Patting Heads

Description

It's Bessie's birthday and time for party games! Bessie has instructed the N (1 < N < 100,000) cows conveniently numbered 1..N to sit in a circle (so that cow i [except at the ends] sits next to cows i-1 and i+1; cow N sits next to cow 1). Meanwhile, Farmer John fills a barrel with one billion slips of paper, each containing some integer in the range 1..1,000,000.

Each cow i then draws a number A_i (1 ≤ A_i ≤ 1,000,000) (which is not necessarily unique, of course) from the giant barrel. Taking turns, each cow i then takes a walk around the circle and pats the heads of all other cows j such that her number A_i is exactly divisible by cow j's number A_j; she then sits again back in her original position.

The cows would like you to help them determine, for each cow, the number of other cows she should pat.

Input

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains a single integer: A_i

Output

* Lines 1..N: On line i, print a single integer that is the number of other cows patted by cow i.

Sample Input

5
2
1
2
3
4

Sample Output

2
0
2
1
3

Hint

The 5 cows are given the numbers 2, 1, 2, 3, and 4, respectively. The first cow pats the second and third cows; the second cows pats no cows; etc.

题目大概的意思就是说牛1手上的数字可以去整除其他牛的数字，则可以去爆他狗头。

一开始我是想说一一去枚举的，从一找到牛手上的数可以整除的数，去加他，但发现超时。

于是有了这种思路，我从1到牛手上的最大的数枚举，if这个数字是有的，那这个数的倍数都加上他的次数；

AC 代码

#include<stdio.h>
 int num1[]={};
    int num2[]={};
    int a[];
int main()
{
    int n,i,mix=,j;

    scanf("%d",&n);
    for(i=;i<=n;i++)
    {
        scanf("%d",&a[i]);
        num1[a[i]]++;

    }

    for(i=;i<=n;i++)
        if(mix<a[i])
        mix=a[i];

    for(i=;i<=mix;i++)
    {    if(num1[i]!=)
        for(j=i;j<=mix;j=j+i)
        {
            num2[j]=num2[j]+num1[i];

        }
    }

   for(i=;i<=n;i++)
        printf("%d\n",num2[a[i]]-);
}