在c ++中将单词转换为数字

Posted 2021-04-15

我有问题将单词转换为数字

五千六百三十二 - 5632

如果有人这样做我甚至不知道怎么开始请指导我怎么做...

答案

在这里，我在python中做到了，它将从算法的角度帮助你或其他人。

#!/usr/bin/python

__author__ = 'tomcat'

all = {
        "one" : 1,
        "two" : 2,
        "three" : 3,
        "four" : 4,
        "five" : 5,
        "six" : 6,
        "seven" : 7,
        "eight" : 8,
        "nine" : 9,
        "ten" : 10,
        "eleven": 11,
        "twelve": 12,
        "thirteen": 13,
        "fourteen": 14,
        "fifteen": 15,
        "sixteen": 16,
        "seventeen": 17,
        "eighteen": 18,
        "nineteen": 19,
        "twenty" : 20,
        "thirty" : 30,
        "forty" : 40,
        "fifty" : 50,
        "sixty" : 60,
        "seventy" : 70,
        "eighty" : 80,
        "ninety" : 90,
        "hundred" : 100,
        "thousand" : 1000,
        "million" : 1000000,
        "billion" : 1000000000,
        "trillion" : 1000000000000,
        "quadrillion" : 1000000000000000,
        "quintillion" : 1000000000000000000,
        "sextillion" : 1000000000000000000000,
        "septillion" : 1000000000000000000000000,
        "octillion" : 1000000000000000000000000000,
        "nonillion" : 1000000000000000000000000000000
        };


spliter = {
        "thousand" : 1000,
        "million" : 1000000,
        "billion" : 1000000000,
        "trillion" : 1000000000000,
        "quadrillion" : 1000000000000000,
        "quintillion" : 1000000000000000000,
        "sextillion" : 1000000000000000000000,
        "septillion" : 1000000000000000000000000,
        "octillion" : 1000000000000000000000000000,
        "nonillion" : 1000000000000000000000000000000
        };

inputnumber = raw_input("Please enter string number : ");

tokens = inputnumber.split(" ");

result = 0;
partial_result = 0;
for index in range(len(tokens)):
    if tokens[index] in spliter :
        if partial_result == 0:
            partial_result = 1;
        partial_result *= all[tokens[index]];
        result += partial_result;
        partial_result = 0;
    else:
        if tokens[index] == "hundred" :
            if partial_result == 0:
                partial_result = 1;
            partial_result *= all[tokens[index]];

        else:
            partial_result += all[tokens[index]];


result += partial_result;

print result;

另一答案

一般来说，如果你不需要编写代码，你会怎么做？在这个例子中，你的单词集合是：

Five Thousand Six Hundred Thirty two

我们可以将每个转换为数字以获得以下集合：

5 1000 6 100 30 2

从5开始（提示：5 <1000在1000的左边。这表明......！??）在获得数字5632时你会采取什么步骤？

如果这个号码怎么办？

六三十三亿五十四百万二万二千三四？

你能想出某种规则（或更好的算法）吗？

一旦你将大问题分解成一系列小问题，那么接下来的战斗就是找到正确解决每个小问题的正确编码方法

另一答案

希望this给你一些开始： -

#include <iostream>
#include <string>
#include <map>
using namespace std;

map<string, int> reference;

string ones[] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"};

void storeOnes(){
    for(int i = 0; i < 11; i++){
        reference[ones[i]] = i;
    }
}

int main(){
    //set up
    storeOnes();
    string test = "onetwothreetwofour";
    string buffer;


    for(int i = 0; i < test.length(); i++){
        buffer.push_back(test.at(i));
        map<string, int>::iterator it = reference.find(buffer);
        if(it != reference.end()){
            cout << (*it).second;
            buffer = "";
        }
    }
    cout << endl << endl;
    system("pause");
    return 0;
}

另一答案

考虑使用map。说five thousand six hundred three ten two。

#include<iostream>
#include<map>
using namespace std;

int main()
{
    map<string,int> digits;
    digits["one"] = 1;
    digits["two"] = 2;
    digits["three"] = 3;
    digits["four"] = 4;
    digits["five"] = 5;
    digits["six"] = 6;
    digits["seven"] = 7;
    digits["eight"] = 8;
    digits["nine"] = 9;
    digits["ten"] = 10; 
    digits["hundred"] = 10; 
    digits["thousand"] = 1000;

    const int num_len = 7;
    string num_str[num_len]={"five", "thousand", "six", "hundred", "three", "ten", "two"}; 

    int number = digits[num_str[0]]*digits[num_str[1]] + 
                 digits[num_str[2]]*digits[num_str[3]] +
                 digits[num_str[4]]*digits[num_str[5]] +
                 digits[num_str[6]];

    cout << number;    
}

另一答案

另一种方法是通过递归来完成（在java和c ++中）

https://github.com/jman27182818/words_to_numbers

大多数代码都是关于解析字符串以获得可以递归操作的字符串向量。基本上算法是

A[] = String array //example {"three","hundred"}
Integer converter(A){
    if(length(A) <= 4)
        handle_base_case; 
        //either A only has small values or is of the form 
        //{"three","billion"}  

    //if length is greater than 4 the array must have a large value
    index = find_first_value_greater_than_100(A);
    arrayl = A[1:index-1];
    arrayr = A[index+1:A.end()];
    return (convert_hundreds(arrayl) * Value_of(A[index])+ converter(arrayr) );
    }

其中“convert_hundreds”接受一个数组的数组，每个字符串的值不大于100（或西班牙语为1,000）并返回数值。这个算法比前一个算法更加耗费内存，但我喜欢它，因为它似乎更好地推广到其他语言。