Codeforces Beta Round #57 (Div. 2)E---Enemy is weak(树状数组+离散化)

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: “A lion is never afraid of a hundred sheep”.

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur’s opinion the weakness of an army is equal to the number of triplets i, j, k such that i < j < k and ai > aj > ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.
Input

The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains n different positive integers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army.
Output

A single integer number, the weakness of the Roman army.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Sample test(s)
Input

3
3 2 1

Output

1

Input

3
2 3 1

Output

0

Input

4
10 8 3 1

Output

4

Input

4
1 5 4 3

Output

1

两次树状数组维护，第一次维护求出每一个位置上的逆序数的对数
第二次维护到目前为止，插入到树状数组上的逆序对的总数目
数据较大，离散化一下

/************************************************************************* > File Name: CF-57-E.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年03月23日 星期一 21时00分16秒 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 1010000;
LL tree[N];
LL inver[N];
int val[N];
LL pre[N];
int xis[N];
int cnt;

int binsearch(int x)
{
    int l = 1;
    int r = cnt;
    int mid;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        if (xis[mid] > x)
        {
            r = mid - 1;
        }
        else if (xis[mid] < x)
        {
            l = mid + 1;
        }
        else
        {
            break;
        }
    }
    return mid;
}

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

void add(int x, int cnt)
{
    for (int i = x; i <= N; i += lowbit(i))
    {
        tree[i] += cnt;
    }
}

LL sum(int x)
{
    LL ans = 0;
    for (int i = x; i; i -= lowbit(i))
    {
        ans += tree[i];
    }
    return ans;
}

int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        cnt = 0;
        LL ans = 0;
        memset(pre, 0, sizeof(pre));
        memset(inver, 0, sizeof(inver));
        memset(tree, 0, sizeof(tree));
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &val[i]);
            xis[++cnt] = val[i];
        }
        sort(xis + 1, xis + 1 + cnt);
        cnt = unique(xis + 1, xis + 1 + cnt) - xis - 1;
        for (int i = 1; i <= n; ++i)
        {
            int x = binsearch(val[i]);
            add(x, 1);
            inver[i] = i - sum(x);
            pre[i] = pre[i - 1] + inver[i];
        }
        memset(tree, 0, sizeof(tree));
        for (int i = 1; i <= n; ++i)
        {
            int x = binsearch(val[i]);
            add(x, inver[i]);
            ans += pre[i] - sum(x);
        }
        printf("%lld\n", ans);
    }
    return 0;
}