问题是：
如果数字1到5用文字写出：一，二，三，四， 五，然后共有3 + 3 + 5 + 4 + 4 = 19个字母。
如果包括1到1000（一千）的所有数字 用文字写出，会用多少个字母？
注意：不要计算空格或连字符。例如，342（三百 和四十二个）包含23个字母和115个（一百一十五个） 包含20个字母。在写出数字时使用“和”是 符合英国用法。
我写的代码是：
int getCount(map<int, int> numWords, int i)
{
if (i <= 20) // one to twenty
return numWords[i];
else if (i <= 99 && i % 10 == 0) // thirty, forty, fifty etc.
return numWords[i];
else if (i <= 99) // thirty one, seventy eight etc.
{
int md = i % 10;
return numWords[i-md] + numWords[md]; // (1) -> if its 55, then take 50; (2) -> take 5
}
else if (i <= 999 && i % 100 == 0) // two hundred, five hundred etc.
{
int md = i / 100;
return numWords[md] + numWords[100]; // number + hundred
}
else if(i <= 999 && i % 10 == 0) // 340 three hundred forty
{
int hunsDig = (i - (i % 100)) / 100; // (340 - 40)/100 = 3
int tens = i - hunsDig*100; // 340-300=40
return numWords[hunsDig] + numWords[100] + numWords[0] + numWords[tens]; // three hundred and forty
}
else if(i <= 999) // 342
{
int hunsDig = (i - (i % 100)) / 100; // (342 - 42)/100 = 3
int units = i % 10; // 342 % 10 = 2
int tens = (i % 100) - units; // (342 % 100=42) - 2 = 40
int tensCount = tens == 0 ? 0 : numWords[tens];
return numWords[hunsDig] + numWords[100] + numWords[0] + tensCount + numWords[units]; // three hundred and forty two
}
else if(i==1000)
return numWords[1] + numWords[1000];
}
void problem17()
{
// make a map of all the words and corresponding word lengths
map<int, int> numWords;
numWords[1] = 3; // one
numWords[2] = 3; // two
numWords[3] = 5; // three
numWords[4] = 4; // four
numWords[5] = 4; // five
numWords[6] = 3; // six
numWords[7] = 5; // seven
numWords[8] = 5; // eight
numWords[9] = 4; // nine
numWords[10] = 3; // ten
numWords[11] = 6; // eleven
numWords[12] = 6; // twelve
numWords[13] = 8; // thirteen
numWords[14] = 8; // fourteen
numWords[15] = 7; // fifteen
numWords[16] = 7; // sixteen
numWords[17] = 9; // seventeen
numWords[18] = 8; // eighteen
numWords[19] = 8; // nineteen
numWords[20] = 6; // twenty
numWords[30] = 6; // thirty
numWords[40] = 5; // forty
numWords[50] = 5; // fifty
numWords[60] = 5; // sixty
numWords[70] = 7; // seventy
numWords[80] = 6; // eighty
numWords[90] = 6; // ninety
numWords[100] = 7; // hundred
numWords[1000] = 8; // thousand
numWords[0] = 3; // and
int totalCount = 0; // total num of words
for (int i=1; i <= 1000; i++)
{
totalCount += getCount(numWords, i);
}
cout << "Total number of letters required: " << totalCount;
}
然而，这并没有给我正确的答案。我究竟做错了什么？输出21088，答案是21124。
答案 0 :(得分：2)
你不是在考虑数十个权利，即111-120 211-220等等。
int tensCount = 0;
if(tens==1) {tenscount = words[10+units]; units=0;}
else if(tens>1) {tenscount = words[tens]; units = words[units];}
else units = words[units];
并在回报中添加单位
答案 1 :(得分：1)
我认为你的问题是像215这样的数字。你认为它是“二百零五”。
顺便说一句，这是我写的python中的正确解决方案（抱歉没有在C中执行）：numWords = {1: 3, 2: 3, 3: 5, 4: 4, 5: 4, 6: 3, 7: 5, 8: 5, 9: 4, 10: 3, 11: 6, 12: 6, 13: 8, 14: 8, 15: 7, 16: 7, 17: 9, 18: 8, 19: 8, 20: 6, 30: 6, 40: 5, 50: 5, 60: 5, 70: 7, 80: 6, 90: 6, 100: 7, 1000: 8, 0: 3}
def get_count(number):
if number==0:
return 0
if number > 99:
return get_count(number / 100) + len("hundred") + ((len('and') + get_count(number % 100)) if (get_count(number % 100)) else 0)
if number > 20:
return numWords[(number / 10)*10] + get_count(number % 10)
return numWords[number]
print sum(get_count(x) for x in xrange(1,1000))+len('one thousand')
你可以看到它比你的短得多。我认为主要的改进是使用对该方法的递归调用。这会阻止你陷入215的问题，因为递归分别处理200和15。
答案 2 :(得分：0)
我的Python代码
ones = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]
tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
def to_words(i):
if 0 <= i < 20:
return ones[i]
elif 20 <= i < 100:
return tens[i//10]+ (ones[i%10] if(i%10!=0) else "")
elif 100 <= i <1000:
return ones[i//100] + "hundred" + ("and"+to_words(i%100) if(i%100!=0) else "")
s=0
for k in range(1,1000):
s+=len(to_words(k))
print(s+11)
答案 3 :(得分：-1)
//这是一个计算数字的单词表示中的字母的函数
int countLetters(int num)
{
int count = 0;
if( num >= 1000000000 )
count += countLetters( num / 1000000000 ) + 7 + countLetters( num % 1000000000);
else if ( num >= 1000000 )
count += countLetters( num / 1000000 ) + 7 + countLetters( num % 1000000);
else if( num >= 1000 )
count += countLetters( num / 1000 ) + 8 + countLetters( num % 1000);
else if( num >= 100 )
count += countLetters( num / 100 ) + 7 + countLetters( num % 100) + (num % 100 > 0 ? 3 : 0);
else if( num >= 20 )
{
switch( num / 10 )
{
case 2: count += 6; break;
case 3: count += 6; break;
case 4: count += 5; break;
case 5: count += 5; break;
case 6: count += 5; break;
case 7: count += 7; break;
case 8: count += 6; break;
case 9: count += 6; break;
}
count += countLetters( num % 10 );
}
else
{
switch( num )
{
case 1: count += 3; break;
case 2: count += 3; break;
case 3: count += 5; break;
case 4: count += 4; break;
case 5: count += 4; break;
case 6: count += 3; break;
case 7: count += 5; break;
case 8: count += 5; break;
case 9: count += 4; break;
case 10: count += 3; break;
case 11: count += 6; break;
case 12: count += 6; break;
case 13: count += 8; break;
case 14: count += 8; break;
case 15: count += 7; break;
case 16: count += 7; break;
case 17: count += 9;break;
case 18: count += 8; break;
case 19: count += 8; break;
}
}
return count;
}