Codeforces 673C Bear and Colors【暴力枚举】

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 numbered1 through n, from left to right. There aren 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) wheret_i is the color of thei-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

41 2 1 2

Output

7 3 0 0 

Input

31 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 is2 so it's clearly a dominant color.
• An interval [4, 4] contains one ball, with color2 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]);    }}