求逆序对的数量,分三种情况:
#include<iostream>
using namespace std;
typedef long long ll;
const int N = 100010;
int a[N];
int tmp[N];
//返回区间里面从l到r逆序对的数量的
ll merge_sort(int a[], int l, int r) {
if (l >= r) return 0;
int mid = l + r >> 1;
//返回左半边内部的逆序对数量+右半边内部的逆序对数量
ll ans = merge_sort(a, l, mid) + merge_sort(a, mid + 1, r);
//归并的过程
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r) {
if (a[i] <= a[j]) tmp[k++] = a[i++];
else {
ans += mid - i + 1;//逆序对的数量
tmp[k++] = a[j++];
}
}
while (i <= mid) tmp[k++] = a[i++];
while (j <= r) tmp[k++] = a[j++];
for (int i = l, j = 0; i <= r; i++, j++) a[i] = tmp[j];
return ans;
}
int main()
{
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
printf("%lld", merge_sort(a, 0, n - 1));
return 0;
}