HDU5869树状数组+gcd预处理

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)

 #include <bits/stdc++.h>
 using namespace std;
 const int N = 1e5+;
 int n,q,i,j,a[N],l[N],v[N];
 int fun(int x,int y){return __gcd(x,y);}

 struct p{
     int l, r, ans, k;
 };
 p query[N];
 bool cmp1(const p& a, const p& b){
     return a.r < b.r;
 }
 bool cmp2(const p& a, const p& b){
     return a.k < b.k;
 }

 int c[N], last[N*], maxn;
 int lowbit(int x){ return x&-x;}
 void add(int x, int d){
     for(int i = x; i <= n; i += lowbit(i))
         c[i] += d;
 }
 int sum(int x){
     int ret = ;
     while(x){
         ret += c[x];
         x -= lowbit(x);
     }
     return ret;
 }

 int main(){
     while(~scanf("%d%d", &n, &q)){
         maxn = -;
         for(i = ; i <= n; i++) scanf("%d", a+i), maxn = max(maxn, a[i]);
         for(i = ; i <= q; i++){
             scanf("%d%d", &query[i].l, &query[i].r);
             query[i].k = i;
         }
         sort(query+, query+q+, cmp1);
         memset(last, , sizeof(int)*(maxn+));
         memset(c, , sizeof(int)*(n+));

         int qq  = ;
         for(i = ; i <= n; i++){
             for(v[i] = a[i], j = l[i] = i; j; j = l[j]-){
                 v[j] = fun(v[j], a[i]);
                 while(l[j] > &&fun(a[i], v[l[j]-]) == fun(a[i], v[j])) l[j] = l[l[j]-];
                 //[l[j]..j,i]区间内的值求fun均为v[j]
                 if(last[ v[j] ]){
                     if(last[ v[j] ] < j){
                         add(last[ v[j] ], -);
                         add(j, );
                         last[ v[j] ] = j;
                     }
                 }
                 else{//v[j] 没有
                     add(j, );
                     last[ v[j] ] = j;
                 }
             }
             while(qq <= q&&query[qq].r == i){
                 query[qq].ans = sum(query[qq].r) - sum(query[qq].l-);
                 qq++;
             }
         }
         sort(query+, query+q+, cmp2);
         for(int i = ; i <= q; i++)
             printf("%d\n", query[i].ans);
     }
 }