题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5792

Description

Given a sequence A with length n,count how many quadruple (a,b,c,d) satisfies: a≠b≠c≠d,1≤a < b≤n,1≤c < d≤n,Aa < Ab,Ac > Ad.

Input

The input consists of multiple test cases.
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An.
1≤n≤50000
0≤Ai≤1e9

Output

For each test case,output a line contains an integer.

Sample Input

4
2 4 1 3
4
1 2 3 4

Sample Output

1
0

Source

2016 Multi-University Training Contest 5

题意：

找出有多少个四元组(a,b,c,d)满足a≠b≠c≠d,a < b,c < d,Aa < Ab,Ac > Ad.

题解：

下面将题意中的Aa < Ab称为升序对，Ac > Ad称为降序对.
首先会想到如果没有a≠b≠c≠d的限制，那么可以分别找到所有升序对和降序对数目相乘. 所以在有a≠b≠c≠d限制的情况下，需要考虑去重.
枚举元素num[i]，考虑以num[i]为升序对右边界的情况：

1. 以num[i]为右边界的升序对个数为：1~i-1中比i小的个数.
2. 降序对：所有降序对 - 包含升序对中元素的降序对.
3. 包含升序对中元素的降序对：Sum((右边比num[j]小 + 左边比num[j]大) + (右边比num[i]小 + 左边比num[i]大)). (j为1~i-1中比num[i]小的数).

接下来的任务是：求出任一数左边比它大、小，右边比它大、小的数的个数各是多少.
这里可以用求逆序数的方法来求. 先将数据离散化.
注意TLE：尽量使用一次树状数组或线段树操作(O(nlogn)) + 多个O(n)的遍历来求得上述四个数组.
若分别使用2次甚至更多次的的线段树操作，将会由于常数太大而TLE(TLE代码附后，使用三次线段树操作).

分别使用left_large,left_small,right_small,right_large来表示上述的四个量. all_less为全部降序对数.
那么结果为：
left_small[i] * all_less - Σ(left_large[i]+right_small[i], left_large[j]+right_small[j]); (其中j为i左边比num[i]小的数，其个数为left_small[i]).
优化：若是直接用上述式子来求，则要再维护一次树状数组求比num[i]小的数的所涉及的重复降序对.
实际上，对于每个num[i]，在减号右边left_large[i]+right_small[i]被计数的次数应该是i的右边比num[i]大的数的个数.
所以上式可优化为代码所示的式子：(总共仅需一次树状数组操作)
ans += left_small[i] * (all_less - left_large[i] - right_small[i]);
ans -= right_large[i] * (left_large[i] + right_small[i]);

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define mid(a,b) ((a+b)>>1)
#define eps 1e-8
#define maxn 55000
#define mod 1000000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;int n;
LL num[maxn], tmpa[maxn];
LL c[maxn];
LL left_large[maxn], left_small[maxn];
LL right_small[maxn], right_large[maxn];
LL tol_cnt[maxn], tol_small[maxn], tol_large[maxn];LL lowbit(LL x) {return x & (-x);
}void update(LL x, LL d) {while(x <=n) {c[x] += d;x += lowbit(x);}
}LL get_sum(LL x) {LL ret = 0;while(x > 0) {ret += c[x];x -= lowbit(x);}return ret;
}int main(int argc, char const *argv[])
{//IN;while(scanf("%d", &n) != EOF){for(int i=1; i<=n; i++)scanf("%d", &num[i]), tmpa[i] = num[i];sort(tmpa+1, tmpa+1+n);map<LL, LL> mymap;LL sz = 0; mymap.clear();for(int i=1; i<=n; i++) {if(!mymap.count(tmpa[i])) {sz++;mymap.insert(make_pair(tmpa[i], sz));}}for(int i=1; i<=n; i++) {num[i] = mymap[num[i]];}memset(left_large, 0, sizeof(left_large));memset(right_large, 0, sizeof(right_large));memset(right_small, 0, sizeof(right_small));memset(left_small, 0, sizeof(left_small));memset(tol_small, 0, sizeof(tol_small));memset(tol_large, 0, sizeof(tol_large));memset(tol_cnt, 0, sizeof(tol_cnt));memset(c, 0, sizeof(c));for(int i=1; i<=n; i++) {tol_cnt[num[i]]++;}for(int i=1; i<=sz; i++) {tol_small[i] = tol_small[i-1] + tol_cnt[i-1];}for(int i=sz; i>=1; i--) {tol_large[i] = tol_large[i+1] + tol_cnt[i+1];}LL all_less = 0;for(int i=1; i<=n; i++) {left_large[i] = get_sum(sz) - get_sum(num[i]);right_large[i] = tol_large[num[i]] - left_large[i];left_small[i] = get_sum(num[i]-1);right_small[i] = tol_small[num[i]] - left_small[i];all_less += right_small[i];update(num[i], 1);}LL ans = 0;for(int i=1; i<=n; i++) {ans += left_small[i] * (all_less - left_large[i] - right_small[i]);ans -= right_large[i] * (left_large[i] + right_small[i]);}printf("%I64d\n", ans);}return 0;
}

TLE代码：(三次线段树操作)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define mid(a,b) ((a+b)>>1)
#define eps 1e-8
#define maxn 55000
#define mod 1000000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;int n;
LL num[maxn], tmpa[maxn];
LL pre[maxn];
LL last[maxn];struct Tree
{int left,right;LL cur;LL sum;
}tree[maxn<<2];void build(int i,int left,int right)
{tree[i].left=left;tree[i].right=right;if(left==right){tree[i].sum = 0;tree[i].cur = 0;return ;}int mid=mid(left,right);build(i<<1,left,mid);build(i<<1|1,mid+1,right);tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;tree[i].cur=tree[i<<1].cur+tree[i<<1|1].cur;
}void update(int i,int x,LL d)
{if(tree[i].left==tree[i].right){tree[i].sum+=1;tree[i].cur+=d;return;}int mid=mid(tree[i].left,tree[i].right);if(x<=mid) update(i<<1,x,d);else update(i<<1|1,x,d);tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;tree[i].cur=tree[i<<1].cur+tree[i<<1|1].cur;
}LL query(int i,int left,int right)
{if(tree[i].left==left&&tree[i].right==right)return tree[i].sum;int mid=mid(tree[i].left,tree[i].right);if(right<=mid) return query(i<<1,left,right);else if(left>mid) return query(i<<1|1,left,right);else return query(i<<1,left,mid)+query(i<<1|1,mid+1,right);
}LL query2(int i,int left,int right)
{if(tree[i].left==left&&tree[i].right==right)return tree[i].cur;int mid=mid(tree[i].left,tree[i].right);if(right<=mid) return query2(i<<1,left,right);else if(left>mid) return query2(i<<1|1,left,right);else return query2(i<<1,left,mid)+query2(i<<1|1,mid+1,right);
}map<LL, LL> mymap;int main(int argc, char const *argv[])
{//IN;while(scanf("%d", &n) != EOF){for(int i=1; i<=n; i++)scanf("%d", &num[i]), tmpa[i] = num[i];sort(tmpa+1, tmpa+1+n);LL sz = 0; mymap.clear();for(int i=1; i<=n; i++) {if(!mymap.count(tmpa[i])) {sz++;mymap.insert(make_pair(tmpa[i], sz));}}for(int i=1; i<=n; i++) {num[i] = mymap[num[i]];}fill(pre, pre+n+1, 0);fill(last, last+n+1, 0);build(1, 1, sz);for(int i=1; i<=n; i++) {update(1, num[i], 1);if(num[i] == sz) continue;pre[i] = query(1, num[i]+1, sz);}build(1, 1, sz);for(int i=n; i>=1; i--) {update(1, num[i], 1);if(num[i] == 1) continue;last[i] = query(1, 1, num[i]-1);}LL all_less  = 0;for(int i=1; i<=n; i++) {all_less += last[i];}build(1, 1, sz);LL ans = 0;for(int i=1; i<=n; i++) {update(1, num[i], pre[i]+last[i]);if(num[i] == 1) continue;LL pre_num = query(1, 1, num[i]-1);LL pre_sum = query2(1, 1, num[i]-1);ans += pre_num * (all_less - pre[i] - last[i]) - pre_sum;}printf("%I64d\n", ans);}return 0;
}