题意:
给一个长为n的序列,a1到an,q次询问,x1到xq,问gcd(al, al + 1, ..., ar) = xi的[l,r]有多少个。
思路:
简单的统计需要O(n^2)统计gcd,容易发现很多区间的gcd都是一样的,因为假设固定左端点,那个右端点移动过程中,gcd最多变化al的因子次。
那么我们从后往前枚举左端点,将相同的区间合并在一起,通过类似静态链表的结构实现下标的跳转,枚举到前面的数字时,可以通过后面的静态链表直接跳过gcd相同的区间,从而实现O(nlog(max(a)))的复杂度,枚举出区间gcd个数后用map<int,long long>统计输出即可。
代码:
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll INFF=0x3f3f3f3f3f3f3f3f;
const ull mod=212370440130137957;
const double eps=1e-5;
const int maxn=100005;
const int maxm=20005;
const int base=31;
int n,m;
int a[maxn];
int nx[maxn];
map<int,ll>maps;
void solve() {
sd(n);
for(int i=1; i<=n; i++) {
sd(a[i]);
nx[i]=i+1;
}
for(int i=n; i>0; i--) {
int rg,lg,l;
lg=rg=a[i];
l=i;
for(int j=i; j<=n+1; j=nx[j]) {
if(j==n+1) {
maps[lg]+=(j-l);
nx[l]=j;
break;
}
rg=__gcd(a[j],lg);
if(rg!=lg) {
maps[lg]+=(j-l);
nx[l]=j;
l=j;
lg=rg;
}
}
}
sd(m);
for(int i=0; i<m; i++) {
int now;
sd(now);
printf("%lld\n",maps[now]);
}
return ;
}
int main() {
#ifdef LOCAL
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#else
// freopen("","r",stdin);
// freopen("","w",stdout);
#endif
solve();
return 0;
}