我不知道我的命名是否正确！无论如何，这些是我所拥有的整数，例如：
76
121
9660
我想将它们围绕到近百，这样他们必须成为：
100
100
9700
如何在C＃中更快地完成？我想一个算法，但也许在C＃上有一些实用程序？
答案 0 :(得分：65)
尝试使用Math.Round方法。方法如下：
Math.Round(76d / 100d, 0) * 100;
Math.Round(121d / 100d, 0) * 100;
Math.Round(9660d / 100d, 0) * 100;
答案 1 :(得分：20)
我写了一个简单的扩展方法来概括这种舍入：
public static class MathExtensions
{
public static int Round(this int i, int nearest)
{
if (nearest <= 0 || nearest % 10 != 0)
throw new ArgumentOutOfRangeException("nearest", "Must round to a positive multiple of 10");
return (i + 5 * nearest / 10) / nearest * nearest;
}
}
它利用整数除法来找到最接近的舍入。
使用示例：
int example = 152;
Console.WriteLine(example.Round(100)); // round to the nearest 100
Console.WriteLine(example.Round(10)); // round to the nearest 10
在你的例子中：
Console.WriteLine(76.Round(100)); // 100
Console.WriteLine(121.Round(100)); // 100
Console.WriteLine(9660.Round(100)); // 9700
答案 2 :(得分：14)
试试这个表达式：
(n + 50) / 100 * 100
答案 3 :(得分：7)
只是@krizzzn已接受答案的一些补充......
请注意以下内容将返回0：
Math.Round(50d / 100d, 0) * 100;
考虑使用以下内容并使其返回100：
Math.Round(50d / 100d, 0, MidpointRounding.AwayFromZero) * 100;
根据您正在做的事情，使用小数可能是更好的选择（请注意 m ）：
Math.Round(50m / 100m, 0, MidpointRounding.AwayFromZero) * 100m;
答案 4 :(得分：5)
我知道这是一个老线程。我写了一个新方法。希望这对某些人有用。
public static double Round(this float value, int precision)
{
if (precision < -4 && precision > 15)
throw new ArgumentOutOfRangeException("precision", "Must be and integer between -4 and 15");
if (precision >= 0) return Math.Round(value, precision);
else
{
precision = (int)Math.Pow(10, Math.Abs(precision));
value = value + (5 * precision / 10);
return Math.Round(value - (value % precision), 0);
}
}
示例：
float value = F6666.677777;
Console.Write(value.Round(2)) // = 6666.68
Console.Write(value.Round(0)) // = 6667
Console.Write(value.Round(-2)) // = 6700
答案 5 :(得分：1)
嗨我写这个扩展名，这会为你传递的每个号码获得下一百个
/// <summary>
/// this extension gets the next hunfìdred for any number you whant
/// </summary>
/// <param name="i">numeber to rounded</param>
/// <returns>the next hundred number</returns>
/// <remarks>
/// eg.:
/// i = 21 gets 100
/// i = 121 gets 200
/// i = 200 gets 300
/// i = 1211 gets 1300
/// i = -108 gets -200
/// </remarks>
public static int RoundToNextHundred(this int i)
{
return i += (100 * Math.Sign(i) - i % 100);
//use this line below if you want RoundHundred not NEXT
//return i % 100 == byte.MinValue? i : i += (100 * Math.Sign(i) - i % 100);
}
//and for answer at title point use this algoritm
var closeHundred = Math.Round(number / 100D)*100;
//and here the extension method if you prefer
/// <summary>
/// this extension gets the close hundred for any number you whant
/// </summary>
/// <param name="number">number to be rounded</param>
/// <returns>the close hundred number</returns>
/// <remarks>
/// eg.:
/// number = 21 gets 0
/// number = 149 gets 100
/// number = 151 gets 200
/// number = -149 gets -100
/// number = -151 gets -200
/// </remarks>
public static int RoundCloseHundred(this int number)
{
return (int)Math.Round(number / 100D) * 100;
}
答案 6 :(得分：0)
int num = 9660;
int remainder = num % 100;
Console.WriteLine(remainder < 50 ? num - remainder : num + (100 -remainder));
注意：我没有彻底测试过。
答案 7 :(得分：0)
如果您只想向上舍入整数（就像OP实际上那样），那么您可以求助于此解决方案：
public static class MathExtensions
{
public static int RoundUpTo(this int number, int nearest)
{
if (nearest < 10 || nearest % 10 != 0)
throw new ArgumentOutOfRangeException(nameof(nearest), $"{nameof(nearest)} must be a positive multiple of 10, but you specified {nearest}.");
int modulo = number % nearest;
return modulo == 0 ? number : modulo > 0 ? number + (nearest - modulo) : number - modulo;
}
}
如果你想进行浮点（或十进制）舍入，那么求助于@krizzzn和@Jim Aho的答案。
答案 8 :(得分：0)
我内部有一个类似的项目，业务需求是在给定数字的100范围内进行搜索，并查找重复的数据库记录。因此，如果用户使用的是856行，我将搜索800-899。如果用户使用的是8567，我将搜索8500-8599。不是精确地四舍五入到100，但是我想我将独特的方法作为这些基本编码问题中的一部分被嵌入到更大的业务项目中。为了对此进行测试，我从1-99999的十进制列表中植入了种子，并将结果吐出到文件中。
/// <summary>
/// This method accepts an inbound Line Number and returns the line range
/// in the form of lowest to highest based on 100's
/// Example would be 9122 returns 9100 - 9199
/// It's used for generating some additional BOM Temp functionality.
/// </summary>
/// <param name="inboundNumber"></param>
/// <returns></returns>
public static ProjectLineRange CalculateLineRange(decimal inboundNumber)
{
var lineRange = new ProjectLineRange();
var numberLength = inboundNumber.ToString(CultureInfo.InvariantCulture).Length;
switch (numberLength)
{
case 0: //NULL?
break;
case 1: //Represents 1 - 9
lineRange.LineBottom = 1;
lineRange.LineTop = 99;
break;
case 2: //Represents 10 - 99
lineRange.LineBottom = 1;
lineRange.LineTop = 99;
break;
case 3: //Represents 100 - 999
lineRange = CalculateHundredsRange((int)(inboundNumber / 100));
break;
case 4: //Represents 1000 - 9999
lineRange = CalculateThousandsRange(
Convert.ToInt32(inboundNumber.ToString(CultureInfo.InvariantCulture).Substring(1, 1)),
Convert.ToInt32(inboundNumber.ToString(CultureInfo.InvariantCulture).Substring(0, 1)));
break;
case 5: //Represents 10000 - 99999
lineRange = CalculateTenThousandsRange(
Convert.ToInt32(inboundNumber.ToString(CultureInfo.InvariantCulture).Substring(2, 1)),
Convert.ToInt32(inboundNumber.ToString(CultureInfo.InvariantCulture).Substring(1, 1)),
Convert.ToInt32(inboundNumber.ToString(CultureInfo.InvariantCulture).Substring(0, 1)));
break;
}
return lineRange;
}
public class ProjectLineRange
{
public decimal LineBottom { get; set; }
public decimal LineTop { get; set; }
}
/// <summary>
/// Helper method to return the range for the 100's place
/// </summary>
/// <param name="hundredsPlaceValue"></param>
/// <returns></returns>
public static ProjectLineRange CalculateHundredsRange(int hundredsPlaceValue)
{
var tempLineRange = new ProjectLineRange();
tempLineRange.LineBottom = hundredsPlaceValue * 100;
tempLineRange.LineTop = tempLineRange.LineBottom + 99;
return tempLineRange;
}
/// <summary>
/// Helper method to return the range for the 100's place when factoring a 1000's number
/// </summary>
/// <param name="hundredsPlaceValue"></param>
/// <param name="thousandsPlaceValue"></param>
/// <returns></returns>
public static ProjectLineRange CalculateThousandsRange(int hundredsPlaceValue, int thousandsPlaceValue)
{
var tempLineRange = new ProjectLineRange();
var tempThousands = thousandsPlaceValue * 1000;
var hundredsRange = CalculateHundredsRange(hundredsPlaceValue);
tempLineRange.LineBottom = tempThousands + hundredsRange.LineBottom;
tempLineRange.LineTop = tempThousands + hundredsRange.LineTop;
return tempLineRange;
}
/// <summary>
/// Helper method to return the range for the 100's place when factoring a 10000's number
/// </summary>
/// <param name="hundredsPlaceValue"></param>
/// <param name="thousandsPlaceValue"></param>
/// <param name="tenThousandsPlaceValue"></param>
/// <returns></returns>
public static ProjectLineRange CalculateTenThousandsRange(int hundredsPlaceValue, int thousandsPlaceValue, int tenThousandsPlaceValue)
{
var tempLineRange = new ProjectLineRange();
var tempThousands = thousandsPlaceValue * 1000;
var tempTenThousands = tenThousandsPlaceValue * 10000;
var hundredsRange = CalculateHundredsRange(hundredsPlaceValue);
tempLineRange.LineBottom = tempTenThousands + tempThousands + hundredsRange.LineBottom;
tempLineRange.LineTop = tempTenThousands + tempThousands + hundredsRange.LineTop;
return tempLineRange;
}