复杂度O(NlogN)
求逆序对:
#include<iostream> #include<cstdio> #include<string> #include<algorithm> #include<queue> #include<deque> #include<set> #include<map> #include<cmath> #include<cstring> #include<stack> #include<cctype> const double PI = acos(-1.0); #define eps 1e-6 #define INF 0x3f3f3f3f typedef long long ll; using namespace std; const int maxn = 5000010; int n; int a[maxn], c[maxn]; ll ans; //存储逆序对个数 //归并排序,分而治之 void msort(int l, int r) { if (l == r) return; int mid = l + r >> 1; int i = l, j = mid + 1, k = l; msort(l, mid); msort(mid + 1, r); while (i <= mid && j <= r) { if (a[i] <= a[j]) c[k++] = a[i++]; else c[k++] = a[j++], ans += mid - i + 1; } while (i <= mid) { c[k++] = a[i++]; } while (j <= r) { c[k++] = a[j++]; } for (int h = l; h <= r; h++) { a[h] = c[h]; } } int main() { scanf("%d", &n); for (int i = 1; i <=n; i++) scanf("%d", &a[i]); msort(1, n); printf("%lld", ans); return 0; }