我试图将数字输入转换为单词。我是初学者，所以我的代码看起来很笨重。我也为张贴这么大的代码而道歉。
#include "stdafx.h"
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
int main()
{
int i = 0, number = 0;
string num_name = "";
cout << "Enter a number (maximum value is 1,000,000,000)" << endl;
cin >> number;
//To find the maximum magnitude
while ((number / pow(10, i)) >= 1)
{
i++;
}
i--;
int j = i;
cout << j << endl; //Check
//To find which multiple of 3 i is
int n = 0;
while (j >= 3)
{
if (j % 3 == 0)
n++;
j--;
}
j = i;
cout << j << endl; //Check
cout << n << endl; //Check
//String construction
int k = n;
int q = 0;
while (k >= 1)
{
while (j >= (3 * k))
{
q = (int)(number / pow(10, j));
if (j == 2 + (3 * k))
{
if (q == 1) num_name += "one";
else if (q == 2) num_name += "two";
else if (q == 3) num_name += "three";
else if (q == 4) num_name += "four";
else if (q == 5) num_name += "five";
else if (q == 6) num_name += "six";
else if (q == 7) num_name += "seven";
else if (q == 8) num_name += "eight";
else if (q == 9) num_name += "nine";
num_name += "hundred";
}
if (j == 1 + (3 * k))
continue;
if (j == 3 * k)
{
if (q == 1) num_name += "one";
else if (q == 2) num_name += "two";
else if (q == 3) num_name += "three";
else if (q == 4) num_name += "four";
else if (q == 5) num_name += "five";
else if (q == 6) num_name += "six";
else if (q == 7) num_name += "seven";
else if (q == 8) num_name += "eight";
else if (q == 9) num_name += "nine";
else if (q == 10) num_name += "ten";
else if (q == 11) num_name += "eleven";
else if (q == 12) num_name += "twelve";
else if (q == 13) num_name += "thirteen";
else if (q == 14) num_name += "fourteen";
else if (q == 15) num_name += "fifteen";
else if (q == 16) num_name += "sixteen";
else if (q == 17) num_name += "seventeen";
else if (q == 18) num_name += "eighteen";
else if (q == 19) num_name += "nineteen";
if (q >= 20 && q <= 99)
{
int r = 0;
r = q % 10;
(int)(q /= 10);
if (q == 2) num_name += "twenty";
else if (q == 3) num_name += "thirty";
else if (q == 4) num_name += "forty";
else if (q == 5) num_name += "fifty";
else if (q == 6) num_name += "sixty";
else if (q == 7) num_name += "seventy";
else if (q == 8) num_name += "eighty";
else if (q == 9) num_name += "ninety";
if (r == 1) num_name += "one";
else if (r == 2) num_name += "two";
else if (r == 3) num_name += "three";
else if (r == 4) num_name += "four";
else if (r == 5) num_name += "five";
else if (r == 6) num_name += "six";
else if (r == 7) num_name += "seven";
else if (r == 8) num_name += "eight";
else if (r == 9) num_name += "nine";
}
if (k == 3) num_name += "billion";
else if (k == 2) num_name += "million";
else if (k == 1) num_name += "thousand";
}
number = number % (int)pow(10, j);
j--;
}
k--;
}
cout << num_name << endl;
cin.ignore();
cin.get();
return 0;
}
＆＃34; String Construction＆＃34;部分崩溃并打印变量q（商）的无限值。我也尝试在q之前打印一个像flag = 1这样的简单变量，这也是无限的1。我无法弄清楚造成这种情况的原因。任何输入都会对我的学习有很大帮助！
答案 0 :(得分：0)
嗯，你能做的最好的事情之一就是不要弄乱主要功能。你弄乱了主要功能：c
创建一些功能可以让您用更少的代码解决问题。要创建的函数的数量是主观的，imho它可以很好地创建2个函数：toDigit和convert，都返回字符串。
另外我会使用一个函数用于logaritm和电源而不使用库（主要是因为power返回一个double，你必须将它转换为int，数学库是重量级和其他原因）
让我们开始:(所有功能必须高于主要功能，而数字必须高于转换）
你可以通过基本的logaritmic函数得到数字的大小，这会给你一个不好的约束，因为我称之为sillyLog：
int sillyLog(int n, int b){
return !(n/b < 1)? 1 + sillyLog(n/b, b) : 1;
}
然后我们为权力创建我们的功能：
int power(int b, int e){
return (e < 1)? 1 : b * power(b, e-1);
}
现在让我们看看主要的：
int main(){
int number = 0, magnitude = 0, counter = 1;
string num_name = "";
cout << "Enter a number (maximum value is 2,147,483,647)" << endl;
cin >> number;
magnitude = sillyLog(number, 10);
while (magnitude - 3 > 0){
magnitude = magnitude - 3;
counter++;
}
while (number){
int tmp;
tmp = (number - (number % power(1000,--counter)))/ power(1000, counter);//remove all digit after the first three and decrease counter
num_name += convert(tmp, counter + 1);
number = number - tmp * power(1000, counter); // we cut the first 3 digit
}
cout << num_name << endl;
return 0;
}
计数器是计算三元组的变量，
同样2,147,483,647是整数类型的最大长度，如果你想要一个更大的数字，你必须使用long / long long / unsigned / other类型。 注意：负数不是一个问题，因为你只是忽略了它们但是它们会吃掉＃34;记忆的一半用于数字。
如何看到我们生成字符串只需调用函数convert
string convert(int n, int m){
string s = "";
s = ((n-n%100)/100)? toDigit((n-n%100)/100) + " hundred " : "";
s += toDigit(n - (n - n%100));
switch (m){
case 4: return s + " billion ";
case 3: return s + " million ";
case 2: return s + " thousand ";
default: return s;
}
}
我使用了条件运算符（三元组）和switch case而不是if只是为了让你知道它们的存在
现在让我们看看最简单的功能：c
string toDigit(int n){
if (n < 20)
switch (n){
case 1: return "one ";
case 2: return "two ";
case 3: return "three ";
case 4: return "four ";
case 5: return "five ";
case 6: return "six ";
case 7: return "seven ";
case 8: return "eight ";
case 9: return "nine ";
case 10: return "ten ";
case 11: return "eleven ";
case 12: return "twelve ";
case 13: return "thirteen ";
case 14: return "fourteen ";
case 15: return "fifteen ";
case 16: return "sixteen ";
case 17: return "seventeen ";
case 18: return "eighteen ";
case 19: return "nineteen ";
default: return "";
}
switch ((n - n%10) / 10){
case 2: return "twenty " + toDigit(n - (n - n % 10));
case 3: return "thirty " + toDigit(n - (n - n % 10));
case 4: return "forty " + toDigit(n - (n - n % 10));
case 5: return "fifty " + toDigit(n - (n - n % 10));
case 6: return "sixty " + toDigit(n - (n - n % 10));
case 7: return "seventy " + toDigit(n - (n - n % 10));
case 8: return "eighty " + toDigit(n - (n - n % 10));
case 9: return "ninety " + toDigit(n - (n - n % 10));
}
}
这是必须使用开关盒使你的代码可读
如果您有任何疑问，我会很乐意回答您