Codeforces 673C Bear and Colors【暴力枚举】

最新推荐文章于 2018-11-10 20:59:00 发布

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本篇讲解了一道关于区间内主色调计数的问题，通过O(n²)枚举区间左右端点的方法，在每次增加右端点时维护当前区间的主色调，并最终统计出每种颜色作为主色调的次数。

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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 $\text{[math]}$ 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.

题目大意：

现在有N个数排咧，每一个数表示一种颜色。

对于区间【l，r】来讲，主色调就是最多出现次数的那个颜色，如果多种颜色出现次数都是最多的，那么取编号最小的作为主色调。

问每种颜色作为主色调的次数。

思路：

观察到数据范围N并不大，我们O（n^2）来枚举区间的左右端点。

接下来的任务就是对于区间【l，r】判定主色调。

如果直接暴力判断，那么需要O(n^3)的时间复杂度。

其实这个题就是一个脑筋急转弯的题。

我们在确定了区间左端点，而枚举右端点的过程中，r是逐渐增加的，其实在枚举右端点的过程中，我们就可以维护主色调的编号了。。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[5005];
int vis[5005];
int ans[5005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++)
        {
            int maxn=0;
            int color=-1;
            memset(vis,0,sizeof(vis));
            for(int j=i;j<=n;j++)
            {
                vis[a[j]]++;
                if(vis[a[j]]>maxn)maxn=vis[a[j]],color=a[j];
                if(vis[a[j]]==maxn)
                {
                    if(a[j]<color)color=a[j];
                }
                ans[color]++;
            }
        }
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);
    }
}