codeforces 673C C. Bear and Colors(暴力)

题目链接：

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 ti.

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.

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

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ n) where ti 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:

给出这么多颜色,在一个序列中,dominant是出现次数最多的数,如果出现次数最多的不止一个,那么就是数值最小的那个;

思路：

暴力跑出[i,j]中每个数出现的次数,同时更新这里面的的dominant;

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+5;
int n,flag[5002][5002],a[5002],ans[5002];
int main()
{
   scanf("%d",&n);
   for(int i=1;i<=n;i++)
   {
       scanf("%d",&a[i]);
   }
   for(int i=1;i<=n;i++)
   {
       int num=0,temp;
       for(int j=i;j<=n;j++)
       {
           flag[i][a[j]]++;
           if(flag[i][a[j]]>num)
           {
               num=flag[i][a[j]];
               temp=a[j];
           }
           else if(flag[i][a[j]]==num)
           {
               if(a[j]<temp)
               {
                   temp=a[j];
               }
           }
           ans[temp]++;
       }

   }
   for(int i=1;i<=n;i++)
   {
       printf("%d ",ans[i]);
   }

    return 0;
}