Pashmak and Parmida's problem（树状数组）

题目链接：http://codeforces.com/contest/459/problem/D

题意： 数列A， ai表示 i-th 的值， f（i，j, x） 表示【i，j】之间x的数目， 问：当 1 <= i < j <= n， 满足 f(1, i, ai) < f（j，n, aj）的对数。

题解：分别求出两组数列  Ii = f(1, i, ai)（1<=i <= n） Jj =  f（j，n, aj）(1 <= j <= n), 然后， 从n开始， 在数列   Ij（i < j <= n）中求出   小于 Ii的个数和。

/***Good Luck***/
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <map>
#include <queue>
#include <vector>
#include <set>
#include <functional>
#include <cmath>

#define Zero(a) memset(a, 0, sizeof(a))
#define Neg(a)  memset(a, -1, sizeof(a))
#define All(a) a.begin(), a.end()
#define PB push_back
#define inf 0x3f3f3f3f
#define inf2 0x7fffffffffffffff
#define ll long long
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")
void get_val(int &a) {
    int value = 0, s = 1;
    char c;
    while ((c = getchar()) == ' ' || c == '
');
    if (c == '-') s = -s; else value = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        value = value * 10 + c - 48;
    a = s * value;
}
const int maxn = 1000010;
int arri[maxn], arrj[maxn];
int arr2[maxn];
int arr1[maxn];
int A[maxn];
int n;
int  lowbit(int x) {
    return x & (-x);
}

int sum(int i) {
    int ret = 0;
    while (i > 0) {
        ret += A[i];
        i -= lowbit(i);
    }
    return ret;
}

void update(int pos, int val) {
    while (pos <= n) {
        A[pos] += val;
        pos += lowbit(pos);
    }
}

int main() {
    //freopen("data.out", "w", stdout);
    //freopen("data.in", "r", stdin);
    //cin.sync_with_stdio(false);

    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", arr1 + i);
        arr2[i] = arr1[i];
    }

    sort(arr2, arr2 + n + 1);           // 离散化
    int len = unique(arr2, arr2 + n) - arr2;
    for (int i = 1; i <= n; ++i) {
        arr1[i] = lower_bound(arr2, arr2 + len, arr1[i]) - arr2;
    }
    Zero(arr2);
    for (int i = 1; i <= n; ++i) {
        arri[i] = ++arr2[arr1[i]];
        
    }
    Zero(arr2);
    for (int i = n ; i > 0; --i) {
        arrj[i] = ++arr2[arr1[i]];
    }
    ll count = 0;
    for (int i = n; i >= 1; --i) {

        count += sum(arri[i] - 1) ;
        update(arrj[i], 1);
    }
    cout << count << endl;
    return 0;
}