题目描述

本题注意的是逆序对的个数可能超过int的范围，需要用long long 类型来做。

    while(i <= mid && j <= r)
    {
        if(q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
        else {
            t += mid - i + 1;
            tmp[k ++ ] = q[j ++ ];
        }
    }

解释下这段，如果q[i]>q[j]，意味着q[i]及其之后的值都大于q[j]，所以t = mid - i + 1.

算法1

(归并排序) $O(nlogn)$

C++ 代码

#include <iostream>
#include <vector>
using namespace std;
typedef long long LL;

LL res = 0;

LL  MergeSort(int* q, int l, int r)
{
    if(l >= r) return 0;

    int mid = l + r >> 1;

    LL t = 0;

    t = MergeSort(q, l, mid) +  MergeSort(q, mid + 1, r);

    int i = l, j = mid + 1, k = 0, as = r - l + 1;
    vector<int> tmp(as, 0);
    while(i <= mid && j <= r)
    {
        if(q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
        else {
            t += mid - i + 1;
            tmp[k ++ ] = q[j ++ ];
        }
    }
    while(i <= mid) tmp[k ++ ] = q[i ++ ];
    while(j <= r) tmp[k ++ ] = q[j ++ ];

    for(int i = 0; i < as; i ++ ) q[i + l] = tmp[i];
    return t;
}

int main()
{
    int n; 
    cin >> n;
    int *array = new int[n];
    for(int i = 0; i < n; i ++ ) scanf("%d", &array[i]);

    res = MergeSort(array, 0, n - 1);

    delete[] array;

    cout << res << endl;

    return 0;
}