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

Output

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

Examples

4
1 2 1 2

7 3 0 0 

3
1 1 1

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.

題意：輸入N,接下來一行輸入N個數，每個數代表一個顏色，在每一個子區間裡面，如果某種顏色的個數最多或者有和這種顏色個數一樣多的顏色，但是代表該顏色的序號小，那麼該顏色就是該區間的主色。問以每種顏色為主色的區間數是多少。

思路：

　　純思維。 兩層for迴圈列舉每個區間，開一個num陣列代表每個區級裡面每種顏色的個數。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=5010;
 7 int n;
 8 int a[maxn],ans[maxn],num[maxn];
 9 int main()
10 {
11     scanf("%d",&n);
12     for(int i=1;i<=n;i++)
13     {
14         scanf("%d",&a[i]);
15     }
16     int pos;
17     int t=0;
18     for(int i=1;i<=n;i++)
19     {
20         memset(num,0,sizeof(num));
21         t=0;
22         for(int j=i;j<=n;j++)
23         {
24             num[a[j]]++;
25             if(num[a[j]]>t||(num[a[j]]==t&&a[j]<pos))
26             {
27                 pos=a[j];
28                 t=num[a[j]];
29                 ans[pos]++;
30             }
31             else
32             {
33                 ans[pos]++;
34             }
35         }
36     }
37     for(int i=1;i<n;i++)
38     {
39         printf("%d ",ans[i]);
40     }
41     printf("%d\n",ans[n]);
42 }