Codeforces Round #351 Div2 C 简单枚举

C. Bear and Colors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bear Limak has n colored balls, arranged in one long row. Balls are numbered 1 through 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 dominant color. 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 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：

每个球对应一种颜色，一共有n个球，所以有个区间，问每种颜色在多少个区间中是dominant？

dominant的意思是在这个区间中这种颜色数最多，如果有相同个数的颜色，编号最小的是dominant。

分析：

枚举个区间，对于每个区间统计那种颜色是dominant即可。

#include<bits/stdc++.h>
using namespace std;
const int N=5009;
int ans[N],num[N],a[N];
int main()
{
    int n;scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",&a[i]);
    int tot=0;
    for(int i=0;i<n;i++){
        memset(num,0,sizeof(num));
        int maxn=0,maxp=i;
        for(int j=i;j<n;j++){
            tot++;
            num[a[j]]++;
            if(num[a[j]]>maxn||(a[j]<maxp&&num[a[j]]==maxn)){
                maxn=num[a[j]];
                maxp=a[j];
                //ans[a[j]]++;
            }
            ans[maxp]++;
        }
    }
   // cout<<tot<<endl;
    for(int i=1;i<=n;i++)printf("%d ",ans[i]);
    return 0;
}