Bear and Colors (思维)

 Bear and Colors 

CodeForces - 673C

Bear Limak has n colored balls, arranged in one long row. Balls are numbered 1through n, from left to right. There are n possible colors, also numbered 1 through n. The i-th ball has color t_i.

For a fixed interval (set of consecutive elements) of balls we can define a dominantcolor. It's a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant.

There are n(n+1)/2 non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — the number of balls.

The second line contains n integers t₁, t₂, ..., t_n (1 ≤ t_i ≤ n) where t_i is the color of the i-th ball.

Output

Print n integers. The i-th of them should be equal to the number of intervals where i is a dominant color.

Examples

Input

4
1 2 1 2

Output

7 3 0 0 

Input

3
1 1 1

Output

6 0 0 

Note

In the first sample, color 2 is dominant in three intervals:

• An interval [2, 2] contains one ball. This ball's color is 2 so it's clearly a dominant color.
• An interval [4, 4] contains one ball, with color 2 again.
• An interval [2, 4] contains two balls of color 2 and one ball of color 1.

There are 7 more intervals and color 1 is dominant in all of them.

题意：输入N,接下来一行输入N个数，每个数代表一个颜色，在每一个子区间里面，如果某种颜色的个数最多或者有和这种颜色个数一样多的颜色，但是代表该颜色的序号小，那么该颜色就是该区间的主色。问以每种颜色为主色的区间数是多少。

思路：

　　纯思维。 两层for循环枚举每个区间，开一个num数组代表每个区级里面每种颜色的个数。

　　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=5010;
int n;
int a[maxn],ans[maxn],num[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int pos;
    int t=0;
    for(int i=1;i<=n;i++)
    {
        memset(num,0,sizeof(num));
        t=0;
        for(int j=i;j<=n;j++)
        {
            num[a[j]]++;
            if(num[a[j]]>t||(num[a[j]]==t&&a[j]<pos))
            {
                pos=a[j];
                t=num[a[j]];
                ans[pos]++;
            }
            else
            {
                ans[pos]++;
            }
        }
    }
    for(int i=1;i<n;i++)
    {
        printf("%d ",ans[i]);
    }
    printf("%d\n",ans[n]);
}