PAT (Top Level) Practise 1009 Triple Inversions (35)

1009. Triple Inversions (35)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CAO, Peng

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:

221 2 3 4 5 16 6 7 8 9 10 19 11 12 14 15 17 18 21 22 20 13

Sample Output:

8

求三个数的逆序对数，只要求出一个数前后的逆序数相乘即可，直接树状数组都不用离散。

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<stack>#include<algorithm>#include<bitset>#include<functional>using namespace std;typedef unsigned long long ull;typedef long long LL;const int maxn = 1e5 + 10;const int low(int x){ return x&-x; }int n, a[maxn], f[maxn], L[maxn], R[maxn];void add(int x){  for (int i = x; i <= n; i += low(i)) f[i]++;}int get(int x){  int ans = 0;  for (int i = x; i; i -= low(i)) ans += f[i];  return ans;}int main(){  while (scanf("%d", &n) != EOF)  {    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);    for (int i = 1; i <= n; i++) f[i] = 0;    for (int i = 1; i <= n; i++) L[i] = i - 1 - get(a[i]), add(a[i]);    for (int i = 1; i <= n; i++) f[i] = 0;    for (int i = n; i; i--) R[i] = get(a[i]), add(a[i]);    LL ans = 0;    for (int i = 1; i <= n; i++) ans += (LL)L[i] * R[i];    printf("%lld\n", ans);  }  return 0;}