PAT 1009. Triple Inversions (35) 数状数组

Given a list of N integers A₁, A₂, A₃,...A_N, there‘s a famous problem to count the number of inversions in it. An inversion is defined as a pair of indices i < j such that A_i > A_j.

Now we have a new challenging problem. You are supposed to count the number of triple inversions in it. As you may guess, a triple inversion is defined as a triple of indices i < j < k such that A_i > A_j > A_k. For example, in the list {5, 1, 4, 3, 2} there are 4 triple inversions, namely (5,4,3), (5,4,2), (5,3,2) and (4,3,2). To simplify the problem, the list A is given as a permutation of integers from 1 to N.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N in [3, 10⁵]. The second line contains a permutation of integers from 1 to N and each of the integer is separated by a single space.

Output Specification:

For each case, print in a line the number of triple inversions in the list.

Sample Input:

22
1 2 3 4 5 16 6 7 8 9 10 19 11 12 14 15 17 18 21 22 20 13

Sample Output:

8

题意：给定1~n的无序数列，求其中长度=3连续递减的字串个数，如样例：(16,14,13)。思路：愚蠢！愚蠢！刚开始找的是三个数中的最前一个，然而判断连续递减就有点困难（这题时间限制为300ms）。其实可以找中间的那个数，再向左查询比中间数大的数的个数，向右查询比中间数小的个数, 两侧相乘就是解了。最大的问题是如何记录大小，因为给定的是1~n连续的数，甚至不需用离散化，通过遍历到某个数，找比它小或大的是否已经被标记了，再标记这个数。快速求和操作显然要用到树状数组。噢，还有这题注意结果使用long long 为此而WA。

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define LL long long
 6 using namespace std;
 7
 8
 9 int n, c[100010], a[100010];
10 LL l[100010], r[100010];
11 void add(int x)  //增加操作
12 {
13     while(x <= n)
14     {
15         c[x]++;
16         x += x & (-x);
17     }
18 }
19
20 int get(int x) //获取和
21 {
22     int ans = 0;
23     while(x)
24     {
25         ans +=c[x];
26         x -= x & (-x);
27     }
28     return ans;
29 }
30 int main()
31 {
32     LL ans;
33     while(cin >> n)
34     {
35         ans = 0;
36         for(int i = 1; i <= n; i++)
37         {
38             scanf("%d", a + i);
39             c[i] = 0;
40         }
41         for(int i = 1; i <= n; i++)
42         {
43             l[i] = i - 1 - get(a[i]);
44             add(a[i]);
45         }
46         for(int i = 1; i <= n; i++)
47             c[i] = 0;
48         for(int i = 1; i <= n; i++)
49         {
50             r[i] = a[i] - 1 - get(a[i]);//为value - 左侧大于它的个数
51             add(a[i]);
52         }
53
54         for(int i = 1; i <= n; i++)
55             ans += l[i]*r[i];
56
57         printf("%lld\n", ans);
58     }
59 }