CodeForces - 459D - Pashmak and Parmida's problem

先上題目：

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 denote f(l, r, x) the number of indices k such that: l ≤ k ≤ r andak = x. His task is to calculate the number of pairs of indicies i, j (1 ≤ i < j ≤ n) such that f(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 contains n 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

 　　題意：給你n個數，定義一個函數f(l,r,x)表示[l,r]裏面有多少個數是等於x的。如今要求有多少對i,j符合f(1,i,ai)>f(j,n,aj)

　　作法：求出某個ak它的前面和後面各有多少個數等於它自己。而後將這些數用樹狀數組計數，這裏須要先對前面的數離散化(固然，這裏用map來保存也沒有問題)，而後像統計逆序對那樣統計就能夠獲得答案了。

 

上代碼：

 

 1 #include <bits/stdc++.h>
 2 #define lowbit(x) (x&(-x))
 3 #define MAX 1000002
 4 #define ll long long
 5 using namespace std;
 6 
 7 int a[MAX],h[MAX],tot;
 8 ll num[MAX],anum[MAX],f[MAX],r[MAX];
 9 ll c[MAX];
10 int n;
11 
12 void add(int i){
13     for(;i<=MAX-1;i+=lowbit(i)) c[i]++;
14 }
15 
16 ll sum(int i){
17     ll ans=0;
18     for(;i>0;i-=lowbit(i)) ans+=c[i];
19     return ans;
20 }
21 
22 int main()
23 {
24     ll ans,e;
25     //freopen("data.txt","r",stdin);
26     while(scanf("%d",&n)!=EOF){
27         for(int i=0;i<n;i++){
28             scanf("%d",&a[i]);
29             h[i]=a[i];
30             num[i+1]=0;
31         }
32         sort(h,h+n);
33         tot=unique(h,h+n)-h;
34         for(int i=0;i<n;i++){
35             a[i]=lower_bound(h,h+tot,a[i])-h+1;
36         }
37         memset(anum,0,sizeof(anum));
38         for(int i=0;i<n;i++){
39             anum[a[i]]++;
40             f[i]=anum[a[i]];
41         }
42         for(int i=0;i<n;i++){
43             r[i]=anum[a[i]]-f[i]+1;
44         }
45         ans=0;
46         memset(c,0,sizeof(c));
47         for(int i=n-1;i>=0;i--){
48             e=sum(f[i]-1);
49             ans+=e;
50             add(r[i]);
51         }
52         printf("%I64d\n",ans);
53     }
54     return 0;
55 }

/*459D*/