找每个数的约数(暴力就够了...1~x^0.5)....看这约数的倍数最后是哪个数...若距离大于了y..统计++...然后将这个约数的最后倍数赋值为当前位置...好叼的想法题....

Program:

#include<iostream>

#include<stack>

#include<queue>

#include<stdio.h>

#include<algorithm>

#include<string.h>

#include<cmath>

#define ll long long

#define oo 1000000007

#define MAXN 100005

using namespace std;

int last[MAXN];

int main()

{

      int T,t;

      scanf("%d",&T);

      memset(last,-1,sizeof(last));

      for (t=1;t<=T;t++)

      {

             int x,y,ans,i,p;

             ans=0;

             scanf("%d%d",&x,&y);

             for (i=1;i*i<=x;i++)

               if (x%i==0)

               {

                     if (t-last[i]>y) ans++;

                     if (x-i*i && t-last[x/i]>y) ans++;

                     last[i]=last[x/i]=t;

               }

             printf("%d\n",ans);

      }

      return 0;

}