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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N];
int ans[N];
int n;
// 树状数组模板
typedef long long LL;
#define lowbit(x) (x & -x)
int c[N];
void add(int x, int d) {
for (int i = x; i < N; i += lowbit(i)) c[i] += d;
}
LL sum(int x) {
LL res = 0;
for (int i = x; i; i -= lowbit(i)) res += c[i];
return res;
}
/*
2
4
5
3
1
*/
int main() {
#ifndef ONLINE_JUDGE
freopen("244.in", "r", stdin);
#endif
scanf("%d", &n);
for (int i = 2; i <= n; i++) scanf("%d", &a[i]); // 读入每头奶牛前面有几头奶牛比自己矮
// 这里是区别!
// 利用树状数组把1~n之间每个位置设置为1,方便后面用来求前缀和
// for (int i = 1; i <= n; i++) add(i, 1);
for (int i = n; i; i--) {
int l = 1, r = n;
while (l < r) { // 用二分找出前缀和恰好为a[i] + 1的数
// 这里是区别!
int mid = (l + r) >> 1; // mid枚举的是下标
if (mid - sum(mid) >= a[i] + 1)
r = mid; // 再小点试试
else
l = mid + 1; // 再大点~
}
ans[i] = l; // 记录答案
add(l, 1); // 找到后,这个数-1,标识为0方便下次求前缀和
}
// 输出结果
for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
return 0;
}