条形图轴的正确通用计算[重复]

我已经在这里添加了225的输出将是该算法的230,而不是OP要求的300。理由:我的实现就是这样。这个问题真的没有正确的答案。这不仅仅是一个设计问题。

想法:使用int到int的映射,如果key是数字,那么value是应该添加到其中以获得适当数字的值。从右端提取输入的数字,并继续改进从右端向左移动的数字,每次一个数字。

添加剂 键,值 }配对如下:

请注意,您不需要将每个整数都添加到其中。如果正确使用映射,只添加一些整数就足够了。

以下是代码/伪代码,您可以从输入数字中获得适当的值:

int GetProperNumber(int input) {
    //Init a variable to keep track of the sign
    int multiplier = 1;

    //For negative numbers, taking some precaution...
    if (input < 0) {
        input *= -1;
        multiplier *= -1;
    }

    //For cases where the input number is only 1 or 2 digits, some quick checks...
    int numberOfDigits = GetNumberOfDigitsInInput(input);
    if (numberOfDigits == 1) return (input + additive[input]) * multiplier;
    if (numberOfDigits == 2) return (input + additive[input - input/10]) * multiplier;

    //Now, coming to the core of the method
    int divisor = 10; //We'll use it to get the left part from the input
    while (true) {

        //First, get right part of the number
        int inputWithDigitsRemoved = input / divisor;

        //If the leftover part is too small, i.e. we have reached the last digit,
        //then break as we have now rounded the number off pretty well.
        if (inputWithDigitsRemoved <= 9) break;

        int inputWithDigitsMasked = inputWithDigitsRemoved * divisor;
        int right = input - inputWithDigitsMasked;

        //Since the number is still not rounded to the right magnitude, 
        //the result should be further improved.
        if (additive.Contains(right)) input += additive[right];
        divisor *= 10;
    }
    return input * multiplier;
}

一些样本结果是:

输入=99,输出=100

输入=2541,输出=2600