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
",ans);
}
return 0;
}