[CF538F]A Heap of Heaps && 持久化线段树

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直接枚举k 每个有儿子的节点和他的儿子区间的起点容易发现这样枚举是调和级数

然后需要一个东西查询某段区间比x小的数上主席树即可

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<queue>#define SF scanf#define PF printfusing namespace std;typedef long long LL;const int MAXN = 200000;int root[MAXN+10], A[MAXN+10], B[MAXN+10], n;struct Node {int ls, rs, sum;} ;struct Seg_Tree {Node tree[MAXN*100+10];int ncnt;void NewNode(int &i) {tree[++ncnt] = tree[i];i = ncnt;}void ins(int &i, int L, int R, int val) {NewNode(i);tree[i].sum++;if(L == R) return ;int mid = (L+R) >> 1;if(val <= mid) ins(tree[i].ls, L, mid, val);else ins(tree[i].rs, mid+1, R, val);}int query(int i, int L, int R, int l, int r) {if(l > R || r < L) return 0;if(l <= L && R <= r) return tree[i].sum;return query(tree[i].ls, L, (L+R)>>1, l, r) + query(tree[i].rs, ((L+R)>>1)+1, R, l, r);}void build(int &i, int L, int R) {NewNode(i);if(L == R) return ;int mid = (L+R) >> 1;build(tree[i].ls, L, mid);build(tree[i].rs, mid+1, R);}} seg;int query(int L, int R, int x) {if(R > n) R = n;return seg.query(root[R], 1, n, 1, x-1) - seg.query(root[L-1], 1, n, 1, x-1);}int main() {SF("%d", &n);seg.build(root[0], 1, n);for(int i = 1; i <= n; i++) SF("%d", &A[i]), B[i] = A[i];sort(B+1, B+1+n);int Len = unique(B+1, B+1+n) - (B+1);for(int i = 1; i <= n; i++) A[i] = lower_bound(B+1, B+1+Len, A[i]) - B;for(int i = 1; i <= n; i++) {root[i] = root[i-1];seg.ins(root[i], 1, n, A[i]);}for(int k = 1; k < n; k++) {LL ans = 0;for(int i = 1, j = 1; j <= n; i++, j += k)ans += query(j+1, j+k, A[i]);cout << ans << ' ';}puts("");return 0;}/*51 21 32 42 5*/

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