原始序列逆序对为ans，新序列的逆序对减少s[i]个，增加n - (s[i] + 1)个，所以增加的逆序对为ans += n - 1 - 2*s[i]；

#include 
     
      using namespace std;typedef long long ll;const int N  = 5e5+10;int s[N], b[N], tmp[N];ll cnt;inline  void merge_sort(int a[], int l, int r){    if(l == r) return;    int mid = (l + r) / 2;    merge_sort(a, l, mid);    merge_sort(a, mid + 1, r);    int i = l, j = mid + 1, k = 0;    while(i <= mid && j <= r){        if(a[i] <= a[j]) tmp[k++] = a[i++];        else tmp[k++] = a[j++], cnt += (ll)mid - i + 1;    }    while(i <= mid) tmp[k++] = a[i++];    while(j <= r) tmp[k++] = a[j++];    for(i = 0; i < k; i++){        a[l + i] = tmp[i];    }}int main(){#ifdef ONLINE_JUDGE#else    freopen("in.txt", "r", stdin);#endif //ONLINE_JUDGE    int n;    while(~scanf("%d", &n)){        cnt = 0;        for(int i = 1; i <= n; i++){            scanf("%d", &s[i]);            b[i] = s[i];        }        merge_sort(b, 1, n);        ll ans = 0x3f3f3f3f;        for(int i = 1; i <= n; i++){            cnt += (n - 1 - 2 * s[i]);            ans = min(cnt, ans);        }        printf("%lld\n", ans);    }    return 0;}