邏輯的頁的數量重疊距離

我需要找到截圖的子頁面的數量，
考慮：，
1）頁的總高度，1子頁面的
2）高度，
3）每個連續截圖高度由「5」個像素重疊在上述屏幕
邏輯的頁的數量重疊距離

Ex : Page height = 100 px 
    1 Screen height (subpage) = 20px 
    Then, 
    1st screenshot : 0px to 20 px 
    2nd screenshot : 15px to 35px.. and so on

由於

來源

2014-01-09 zeetit

private int GetSubPageCount(int heightOfPage) 
{ 
    int heightOfSubPage = 20; 
    int overlap = 5; 
    // that gives you count of completely filled sub pages 
    int subPageCount = (heightOfPage - overlap)/(heightOfSubPage - overlap); 

    // if last sub page is not completely filled, than add one page 
    if ((heightOfPage - overlap) % (heightOfSubPage - overlap) > 0) 
     subPageCount++; 

    return subPageCount; 
}

或者你也可以舍入到最小整數值比完全填充的子頁面，大於或等於使用浮點數計算：

private int GetSubPageCount(int heightOfPage) 
{ 
    int heightOfSubPage = 20; 
    int overlap = 5; 
    return (int)Math.Ceiling((double)(heightOfPage - overlap)/(heightOfSubPage - overlap)); 
}

來源

2014-01-09 08:13:49

謝謝..我還能夠解決同樣使用算術級數： AddedLag =（pageHeight/subpageHeight）*重疊因此，額外的頁面：= Math.ceiling（AddedLag/subpageHeight） – zeetit

你可以用一個簡單的公式：

number = (page_height - overlap)/(subpage_height - overlap)

如果數字是分數一個（例如5.123）你應該把加起來（5.123到6）。

例如，如果page_height = 100，subpage_height = 20和重疊= 5，我們可以推導出

number = (100 - 5)/(20 - 5) = 6.333333 
number = 7 (rounded up)

樣品的編號：

public static int SubPageCount(int pageHeight, int height, int overlap) { 
    int result = (pageHeight - overlap)/(height - overlap); 

    // If there's a fractional part - i.e. remainder in not 0 
    // one should round the result up: add 1 
    if (((pageHeight - overlap) % (height - overlap)) != 0) 
    result += 1; 

    return result; 
}

來源

2014-01-09 08:39:30

邏輯的頁的數量重疊距離

回答

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