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

Input

3
2 3 1

Output

Input

4
10 8 3 1

Output

Input

4
1 5 4 3

Output

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: zchao1995@gmail.com > 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;
}

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