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.
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.
Print n integers. The i-th of them should be equal to the number of intervals where i is a dominant color.
4
1 2 1 2
7 3 0 0
3
1 1 1
6 0 0
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;
}