题目链接：
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.
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:
给出这么多颜色,在一个序列中,dominant是出现次数最多的数,如果出现次数最多的不止一个,那么就是数值最小的那个;
思路：
暴力跑出[i,j]中每个数出现的次数,同时更新这里面的的dominant;
AC代码：
#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+;
int n,flag[][],a[],ans[];
int main()
{
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(int i=;i<=n;i++)
{
int num=,temp;
for(int j=i;j<=n;j++)
{
flag[i][a[j]]++;
if(flag[i][a[j]]>num)
{
num=flag[i][a[j]];
temp=a[j];
}
else if(flag[i][a[j]]==num)
{
if(a[j]<temp)
{
temp=a[j];
}
}
ans[temp]++;
} }
for(int i=;i<=n;i++)
{
printf("%d ",ans[i]);
} return ;
}
