Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) C （用map 超时）

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：给你n个数共n*（n+1）/2个区间求各个区间的众数作为标记数若有多个众数记录最小的数作为标记数输出每个数作为多少个区间的标记数

题解：暴力处理记录各个区间的标记数。

吃了 STL 的亏用map 超时了 xjb用用数组存就是！！！！！

#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<map>
#define ll __int64
#define pi acos(-1.0)
using namespace std;
int n;
int mp[5005];
int mpp[5005];
int a[5005];
int main()
{
 //mp.clear();
 scanf("%d",&n);
 for(int i=1;i<=n;i++)
  {
    scanf("%d",&a[i]);
  }
 for(int i=1;i<=n;i++)
 {
  int maxc=-1;
  int minx=5001;
  int ans=0;
   
  for(int j=1;j<=n;j++)
   mpp[j]=0;
  for(int j=i;j<=n;j++)
  {
   mpp[a[j]]++;
   if(maxc==mpp[a[j]])
     {
       ans=min(ans,a[j]);
     }
   if(maxc<mpp[a[j]])
   {
    maxc=mpp[a[j]];
    ans=a[j]; 
   }
       mp[ans]++;
  }  
 }
 cout<<mp[1];
 for(int i=2;i<=n;i++)
 printf(" %d",mp[i]);
 cout<<endl;
 return 0;
}

巴特西

Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) C （用map 超时）

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