Codeforce P471D CGCDSSQ

D. CGCDSSQ

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r)such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.

 is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.

Input

The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).

The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).

Output

For each query print the result in a separate line.

Sample test(s)

input

32 6 3512346

output

12201

input

710 20 3 15 1000 60 16101234561020601000

output

14022202211

考数据结构的题，也是考数学功底的题，对于一段(l,r)从(l,l+1)开始的gcd严格单调递减，所以可用二分查出(l,r)里和(l,l)公约数一样的最大长度，之后又是新一轮的查找，因为一个数不会超过31个素因数，所以新一轮的查找不会很多，自己可以模拟一下，最多30次。具体做法就是枚举起点，终点始终是n，进行这种查找，找出来的数用map保存，最后的询问直接map回答。

这里还要求区间取GCD，线段树是不行的，因为此题卡了常数，所以得用ST算法，就算是ST，也得用位运算加速，即rmq的时候，不能用log，只能用31-__builtin_clz(v-s+1)，不然还是超时。。。

#include<iostream>#include<cstdio>#include<cstdlib>#include<ctime>#include<string>#include<cstring>#include<algorithm>#include<fstream>#include<queue>#include<stack> #include<vector>#include<cmath>#include<iomanip>#include<map>#define rep(i,n) for(i=1;i<=n;i++)#define MM(a,t) memset(a,t,sizeof(a))#define INF 1e9typedef long long ll;#define mod 1000000007using namespace std;map<int,ll> mp;int dp[100020][20],n,m;int gcd(int a,int b){if(a==0) return b;if(b==0) return a;if(a<b) return gcd(b,a);if(a%b==0) return b;else       return gcd(b,a%b);}void makermq(){int i,j;for(j=1;(1<<j)<=n;j++)  for(i=1;i+(1<<j)-1<=n;i++)    dp[i][j]=gcd(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int rmq(int s,int v){int k=31-__builtin_clz(v-s+1);return gcd(dp[s][k],dp[v-(1<<k)+1][k]); }int main(){int i,j;    while(scanf("%d",&n)!=EOF){      MM(dp,0);  rep(i,n) scanf("%d",&dp[i][0]);  makermq();  mp.clear();  rep(i,n){      int st=i,l,r,tmp,mid,v;while(st<=n){  v=st; l=st; r=n; tmp=rmq(i,l);  while(l<=r){  mid=(l+r)>>1;  if(rmq(i,mid)==tmp){      l=mid+1;  v=mid;    }  else                r=mid-1;    }    mp[tmp]+=v-st+1;    st=v+1;}    }    scanf("%d",&m);    while(m--){    int tmp;    scanf("%d",&tmp);    printf("%I64d\n",mp[tmp]);    }    }return 0;}