FZU 2224 An exciting GCD problem(GCD种类预处理+树状数组维护)同hdu5869

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
; //_pos存的是gcd上一次出现的位置
18 vector <P> ans[N/10 + 5]; //存的是以i为右端点的gcd
19 map <int, int> _pos;
20 struct query {
21     int l, r, pos;
22     bool operator <(const query& cmp) const {
23         return r < cmp.r;
24     }
25 }q[N/10 + 5];
26 int res[N/10 + 5]; //答案
27
28 void init(int n) {
29     _pos.clear();
30     memset(bit, 0, sizeof(bit));
31     for(int i = 1; i <= n; ++i) {
32         ans[i].clear();
33     }
34 }
35
36 int GCD(int a, int b) {
37     return b ? GCD(b, a % b): a;
38 }
39
40 void add(int i, int x) {
41     for( ; i <= N; i += (i&-i))
42         bit[i] += x;
43 }
44
45 int sum(int i) {
46     int s = 0;
47     for( ; i >= 1; i -= (i&-i))
48         s += bit[i];
49     return s;
50 }
51
52 int main()
53 {
54     int n, m, t;
55     scanf("%d", &t);
56     while(t--) {
57         scanf("%d %d", &n, &m);
58         init(n);
59         for(int i = 1; i <= n; ++i) {
60             scanf("%d", a + i);
61         }
62         for(int i = 1; i <= m; ++i) {
63             scanf("%d %d", &q[i].l, &q[i].r);
64             q[i].pos = i;
65         }
66         for(int i = 1; i <= n; ++i) {
67             int x = a[i], y = i;
68             for(int j = 0; j < ans[i - 1].size(); ++j) {
69                 int gcd = GCD(x, ans[i - 1][j].first);
70                 if(gcd != x) {
71                     ans[i].push_back(make_pair(x, y));
72                     x = gcd, y = ans[i - 1][j].second;
73                 }
74             }
75             ans[i].push_back(make_pair(x, y));
76         }
77         sort(q + 1, q + m + 1);
78         int p = 1;
79         for(int i = 1; i <= n; ++i) {
80             for(int j = 0; j < ans[i].size(); ++j) {
81                 if(!_pos[ans[i][j].first]) {
82                     add(ans[i][j].second, 1);
83                     _pos[ans[i][j].first] = ans[i][j].second;
84                 } else {
85                     add(_pos[ans[i][j].first], -1);
86                     _pos[ans[i][j].first] = ans[i][j].second;
87                     add(ans[i][j].second, 1);
88                 }
89             }
90             while(i == q[p].r && p <= m) {
91                 res[q[p].pos] = sum(q[p].r) - sum(q[p].l - 1);
92                 ++p;
93             }
94         }
95         for(int i = 1; i <= m; ++i) {
96             printf("%d\n", res[i]);
97         }
98     }
99     return 0;
100 }