#559 树上逆序对

#includeusing namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)#define per(i,l,r) for(int i=(l);i>=(r);i--)#define ll long long#define mset(s,t) memset(s,t,sizeof(t))#define mcpy(s,t) memcpy(s,t,sizeof(t))#define fi first#define se second#define pb push_back#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define mp make_pairtypedef pair pii;typedef pair pll;typedef vector vi; typedef vector Vll; typedef vector > vpii;typedef vector > vpll; const ll mod = 1e9 + 7;//const ll mod = 998244353;const double pi = acos(-1.0);inline ll qmi (ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans = ans * a%mod; a = a * a %mod; b >>= 1; } return ans;}inline int read () { int x = 0, f = 0; char ch = getchar(); while (!isdigit(ch)) f |= (ch=='-'),ch= getchar(); while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return f?-x:x;}template void print(T x) { if (x < 0) putchar('-'), x = -x; if (x >= 10) print(x/10); putchar(x % 10 + '0');}inline ll sub (ll a, ll b) { return ((a - b ) %mod + mod) %mod;}inline ll add (ll a, ll b) { return (a + b) %mod;}inline ll inv (ll a) { return qmi(a, mod - 2);}int a[200005], T[200005];int n;int lowbit (int x) { return x & -x;}void update(int x, int y) { for(;x <= n; x += lowbit(x)) T[x] += y; }int query (int x) { int ans = 0; for (;x; x-=lowbit(x)) { ans += T[x]; } return ans;}namespace ls { const int maxn = 5e5 + 10; int C[maxn], L[maxn], A[maxn]; void ls (int *f , int *a, int n) { for (int i = 0; i < n; i ++) A[i] = a[i + 1]; memcpy(C, A, sizeof A); sort(C, C + n); int l = unique(C, C + n) - C; for (int i = 0; i < n; i ++) L[i] = lower_bound(C, C + l, A[i]) - C + 1; for (int i = 0; i < n; i ++) f[i + 1] = L[i]; }}namespace nx { const int MAXN = 5e5 + 10; struct node { int k, x, id; }; vectorg[MAXN]; int m = 0; void add_que(int i, int L, int R, int k) { //离线下询问，i为询问的id [L,R]中小于k的个数 g[L - 1].push_back({ -1,k,i }); g[R].push_back({ 1,k,i }); m++; } void solve_(int *ans ,int *a,int n) { //对a的逆序对的结果放到ans中 for (int i = 1; i <= n; i++) { update(a[i], 1);//update为树状数组的单点加 for (auto v : g[i]) { ans[v.id] += v.k * query(v.x - 1);//query为树状数组的前缀和 } } }}int ans[500005], b[500005];void solve() { cin >> n; for (int i = 1; i <= n; i ++) cin >> b[i]; ls::ls(a, b, n); for (int k = 1; k <= n - 1; k ++) { for (int i= 1;; i++) { int L = k * (i - 1) + 2, R = min(n, i * k + 1); if (L > n) break; nx::add_que(k, L, R, a[i]); } } nx::solve_(ans, a, n); for (int k = 1; k <= n- 1; k ++) cout << ans[k] << ' ';}int main () { // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0); int t; t =1; //cin >> t; while (t --) solve(); return 0;}

问题查询某个区间内的<=k的数量

把问题变成查询问题。问题就变得简单了

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