CF474 F. Ant colony
题目传送门:https://codeforces.com/problemset/problem/474/F
题目大意:
给定一串长度为\(n\)的序列\(A\),记子串\([l_i,r_i]\)中,\(v_k+1=\sum\limits_{j=l_i}^{r_i}[A_j\%A_k==0],(l_i\leqslant k\leqslant r_i)\),(\(v_k+1\)是为了包含了自己\(\%\)自己的情况)
现有\(m\)组询问,每次询问子串\([l_i,r_i]\)中,\(v_k\neq r_i-l_i\)的数的个数
考虑什么数才会有\(v_k=r_i-l_i\),显然是\([l_i,r_i]\)所有数的最大公约数
故我们求出\([l_i,r_i]\)的最大公约数后,再求其在\([l_i,r_i]\)中的出现次数即可
一个用线段树维护,一个用主席树维护,码就完事了
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=1e5,M=5e6;
int V[N+10],list[N+10];
int gcd(int a,int b){return !b?a:gcd(b,a%b);}
struct S1{
#define ls (p<<1)
#define rs (p<<1|1)
int Tree[(N<<2)+10];
void update(int p){Tree[p]=gcd(Tree[ls],Tree[rs]);}
void Build(int p,int l,int r){
if (l==r){
Tree[p]=list[V[l]];
return;
}
int mid=(l+r)>>1;
Build(ls,l,mid);
Build(rs,mid+1,r);
update(p);
}
int Query(int p,int l,int r,int L,int R){
if (L<=l&&r<=R) return Tree[p];
int mid=(l+r)>>1,res=0;
if (L<=mid) res=gcd(res,Query(ls,l,mid,L,R));
if (R>mid) res=gcd(res,Query(rs,mid+1,r,L,R));
return res;
}
#undef ls
#undef rs
}ST;//Segment Tree
int Root[N+10];
struct S2{
int ls[M+10],rs[M+10],Cnt[M+10],tot;
void insert(int p,int &k,int l,int r,int x){
Cnt[k=++tot]=Cnt[p]+1;
ls[k]=ls[p],rs[k]=rs[p];
if (l==r) return;
int mid=(l+r)>>1;
if (x<=mid) insert(ls[p],ls[k],l,mid,x);
if (x>mid) insert(rs[p],rs[k],mid+1,r,x);
}
int Query(int p,int k,int l,int r,int x){
if (l==r) return Cnt[k]-Cnt[p];
int mid=(l+r)>>1;
if (x<=mid) return Query(ls[p],ls[k],l,mid,x);
if (x>mid) return Query(rs[p],rs[k],mid+1,r,x);
return 0;
}
}CT;//Chairman Tree
int main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n=read(0);
for (int i=1;i<=n;i++) list[i]=V[i]=read(0);
sort(list+1,list+1+n);
int T=unique(list+1,list+1+n)-list-1;
for (int i=1;i<=n;i++) V[i]=lower_bound(list+1,list+1+T,V[i])-list;
ST.Build(1,1,n);
for (int i=1;i<=n;i++) CT.insert(Root[i-1],Root[i],1,T,V[i]);
int m=read(0);
for (int i=1;i<=m;i++){
int l=read(0),r=read(0),res=ST.Query(1,1,n,l,r);
int pos=lower_bound(list+1,list+1+T,res)-list;
int Cnt=list[pos]!=res?0:CT.Query(Root[l-1],Root[r],1,T,pos);
printf("%d\n",r-l+1-Cnt);
}
return 0;
}