在一个数组中，如果两个元素a[i]和a[j]满足i < j且a[i] > a[j]，则a[i]和a[j]是一个逆序对。以下代码用于统计数组a区间[l,r]内的逆序对总数：

cnt = 0

def merge_count(a, l, m, r):
    global cnt
    i = l
    j = m + 1
    while i <= m and j <= r:
        if a[i] <= a[j]:
            i += 1
        else:
            cnt += (m - i + 1)
            j += 1

if __name__ == "__main__":
    a = [2, 4, 1, 3]
    merge_count(a, 0, 1, 3)
    print(f"跨区间逆序对数量：{cnt}")