求数组逆序对数最快的c++代码

以下是基于归并排序的算法，可以在O(nlogn)的时间复杂度内求出数组的逆序对数：

#include <stdio.h>

long long merge(int arr[], int left, int mid, int right) {
    int i = left, j = mid + 1, k = 0;
    long long count = 0;
    int temp[right - left + 1];

    while (i <= mid && j <= right) {
        if (arr[i] <= arr[j]) {
            temp[k++] = arr[i++];
        } else {
            temp[k++] = arr[j++];
            count += mid - i + 1;
        }
    }

    while (i <= mid) {
        temp[k++] = arr[i++];
    }

    while (j <= right) {
        temp[k++] = arr[j++];
    }

    for (i = left, k = 0; i <= right; i++, k++) {
        arr[i] = temp[k];
    }

    return count;
}

long long mergeSort(int arr[], int left, int right) {
    if (left >= right) {
        return 0;
    }

    int mid = (left + right) / 2;
    long long count = 0;
    count += mergeSort(arr, left, mid);
    count += mergeSort(arr, mid + 1, right);
    count += merge(arr, left, mid, right);
    return count;
}

long long countInversePairs(int arr[], int n) {
    return mergeSort(arr, 0, n - 1);
}

int main() {
    int arr[] = { 7, 5, 6, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    long long count = countInversePairs(arr, n);

    printf("Number of inverse pairs: %lld\n", count);

    return 0;
}