CodeForces 673C - Bear and Colors(模拟)

C. Bear and Colors

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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：

已知一串数字，每种数字代表一种颜色，如果一个区间[i,j]中, ai颜色出现次数最多或者有和ai出现次数一样多的颜色但是颜色编号都比ai大，那么ai就是这个区间的领导颜色，统计各种颜色领导的区间数量。

解题思路：

模拟。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e3+5;
int a[maxn];
int cnt[maxn];
int res[maxn];
int main()
{
int n;
scanf("%d",&n);
for(int i = 1;i <= n;i++)   scanf("%d",&a[i]);
for(int i = 1;i <= n;i++)
{
memset(cnt,0,sizeof(cnt));
cnt[a[i]]++;
res[a[i]]++;
int index = a[i],minn = 1;
for(int j = i+1;j <= n;j++)
{
cnt[a[j]]++;
if(minn < cnt[a[j]])    minn = cnt[a[j]],index = a[j];
else if(minn == cnt[a[j]] && a[j] < index)  index = a[j];
res[index]++;
}
}
for(int i = 1;i <= n;i++)   printf("%s%d",i == 1?"":" ",res[i]);
return 0;
}