2016 ACM/ICPC Asia Regional Dalian Online 1002/HDU 5869

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 681 Accepted Submission(s): 240

Problem Description

This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as it looks, Bob cannot figure it out himself. Now he turns to you for help, and here is the problem:

Given an array

Input

There are several tests, process till the end of input.

For each test, the first line consists of two integers

Output

For each query, output the answer in one line.

Sample Input

5 3

1 3 4 6 9

3 5

2 5

1 5

Sample Output

Source

2016 ACM/ICPC Asia Regional Dalian Online

题意：

题解：

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 #include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<queue>
14 #define LL __int64
15 #define pii pair<int,int>
16 #define MP make_pair
17 const int N=1000006;
18 using namespace std;
19 int gcd(int a,int b)
20 {
21     return b==0 ? a : gcd(b,a%b);
22 }
23 int n,q,a[N],ans[N];
24 vector<pii> G[N];
25 struct QQ{
26   int l,r,id;
27   bool operator < (const QQ &a) const
28   {
29        return a.r>r;
30   }
31 }Q[N];
32 int C[N],vis[N];
33 void update (int x,int c)
34 {
35     for(int i=x;i<N;i+=i&(-i)) C[i]+=c;
36 }
37 int ask(int x)
38 {
39     int s=0;
40     for(int i=x;i;i-=i&(-i)) s+=C[i];
41     return s;
42 }
43 int main()
44 {
45     while(scanf("%d %d",&n,&q)!=EOF)
46     {
47         for(int i=1;i<=n;i++)
48             scanf("%d",&a[i]);
49         for(int i=0;i<=n;i++)
50             G[i].clear();
51         for(int i=1;i<=n;i++)
52         {
53             int x=a[i];
54             int y=i;
55             for(int j=0;j<G[i-1].size();j++)
56             {
57                 int res=gcd(x,G[i-1][j].first);
58                 if(x!=res)
59                 {
60                     G[i].push_back(MP(x,y));
61                     x=res;
62                     y=G[i-1][j].second;
63                 }
64             }
65             G[i].push_back(MP(x,y));
66         }
67         memset(C,0,sizeof(C));
68         memset(vis,0,sizeof(vis));
69         for(int i=1;i<=q;i++)
70         {
71             scanf("%d %d",&Q[i].l,&Q[i].r);
72             Q[i].id=i;
73         }
74         sort(Q+1,Q+q+1);
75         for(int R=0,i=1;i<=q;i++)
76         {
77             while(R<Q[i].r)
78             {
79                 R++;
80             for(int j=0;j<G[R].size();j++)
81             {
82                 int res=G[R][j].first;
83                 int ids=G[R][j].second;
84                 if(vis[res])
85                     update(vis[res],-1);
86                 vis[res]=ids;
87                 update(vis[res],1);
88             }
89             }
90             ans[Q[i].id]=ask(R)-ask(Q[i].l-1);
91         }
92         for(int i=1;i<=q;i++)
93             cout<<ans[i]<<endl;
94     }
95     return 0;
96 }