AcWing 788. 逆序对的数量
原题链接
简单
作者:
我要出去乱说
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2021-01-22 23:19:36
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所有人可见
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阅读 298
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1e6 + 10;
int q[N], tmp[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);
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
{
if (q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
else //q[i]属于左半部分,只有左半部分大于右半部分时才会出现逆序对
{
tmp[k ++ ] = q[j ++ ];
res += mid - i + 1; //从q[i]到q[mid]为止都是比q[j]大的数,故逆序对个数这样算
}
}
while (i <= mid) tmp[k ++ ] = q[i ++ ];
while (j <= r) tmp[k ++ ] = q[j ++ ];
for (int i = l, j = 0; i <= r; i ++ , j ++ ) q[i] = tmp[j];
return res;
}
int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i ++ ) cin >> q[i];
LL t = merge_sort(0, n - 1);
cout << t << endl;
return 0;
}