GCD
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1497 Accepted Submission(s): 483
Problem Description
Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,al+1,...,ar) and count the number of pairs(l′,r′)(1≤l<r≤N)such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).
Input
The first line of input contains a number T, which stands for the number of test cases you need to solve.
The first line of each case contains a number N, denoting the number of integers.
The second line contains N integers, a1,...,an(0<ai≤1000,000,000).
The third line contains a number Q, denoting the number of queries.
For the next Q lines, i-th line contains two number , stand for the li,ri, stand for the i-th queries.
Output
For each case, you need to output “Case #:t” at the beginning.(with quotes, t means the number of the test case, begin from 1).
For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+1,...,ar) and the second number stands for the number of pairs(l′,r′) such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).
Sample Input
1
5
1 2 4 6 7
4
1 5
2 4
3 4
4 4
Sample Output
Case #1:
1 8
2 4
2 4
6 1
Author
HIT
Source
2016 Multi-University Training Contest1
思路：RMQ+二分
一开始我用的线段树去维护各个区间的gcd但是超时，并且我没有优化。
暴力统计的各个区间的gcd；
后来发现gcd的性质，也就是求数的gcd，这些数的个数越多那么gcd越小，所以从左到右，以某个端点开始的gcd的大小随着区间增大而减小，那么二分统计以某个点为端点的gcd
最多每个端点二分30次。
然后还是超时。
然后改用RMQ基于稀疏表的，查询O（n）；
复杂度（n*log(n)）;
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<stdlib.h>
6 #include<queue>
7 #include<map>
8 #include<math.h>
9 using namespace std;
10 typedef long long LL;
11 int RMQ[20][100005];
12 int ans[100005];
13 map<int,LL>my;
14 int gcd(int n,int m)
15 {
16 if(m==0)
17 return n;
18 else if(n%m==0)
19 {
20 return m;
21 }
22 else return gcd(m,n%m);
23 }
24 int main(void)
25 {
26 int i,j,k;
27 int n,m;
28 scanf("%d",&k);
29 int ca=0;
30 while(k--)
31 {
32 my.clear();
33 scanf("%d",&n);
34 for(i=1; i<=n; i++)
35 {
36 scanf("%d",&ans[i]);
37 }
38 for(i=1; i<=n; i++)
39 {
40 RMQ[0][i]=ans[i];
41 }
42 scanf("%d",&m);
43 for(i=1; i<20; i++)
44 {
45 for(j=1; j<=n; j++)
46 {
47 if(j+(1<<i)-1<=n)
48 {
49 RMQ[i][j]=gcd(RMQ[i-1][j],RMQ[i-1][j+(1<<(i-1))]);
50 }
51 }
52 }
53 for(i=1; i<=n; i++)
54 {
55 for(j=i; j<=n;)
56 {
57 int t=log2(j-i+1);
58 int ac=gcd(RMQ[t][i],RMQ[t][j-(1<<t)+1]);
59 int l=j;
60 int r=n;
61 int id=0;
62 while(l<=r)
63 {
64 int mid=(l+r)/2;
65 int c=log2(mid-i+1);
66 int ak=gcd(RMQ[c][i],RMQ[c][mid-(1<<c)+1]);
67 if(ak>=ac)
68 {
69 id=mid;
70 l=mid+1;
71 }
72 else r=mid-1;
73 }
74 my[ac]+=id-j+1;
75 j=id+1;
76 }
77 }
78 printf("Case #%d:\n",++ca);
79 while(m--)
80 {
81 int x,y;
82 scanf("%d %d",&x,&y);
83 if(x>y)
84 {
85 swap(x,y);
86 }
87 int ct=log2(y-x+1);
88 int acc=gcd(RMQ[ct][x],RMQ[ct][y-(1<<ct)+1]);
89 printf("%d %lld\n",acc,my[acc]);
90 }
91 }
92 return 0;
93 }