Log for YSITD
/// <summary>
/// Saves only the specified page of <see cref="Src"/>.
/// </summary>
/// <param name="page">The page to show.</param>
/// <param name="pageSize">How many entries to show in a page.</param>
/// <returns>this <see cref="SortPageUtils{T}"/></returns>
public SortPageUtils<T> Page(
int page, int pageSize
)
{
if (page != 0 && pageSize != 0) {
Src = Src.Skip((page - 1) * pageSize).Take(page * pageSize);
}
return this;
}
怎麼確定自己寫的分頁演算法沒有任何錯誤阿 qwq反正就是
(page, pageSize) = (1, 10) 時會顯示「輸出第 0 到 10 個項目」
(page, pageSize) = (3, 10) 時會顯示「輸出第 30 到 40 個項目」
(page, pageSize) = (1, 10) 時會顯示「輸出第 0 到 10 個項目」
(page, pageSize) = (3, 10) 時會顯示「輸出第 30 到 40 個項目」
or just make it an internal node. the leaf will become the children of the newly added internal node
Proof. See Figure 3. Note that, formally, v might be a master node (and thus
with an edge descending below it) but no codewords below it. This is why v
and its descendants are all removed.
Since T1's left child did not originally have a left child, ai can be moved
there to become the leftmost grandchild of the root. T01
2 T0(n) by de finition.
n > 3 ensures that depth(v) in T1 was at least 3. L(T01
) < L(T1) follows.
with an edge descending below it) but no codewords below it. This is why v
and its descendants are all removed.
Since T1's left child did not originally have a left child, ai can be moved
there to become the leftmost grandchild of the root. T01
2 T0(n) by de finition.
n > 3 ensures that depth(v) in T1 was at least 3. L(T01
) < L(T1) follows.
like turning the original node into an internal node, and extend the tree by making the leaf become the left child of that node
what's more important is the intuition why you modify the tree in the first place