Codeforces Round #261 (Div. 2) D. Pashmak and Parmida's problem(树状数组+逆序数对)

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.

Sample test(s)

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

                                                                             

由于原数过大，可使用map 映射，就不需要离散化了~两个数组，一个记录前面与本数相等的数的个数，一个记录后面与本数相等的数的个数~

题目便转化成求逆序数对，树状数组。答案long long

CODE：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
const int inf=0xfffffff;
typedef long long ll;
using namespace std;

const int Max=1000006;
int num[Max];
int f1[Max],f2[Max];
int c[Max];
map<int,int>m1,m2;
int n;

int lowbit(int x)
{
return x&-x;
}
void add(int x,int d)
{
while(x<=n){
c[x]+=d;
x+=lowbit(x);
}
}
int sum(int x)
{
int ans=0;
while(x>0){
ans+=c[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
//freopen("in","r",stdin);
while(~scanf("%d",&n)){
m1.clear();
m2.clear();
memset(f1,0,sizeof(f1));
memset(f2,0,sizeof(f2));
memset(c,0,sizeof(c));
for(int i=1;i<=n;i++){
scanf("%d",&num[i]);
}
for(int i=1;i<=n;i++){
f1[i]=(++m1[num[i]]);
}
for(int i=n;i>=1;i--){
f2[i]=(++m2[num[i]]);
}
ll ans=0;
for(int i=n-1;i>=1;i--){
add(f2[i+1],1);
ans+=(ll)sum(f1[i]-1);
}
printf("%I64d\n",ans);
}
return 0;
}