codeforces 351 div2 C. Bear and Colors 暴力

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.

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：找出每个区间的重数，将重数的次数输出；

思路：暴力找复杂度o(n*n)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
int a[5010];
int flag[5010];
int ans[5010];
int main()
{
    int x,y,z,i,t;
    scanf("%d",&x);
    for(i=1;i<=x;i++)
    scanf("%d",&a[i]);
    for(i=1;i<=x;i++)
    {
        memset(flag,0,sizeof(flag));
        int maxx=0,ji;
        for(t=i;t<=x;t++)
        {
            //cout<<maxx<<" "<<ji<<" "<<a[t]<<endl;
            flag[a[t]]++;
            if(flag[a[t]]>maxx)
            {
                maxx=flag[a[t]];
                ji=a[t];
                ans[ji]++;
            }
            else if(flag[a[t]]==maxx&&ji>a[t])
            {
                ji=a[t];
                ans[ji]++;
            }
            else
            ans[ji]++;
        }
    }
    for(i=1;i<=x;i++)
    printf("%d ",ans[i]);
    return 0;
}