Codeforces Round #261 (Div. 2) D. 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,?...,?a_n. Let‘s denote
f(l,?r,?x) the number of indices
k such that: l?≤?k?≤?r and
a_k?=?x. His task is to calculate the number of pairs of indicies
i,?j (1?≤?i?<?j?≤?n) such that
f(1,?i,?a_i)?>?f(j,?n,?a_j).

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,?...,?a_n
(1?≤?a_i?≤?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

解析：f(l,?r,?x) 表示
l?≤?k?≤?r and
a_k?=?x的个数，要计算 
f(1,?i,?a_i)?>?f(j,?n,?a_j)，且i,?j
(1?≤?i?<?j?≤?n) 的对数.

先预处理f(1, i, ai)和f(j, n, aj)，然后就特别像逆序数了。用树状数组搞一下就行了。

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;

int a[N], f[N], g[N], c[N];

map<int, int> m;

int lowbit(int x){
    return x & -x;
}

void add(int x, int d){
    for( ; x < N; x += lowbit(x)) c[x] += d;
}

int sum(int x){
    int ret = 0;
    for( ; x > 0; x -= lowbit(x)) ret += c[x];
    return ret;
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    int n;
    while(~scanf("%d", &n)){
        m.clear();
        for(int i=1; i<=n; i++){
            scanf("%d", &a[i]);
            m[ a[i] ] ++;
            f[i] = m[ a[i] ];        //f(1, i, ai)
        }
        m.clear();
        for(int i=n; i>0; i--){
            m[ a[i] ] ++;
            g[i] = m[ a[i] ];        //f(j, n, aj)
        }
        memset(c, 0, sizeof(c));
        long long ans = 0;
        for(int i=n-1; i>0; i--){
            add(g[i+1], 1);
            ans += sum(f[i] - 1);
        }
        printf("%lld\n", ans);

    }
    return 0;
}

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