n; ++i) {
77 for(int j = 0; j <ans[i].size(); ++j) {
78 if(!_pos[ans[i][j].first]) {
79 add(ans[i][j].second, 1);
80 _pos[ans[i][j].first] = ans[i][j].second;
81} else {
82 add(_pos[ans[i][j].first], -1);
83 _pos[ans[i][j].first] = ans[i][j].second;
84 add(ans[i][j].second, 1);
85}
86}
87 while(i == q[p].r && p <= m) {
88 res[q[p].pos] = sum(q[p].r)-sum(q[p].l-1);
89 ++p;
90}
91}
92 for(int i = 1; i <= m; ++i) {
93 printf("%d\n", res[i]);
94}
95}
96 return 0;
97} 

HDU 5869 Different GCD subarray Query (GCD type preprocessing + Tree array maintenance)