我正在尝试制作一个简单的爱情计算器。我已预先定义了所有字符值,如a = 1,b = 2和all。但问题是我不知道如何计算长度并逐个添加所有字符。
答案 0 :(得分:1)
如果a = 1,b = 2,...,z = 26
,你可以做什么的简单例子string name = "Amélie Benoît Françoise Ågot Annbjørg";
// We transform letters with diacritics to "pure" letters (é->e, ç ->c...)
string normalized = name.Normalize(NormalizationForm.FormD);
var onlyLetters = normalized.Where(x => x >= 'A' && x <= 'Z' || x >= 'a' && x <= 'z');
// Note that the ø of Annbjørg will be stripped :-(
string strippedName = new string(onlyLetters.ToArray());
Console.WriteLine("Calculating for {0}", strippedName);
int sum = 0;
foreach (char ch in onlyLetters)
{
char ch2 = char.ToUpper(ch);
// char have a value... 'A' == 65, 'B' == 66 and so on,
sum += ch2 - 'A' + 1;
}
// Done
请注意,此代码在Turky中将以不完整的方式工作,因为它们有4 i :-)挪威的名字甚至存在ø的问题,而没有说出不使用“ø”的名称欧洲“字母: - )
您可以使用此方法(应与Windows应用商店兼容)替换Normalize方法调用:
public enum NORM_FORM
{
NormalizationOther = 0,
NormalizationC = 0x1,
NormalizationD = 0x2,
NormalizationKC = 0x5,
NormalizationKD = 0x6
};
[DllImport("Normaliz.dll", CharSet = CharSet.Unicode, ExactSpelling = true, SetLastError = true)]
public static extern int NormalizeString(NORM_FORM NormForm,
string lpSrcString,
int cwSrcLength,
StringBuilder lpDstString,
int cwDstLength);
public static string NormalizeString(string unnormalized, NORM_FORM form)
{
if (unnormalized == string.Empty)
{
return unnormalized;
}
int bufferSize = NormalizeString(form, unnormalized, unnormalized.Length, null, 0);
if (bufferSize <= 0)
{
throw new Exception(string.Format("Win32 error: {0}", Marshal.GetLastWin32Error()));
}
var sb = new StringBuilder(bufferSize);
int result = NormalizeString(form, unnormalized, unnormalized.Length, sb, bufferSize);
if (result <= 0)
{
throw new Exception(string.Format("Win32 error: {0}", Marshal.GetLastWin32Error()));
}
return sb.ToString();
}
和
string normalized = NormalizeString(name, NORM_FORM.NormalizationD);