CodeForces 111B - Petya and Divisors

Description

Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:

You are given n queries in the form "xiyi".
For each query Petya should count how many divisors of number xi divide none of the numbers xi - yi, xi - yi + 1, ..., xi - 1.
Help him.

Input

The first line contains an integer n (1 ≤ n ≤ 105). Each of the following n lines
contain two space-separated integers xi and yi (1 ≤ xi ≤ 105, 0 ≤ yi ≤ i - 1,
where i is the query's ordinal number; the numeration starts with 1).

If yi = 0 for the query, then the answer to the query will be the number of divisors of the number xi.
In this case you do not need to take the previous numbers x into consideration.

Output

For each query print the answer on a single line: the number of positive integers k such that 


Sample Input

Input

6
4 0
3 1
5 2
6 2
18 4
10000 3

Output

3
1
1
2
2
22

没有了固化的算法来解决问题就不会了。

算法思路：

定义：divisor[i] ：以i为因子的最后一个数的下标；

①对每个输入的数找出它的所有因子，比较该因子是否在给定范围内也是其它数的因子；

②若不是，则累加计数；

③更新当前数的因子的位置为当前下
4000
标。

#include<iostream>

#include<cstring>

using namespace std;

const int N = 100000 + 5;

int divisor
;

int main()

{
int n;
cin>>n;

int x, y;
memset(divisor, -1, sizeof(divisor));

for(int i = 0; i < n; i++)
{
cin>>x>>y;
int res = 0;
for(int j = 1; j * j <= x; j++)
if(x % j == 0)
{
if(i - y > divisor[j])
res++;
if(j * j != x && i - y > divisor[x / j])
res++;
divisor[j] = divisor[x / j] = i;
}
cout<<res<<endl;
}

}