math - 将小数转换为混合基数（基数）

Question

如何将十进制数转换为混合基数表示法？

我猜想给定每个基数的数组和十进制数的输入，它应该输出每列值的数组。

Answer 1

伪代码：

bases = [24, 60, 60]
input = 86462                       #One day, 1 minute, 2 seconds
output = []

for base in reverse(bases)
    output.prepend(input mod base)
    input = input div base          #div is integer division (round down)

Answer 2

数字 -> 设置：

factors = [52,7,24,60,60,1000]
value = 662321
for i in n-1..0
  res[i] = value mod factors[i]
  value = value div factors[i]

反过来：

如果您有 32(52)、5(7)、7(24)、45(60)、15(60)、500(1000) 之类的数字，并且希望将其转换为十进制：

取数字 n，乘以 n-1 的因子，继续 n-1..n=0

values = [32,5,7,45,15,500]
factors = [52,7,24,60,60,1000]

res = 0;
for i in 0..n-1
  res = res * factors[i] + values[i]

你有号码。

Answer 3

在Java你可以做

public static int[] Number2MixedRadix(int[] base, int number) throws Exception {
            //NB if the max number you want @ a position is say 3 then the base@ tha position
            //in your base array should be 4 not 3

            int[] RadixFigures = new int[base.length];
            int[] PositionPowers = new int[base.length];
            PositionPowers[base.length-1] = 1;
            for (int k = base.length-2,pow = 1; k >-1; k--){
                pow*=base[k+1];
                PositionPowers[k]=pow;
            }for (int k = 0; k<base.length; k++){
                RadixFigures[k]=number/PositionPowers[k];
                if(RadixFigures[k]>base[k])throw new Exception("RadixFigure@["+k+"] => ("+RadixFigures[k]+") is > base@["+k+"] => ("+base[k]+") | ( number is Illegal )");
                number=number%PositionPowers[k];
            }return RadixFigures;
        }

例子

//e.g. mixed-radix base for 1day
int[] base = new int[]{1, 24, 60, 60};//max-day,max-hours,max-minutes,max-seconds
int[] MixedRadix = Number2MixedRadix(base, 19263);//19263 seconds
//this would give [0,5,21,3] => as per 0days 5hrs 21mins 3secs

逆转

 public static int MixedRadix2Number(int[] RadixFigures,int[] base) throws Exception {
            if(RadixFigures.length!=base.length)throw new Exception("RadixFigures.length must be = base.length");
            int number=0;
            int[] PositionPowers = new int[base.length];
            PositionPowers[base.length-1] = 1;
            for (int k = base.length-2,pow = 1; k >-1; k--){
                pow*=base[k+1];
                PositionPowers[k]=pow;
            }for (int k = 0; k<base.length; k++){
                number+=(RadixFigures[k]*PositionPowers[k]);
                if(RadixFigures[k]>base[k])throw new Exception("RadixFigure@["+k+"] => ("+RadixFigures[k]+") is > base@["+k+"] => ("+base[k]+") | ( number is Illegal )");
            }return number;
        }

Answer 4

我想出了一个稍微不同的方法，可能不如这里的其他方法好，但我想我还是要分享一下：

    var theNumber = 313732097; 
    
    //             ms   s   m   h    d
    var bases = [1000, 60, 60, 24, 365];
    var placeValues = [];  // initialise an array
    var currPlaceValue = 1;
    
    for (var i = 0, l = bases.length; i < l; ++i) {
        placeValues.push(currPlaceValue);
        currPlaceValue *= bases[i];
    }
    console.log(placeValues);
    // this isn't relevant for this specific problem, but might
    // be useful in related problems.
    var maxNumber = currPlaceValue - 1;
    
    
    var output = new Array(placeValues.length);
    
    for (var v = placeValues.length - 1; v >= 0; --v) {
        output[v] = Math.floor(theNumber / placeValues[v]);
        theNumber %= placeValues[v];
    }
    
    console.log(output);
    // [97, 52, 8, 15, 3] --> 3 days, 15 hours, 8 minutes, 52 seconds, 97 milliseconds

Answer 5

我之前尝试了一些示例，发现了一个他们没有涵盖的边缘情况，如果你最大化你的规模，你需要在最后一步的结果之前

def intToMix(number,radix=[10]):
    mixNum=[]

    radix.reverse()
    for i in range(0,len(radix)):
        mixNum.append(number%radix[i])
        number//=radix[i]
    mixNum.append(number)
    mixNum.reverse()
    radix.reverse()
    return mixNum


num=60*60*24*7

radix=[7,24,60,60]

tmp1=intToMix(num,radix)