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| 1 | +// 1. Problem Statement ------------------------------------------------- |
| 2 | +/* |
| 3 | +Given an array of n distinct integers. |
| 4 | +Define the weakness as the number of triplets (i, j, k) such that: |
| 5 | +i < j < k and a[i] > a[j] > a[k]. |
| 6 | +
|
| 7 | +You need to count all such decreasing triplets. |
| 8 | +Constraints are large, so brute force is not allowed. |
| 9 | +*/ |
| 10 | + |
| 11 | +// 2. Approach ----------------------------------------------------------- |
| 12 | +/* |
| 13 | +Fix j as the middle element. |
| 14 | +For each j: |
| 15 | +- Count how many elements greater than a[j] exist on its left |
| 16 | +- Count how many elements smaller than a[j] exist on its right |
| 17 | +
|
| 18 | +Each combination forms a valid triplet. |
| 19 | +Fenwick Tree (Binary Indexed Tree) is used to compute counts efficiently. |
| 20 | +Coordinate compression is applied because values are large. |
| 21 | +*/ |
| 22 | + |
| 23 | +// 3. Complexity --------------------------------------------------------- |
| 24 | +/* |
| 25 | +Time Complexity: O(n log n) |
| 26 | +Space Complexity: O(n) |
| 27 | +*/ |
| 28 | + |
| 29 | +// 4. Problem Submission Link -------------------------------------------- |
| 30 | +/* |
| 31 | +https://codeforces.com/contest/61/submission/356384283 |
| 32 | +*/ |
| 33 | + |
| 34 | +import java.io.*; |
| 35 | +import java.util.*; |
| 36 | + |
| 37 | +public class Solution2 { |
| 38 | + |
| 39 | + static class Fenwick { |
| 40 | + int n; |
| 41 | + long[] bit; |
| 42 | + |
| 43 | + Fenwick(int n) { |
| 44 | + this.n = n; |
| 45 | + bit = new long[n + 1]; |
| 46 | + } |
| 47 | + |
| 48 | + void update(int i, long v) { |
| 49 | + while (i <= n) { |
| 50 | + bit[i] += v; |
| 51 | + i += i & -i; |
| 52 | + } |
| 53 | + } |
| 54 | + |
| 55 | + long query(int i) { |
| 56 | + long s = 0; |
| 57 | + while (i > 0) { |
| 58 | + s += bit[i]; |
| 59 | + i -= i & -i; |
| 60 | + } |
| 61 | + return s; |
| 62 | + } |
| 63 | + |
| 64 | + long range(int l, int r) { |
| 65 | + return query(r) - query(l - 1); |
| 66 | + } |
| 67 | + } |
| 68 | + |
| 69 | + public static void main(String[] args) throws Exception { |
| 70 | + FastScanner fs = new FastScanner(System.in); |
| 71 | + |
| 72 | + int n = fs.nextInt(); |
| 73 | + int[] a = new int[n]; |
| 74 | + for (int i = 0; i < n; i++) a[i] = fs.nextInt(); |
| 75 | + |
| 76 | + int[] b = a.clone(); |
| 77 | + Arrays.sort(b); |
| 78 | + Map<Integer, Integer> map = new HashMap<>(n); |
| 79 | + int id = 1; |
| 80 | + for (int x : b) if (!map.containsKey(x)) map.put(x, id++); |
| 81 | + |
| 82 | + int[] c = new int[n]; |
| 83 | + for (int i = 0; i < n; i++) c[i] = map.get(a[i]); |
| 84 | + |
| 85 | + long[] L = new long[n]; |
| 86 | + long[] R = new long[n]; |
| 87 | + |
| 88 | + Fenwick fw = new Fenwick(n); |
| 89 | + for (int i = 0; i < n; i++) { |
| 90 | + L[i] = fw.range(c[i] + 1, n); |
| 91 | + fw.update(c[i], 1); |
| 92 | + } |
| 93 | + |
| 94 | + fw = new Fenwick(n); |
| 95 | + for (int i = n - 1; i >= 0; i--) { |
| 96 | + R[i] = fw.query(c[i] - 1); |
| 97 | + fw.update(c[i], 1); |
| 98 | + } |
| 99 | + |
| 100 | + long ans = 0; |
| 101 | + for (int i = 0; i < n; i++) ans += L[i] * R[i]; |
| 102 | + |
| 103 | + System.out.println(ans); |
| 104 | + } |
| 105 | + |
| 106 | + static class FastScanner { |
| 107 | + private final byte[] buf = new byte[1 << 16]; |
| 108 | + private int idx = 0, size = 0; |
| 109 | + private final InputStream in; |
| 110 | + |
| 111 | + FastScanner(InputStream in) { |
| 112 | + this.in = in; |
| 113 | + } |
| 114 | + |
| 115 | + int read() throws IOException { |
| 116 | + if (idx >= size) { |
| 117 | + size = in.read(buf); |
| 118 | + idx = 0; |
| 119 | + if (size <= 0) return -1; |
| 120 | + } |
| 121 | + return buf[idx++]; |
| 122 | + } |
| 123 | + |
| 124 | + int nextInt() throws IOException { |
| 125 | + int c, s = 1, x = 0; |
| 126 | + do c = read(); while (c <= ' '); |
| 127 | + if (c == '-') { |
| 128 | + s = -1; |
| 129 | + c = read(); |
| 130 | + } |
| 131 | + while (c > ' ') { |
| 132 | + x = x * 10 + (c - '0'); |
| 133 | + c = read(); |
| 134 | + } |
| 135 | + return x * s; |
| 136 | + } |
| 137 | + } |
| 138 | +} |
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