AcWing 788. 逆序对的数量
原题链接
简单
作者:
no_moment
,
2024-12-28 07:28:28
,
所有人可见
,
阅读 3
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 100010;
int q[N], tmp[N];
LL merge_sort(int q[], int l, int r) {
if(l >= r) return 0;
int mid = l + r >> 1;
LL res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
int i = l , j = mid + 1, k = 0;
while(i <= mid && j <= r) {
if(q[i] <= q[j]) tmp[k ++] = q[i ++];
else {
tmp[k ++] = q[j ++];
res += mid - i + 1;
}
}
//扫尾
while(i <= mid) tmp[k ++] = q[i ++];
while(j <= r) tmp[k ++] = q[j ++];
//更新q[]
for(int i = l, j = 0; i <= r; i ++, j ++) q[i] = tmp[j];
return res;
}
int main() {
int n;
scanf("%d", &n);
for(int i = 0; i < n; i ++) scanf("%d", &q[i]);
printf("%ld", merge_sort(q, 0, n - 1));
return 0;
}