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CodeForces 673C Bear and Colors

来源:互联网 作者:佚名 时间:2016-05-12 15:56
span style=font-size:18px;思路:暴力出奇迹/span #includebits\stdc++.husing namespace std;const int maxn =5005;int a[maxn];int vis[maxn];int ans[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++
<span style="font-size:18px;">思路:暴力出奇迹</span>
#include<bits\stdc++.h>
using namespace std;
const int maxn =5005;
int a[maxn];
int vis[maxn];
int ans[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(vis,0,sizeof(vis));
		int ans1 = 0;
		int ans2 = 0;
		for (int j = i;j<=n;j++)
		{
			vis[a[j]]++;
			if(ans1<vis[a[j]] || (ans1==vis[a[j]])&&a[j]<ans2)
			{
				ans1=vis[a[j]];
				ans2=a[j];
			}
			ans[ans2]++;
		}
	}
	for (int i = 1;i<=n;i++)
		printf("%d ",ans[i]);
	printf("\n");
}

Description

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.

Sample Input

Input
4
1 2 1 2
Output
7 3 0 0 
Input
3
1 1 1
Output
6 0 0 

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