4.逆序对的数量

注意数据范围，n=100000，从大到小时，逆序对数量最多。

约为5 * 10 ^ 9。超过了int的最大范围，所以需要用long long来存。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 100010;
 4 int a[N];
 5 int tmp[N];
 6 typedef long long ll;
 7 ll merge_sort(int l, int r) {
 8     if (l >= r) {
 9         return 0;
10     }
11     int mid = (l + r) >> 1;
12     ll ans = merge_sort(l, mid) + merge_sort(mid + 1, r);
13     int k = 0, i = l, j = mid + 1;
14     while (i <= mid && j <= r) {
15         if (a[i] <= a[j]) {
16             tmp[k++] = a[i++];
17         } else {
18             tmp[k++] = a[j++];
19             ans += mid - i + 1;
20         }
21     }
22     while (i <= mid) {
23         tmp[k++] = a[i++];
24     }
25     while (j <= r) {
26         tmp[k++] = a[j++];
27     }
28     for (int i = l, j = 0; i <= r; i++, j++) {
29         a[i] = tmp[j];
30     }
31     return ans;
32 }
33 int main () {
34     int n;
35     cin >> n;
36     for (int i = 0; i < n; i++) {
37         cin >> a[i];
38     }
39     cout << merge_sort(0, n - 1) << endl;
40     return 0;
41 }