Codeforces 459D Pashmak and Parmida's problem【树状数组】

D. Pashmak and Parmida's problem

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partner to be clever too (although he's not)! Parmida has prepared the following test problem for Pashmak.

There is a sequence a that consists of
n integers a1, a2, ..., an. Let's denotef(l, r, x)
the number of indicesk such that:
l ≤ k ≤ r andak = x. His task is to calculate the number of pairs of indiciesi, j
(1 ≤ i < j ≤ n) such thatf(1, i, ai) > f(j, n, aj).

Help Pashmak with the test.

Input
The first line of the input contains an integer n(1 ≤ n ≤ 106). The second line containsn space-separated
integers a1, a2, ..., an(1 ≤ ai ≤ 109).

Output
Print a single integer — the answer to the problem.

Examples

Input

7
1 2 1 1 2 2 1

Output

8

Input

3
1 1 1

Output

1

Input

5
1 2 3 4 5

Output

0

题目大意：

给你N个数，定义F（L,R,X）=区间【L,R】内，a【i】==x的个数。

让你求一共有多少对（i，j）使得F（1，i，a【i】）>F（j，n，a【j】）；

思路：

用map统计前缀a【i】个数，设定为pre【i】。表示【1，i】中，a【i】的个数；

以及后缀a【i】个数，设定为back【i】。表示【i，n】中，a【i】的个数。

然后O（n）动态更新树然后查询i后边有多少个back【i】小于pre【i】，这里树状数组优化一下即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
#define ll __int64
int back[1500000];
int pre[1500000];
int a[1500000];
int tree[1500005];//树
int n;
int lowbit(int x)//lowbit
{
return x&(-x);
}
int sum(int x)//求和求的是比当前数小的数字之和，至于这里如何实现，很简单：int sum=sum(a[i])；
{
int sum=0;
while(x>0)
{
sum+=tree[x];
x-=lowbit(x);
}
return sum;
}
void add(int x,int c)//加数据。
{
while(x<=n)
{
tree[x]+=c;
x+=lowbit(x);
}
}
void Get_array()
{
map<int ,int >s;
for(int i=1;i<=n;i++)
{
s[a[i]]++;
pre[i]=s[a[i]];
}
s.clear();
for(int i=n;i>=1;i--)
{
s[a[i]]++;
back[i]=s[a[i]];
}
}
int main()
{
while(~scanf("%d",&n))
{
memset(tree,0,sizeof(tree));
ll output=0;
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
Get_array();
for(int i=1;i<=n;i++)
{
add(back[i],1);
}
for(int i=1;i<=n;i++)
{
add(back[i],-1);
output+=(ll)sum(pre[i]-1);
}
printf("%I64d\n",output);
}
}