D. Prefix Purchase

D. Prefix Purchase

You have an array $a$ of size $n$, initially filled with zeros ($a_1 = a_2 = \ldots = a_n = 0$). You also have an array of integers $c$ of size $n$.

Initially, you have $k$ coins. By paying $c_i$ coins, you can add $1$ to all elements of the array $a$ from the first to the $i$-th element ($a_j \mathrel{+}= 1$ for all $1 \leq j \leq i$). You can buy any $c_i$ any number of times. A purchase is only possible if $k \geq c_i$, meaning that at any moment $k \geq 0$ must hold true.

Find the lexicographically largest array $a$ that can be obtained.

An array $a$ is lexicographically smaller than an array $b$ of the same length if and only if in the first position where $a$ and $b$ differ, the element in array $a$ is smaller than the corresponding element in $b$.

Input

The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. This is followed by a description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of arrays $a$ and $c$.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the array $c$.

The third line of each test case contains a single integer $k$ ($1 \leq k \leq 10^9$) — the number of coins you have.

It is guaranteed that the sum of all $n$ values across all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case, output $n$ integers $a_1, a_2, \ldots, a_n$ — the lexicographically largest array $a$ that can be obtained.

Example

input

4
3
1 2 3
5
2
3 4
7
3
3 2 1
2
6
10 6 4 6 3 4
7

output

5 0 0 
2 1 
2 2 2 
2 2 2 2 2 1 

Note

In the first test case, $a_1$ cannot be greater than $5$, and if we buy $c_1$ five times, we will run out of money, so $a = [5, 0, 0]$.

In the second test case, $a_1$ cannot be greater than $2$, but we can buy $c_1$ and $c_2$ once each (buying $c_2$ twice is not possible), so $a = [2, 1]$.

 

解题思路

　　如果有下标 $i < j$，并且 $c_i \geq c_j$，那么很明显买入 $c_j$ 要比 $c_i$ 更优，因为不仅前缀的长度 $j$ 比 $i$ 大，而且价格也不超过 $c_i$。因此要把所有这样的 $c_i$ 删掉，有点类似于单调栈，最后栈内的元素（存的是下标）是严格递增的，满足 $i < j$，则 $c_i < c_j$。

　　然后就是贪心地选择栈内的哪些元素购买，使得序列 $a$ 的字典序最大，这步卡了我很久。设栈的大小为 $sz$，栈底元素是 $\text{stk}[1]$，由于栈底元素对应的 $c_{\text{stk}[1]}$ 最小，因此要使得 $a$ 的前 $\text{stk}[1]$ 个元素字典序最大，那么在下标 $\text{stk}[1]$ 处最多能买入 $\left\lfloor \dfrac{k}{c_{\text{stk}[1]}} \right\rfloor$ 次。对于前 $\text{stk}[1]$ 个元素而言这是上界，否则如果在其他位置 $\text{stk}[i], \, i \in [2,sz]$ 买入，那么字典序肯定不如这个上界大，因为 $c_{\text{stk}[i]} > c_{\text{stk}[1]}, \, \left\lfloor \dfrac{k}{c_{\text{stk}[i]}} \right\rfloor < \left\lfloor \dfrac{k}{c_{\text{stk}[1]}} \right\rfloor$。

　　买入后会有剩余，记作 $k' = k \bmod c_{\text{stk}[1]}$，在保持 $a$ 的前 $\text{stk}[1]$ 个元素字典序最大的前提下，看看能不能买入其他的 $\text{stk}[i]$ 来替换 $\text{stk}[1]$，使得字典序更大。对于 $\text{stk}[2]$ 的话，如果想要用 $\text{stk}[1]$ 替换，那么需要 $c_{\text{stk}[2]} - c_{\text{stk}[1]}$ 的花费，而最多可以替换 $\left\lfloor \dfrac{k'}{c_{\text{stk}[2]} - c_{\text{stk}[1]}} \right\rfloor$ 次。另外还要考虑到在 $\text{stk}[1]$ 买入的次数 $a_{\text{stk}[1]}$，即替换的次数不能超过这个值，因此实际上最多能替换 $\min \left\{ a_{\text{stk}[1]}, \, \left\lfloor \dfrac{k'}{c_{\text{stk}[2]} - c_{\text{stk}[1]}} \right\rfloor \right\}$ 次。

　　记剩余的 $k'' = k' \bmod c_{\text{stk}[2]} - c_{\text{stk}[1]}$，同理对于 $\text{stk}[3]$，我们应该用 $\text{stk[2]}$ 来替换 $\min \left\{ a_{\text{stk}[2]}, \, \left\lfloor \dfrac{k''}{c_{\text{stk}[3]} - c_{\text{stk}[2]}} \right\rfloor  \right\}$ 次。而不用再考虑用 $\text{stk}[1]$ 来替换，这是因为将 $\text{stk}[1]$ 替换成 $\text{stk}[3]$，等价于先将 $\text{stk}[1]$ 替换成 $\text{stk}[2]$，再将 $\text{stk}[2]$ 替换成 $\text{stk}[3]$，两种替换方式的花费是一样的。而上一步已经通过最大次数来将 $\text{stk}[1]$ 替换成 $\text{stk}[2]$，以使得 $a$ 的字典序尽可能大，因此 $k''$ 是无法再将 $\text{stk}[1]$ 替换成 $\text{stk}[2]$，因此不足以将 $\text{stk}[1]$ 替换成 $\text{stk}[3]$。

　　以此类推，可以发现每一次只需将前一个元素替换即可。

　　AC 代码如下，时间复杂度为 $O(n)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int a[N];
int stk[N], tp;
int ans[N];

void solve() {
    int n, m;
    scanf("%d", &n);
    tp = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", a + i);
        while (tp && a[i] <= a[stk[tp]]) {    // 删掉下标小于i且花费大于等于a[i]的元素
            tp--;
        }
        stk[++tp] = i;    // 栈里存的是下标
    }
    scanf("%d", &m);
    memset(ans, 0, n + 10 << 2);
    ans[0] = 2e9;
    for (int i = 1; i <= tp; i++) {
        int t = min(ans[stk[i - 1]], m / (a[stk[i]] - a[stk[i - 1]]));    // 最多替换的次数
        ans[stk[i]] = t;
        ans[stk[i - 1]] -= t;
        m -= t * (a[stk[i]] - a[stk[i - 1]]);
    }
    for (int i = n; i; i--) {
        ans[i] += ans[i + 1];
    }
    for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
    }
    printf("\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }

    return 0;
}

 

参考资料

　　CodeTON Round 6 (Div. 1 + Div. 2, Rated, Prizes!) Editorial：https://codeforces.com/blog/entry/120524