AcWing 788. 逆序对的数量
原题链接
简单
作者:
冉俊泽
,
2022-02-15 21:27:38
,
所有人可见
,
阅读 133
#include<iostream>
using namespace std;
typedef long long LL;
const int N = 100010;
int q[N], tem[N];
LL merge_sort(int l, int r)
{
if (l >= r) return 0;
int mid = (l + r) >> 1;
LL res = merge_sort(l, mid) + merge_sort(mid + 1, r);
//归并排序,并处理逆序对数
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
{
if (q[i] <= q[j]) tem[k++] = q[i++];
else
{
tem[k++] = q[j++];
res += mid - i + 1;//处理逆序对个数
}
}
//扫尾
while (i <= mid) tem[k++] = q[i++];
while (j <= r) tem[k++] = q[j++];
//物归原主,将排序好的数组放回原数组
for (int i = l, j = 0; i <= r; i++, j++) q[i] = tem[j];
return res;
}
int main()
{
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &q[i]);
cout << merge_sort(0, n - 1) << endl;
return 0;
}