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。

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N];
int tmp[N];
typedef long long ll;
ll merge_sort(int l, int r) {
    if (l >= r) {
        return 0;
    }
    int mid = (l + r) >> 1;
    ll ans = merge_sort(l, mid) + merge_sort(mid + 1, r);
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r) {
        if (a[i] <= a[j]) {
            tmp[k++] = a[i++];
        } else {
            tmp[k++] = a[j++];
            ans += mid - i + 1;
        }
    }
    while (i <= mid) {
        tmp[k++] = a[i++];
    }
    while (j <= r) {
        tmp[k++] = a[j++];
    }
    for (int i = l, j = 0; i <= r; i++, j++) {
        a[i] = tmp[j];
    }
    return ans;
}
int main () {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    cout << merge_sort(0, n - 1) << endl;
    return 0;
}