【题解】CF91B：Queue

原题传送门
非常简单的题目
对于每个数，需要找到比自己小的下标最大的数，直接离散化然后开个以权值为下标的树状数组维护一下实际下标最大值就完事了
Code：

#include <bits/stdc++.h>
#define maxn 100010
#define _ 0 
using namespace std;
struct data{
	int x, id;
}b[maxn];
int n, a[maxn], ans[maxn], tree[maxn], p;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

bool cmp(data x, data y){ return x.x < y.x; }
void update(int x, int y){ for (; x <= p; x += x & -x) tree[x] = max(tree[x], y); }
int query(int x){ int sum = 0; for (; x; x -= x & -x) sum = max(sum, tree[x]); return sum; }

int main(){
	n = read();
	for (int i = 1; i <= n; ++i) b[i] = (data){read(), i};
	sort(b + 1, b + 1 + n, cmp);
	b[0].x = b[1].x + 1;
	for (int i = 1; i <= n; ++i) a[b[i].id] = b[i].x == b[i - 1].x ? p : ++p;
	for (int i = n; i; --i){
		ans[i] = max(-1, query(a[i] - 1) - i - 1);
		update(a[i], i);
	}
	for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
	return 0 ^ _ ^ 0;
}