HDU5869树状数组+gcd预处理

Posted 2020-08-09

比赛的时候知道用树状数组，但有点乱不知道怎么处理。

统计不同的gcd的个数其实就是用树状数组统计区间内不同的数的模板题啊...

复杂度O(nlogn)

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 1e5+10;
 4 int n,q,i,j,a[N],l[N],v[N];
 5 int fun(int x,int y){return __gcd(x,y);}
 6 
 7 struct p{
 8     int l, r, ans, k;
 9 };
10 p query[N];
11 bool cmp1(const p& a, const p& b){
12     return a.r < b.r;
13 }
14 bool cmp2(const p& a, const p& b){
15     return a.k < b.k;
16 }
17 
18 int c[N], last[N*10], maxn;
19 int lowbit(int x){ return x&-x;}
20 void add(int x, int d){
21     for(int i = x; i <= n; i += lowbit(i))
22         c[i] += d;
23 }
24 int sum(int x){
25     int ret = 0;
26     while(x){
27         ret += c[x];
28         x -= lowbit(x);
29     }
30     return ret;
31 }
32 
33 int main(){
34     while(~scanf("%d%d", &n, &q)){
35         maxn = -1;
36         for(i = 1; i <= n; i++) scanf("%d", a+i), maxn = max(maxn, a[i]);
37         for(i = 1; i <= q; i++){
38             scanf("%d%d", &query[i].l, &query[i].r);
39             query[i].k = i;
40         }
41         sort(query+1, query+q+1, cmp1);
42         memset(last, 0, sizeof(int)*(maxn+5));
43         memset(c, 0, sizeof(int)*(n+5));
44 
45         int qq  = 1;
46         for(i = 1; i <= n; i++){
47             for(v[i] = a[i], j = l[i] = i; j; j = l[j]-1){
48                 v[j] = fun(v[j], a[i]);
49                 while(l[j] > 1&&fun(a[i], v[l[j]-1]) == fun(a[i], v[j])) l[j] = l[l[j]-1];
50                 //[l[j]..j,i]区间内的值求fun均为v[j]
51                 if(last[ v[j] ]){
52                     if(last[ v[j] ] < j){
53                         add(last[ v[j] ], -1);
54                         add(j, 1);
55                         last[ v[j] ] = j;
56                     }
57                 }
58                 else{//v[j] 没有
59                     add(j, 1);
60                     last[ v[j] ] = j;
61                 }
62             }
63             while(qq <= q&&query[qq].r == i){
64                 query[qq].ans = sum(query[qq].r) - sum(query[qq].l-1);
65                 qq++;
66             }
67         }
68         sort(query+1, query+q+1, cmp2);
69         for(int i = 1; i <= q; i++)
70             printf("%d\n", query[i].ans);
71     }
72 }

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