减去的数字使用数组 - C ++ [英] Subtract numbers using arrays - C++

问题描述

我要计算两个数字（之间的差异让我们说 v 和 N ，所以 VN <​​/ code>）使用数组（不要问我为什么这样做）。每个号码的阵列以如下方式制造：



• 他们的能力之间的的最大的位数v 和 N （= <$ C $数ç>问：在code）


• VARRAY [I] = I 第v 的数字除了前导零填补了整个阵列


• nArray [I] = - I 日的数字 N 除了前导零填补了整个阵列

例如，选择 v = 10和 N = 2，那么，

  VARRAY = [1,0]
nArray = [0，-2]
 

所以我写了这个code来计算之阵列将等于差的数字（和= [0 9] 上面的例子）：

 长R = 0;
对（INT I = Q-1;我-1个; I  - ）{
    总和[I] = VARRAY [I] + nArray [I]
    如果（总和[1]  - ; 0）{
        R =地板（总和[I] / 10）;
        总和[I-1]  -  = R;
        总和[I] =总和[I] +10;
    }其他{
        R = 0;
    }    的NSLog（@％礼，总和[I]）;
}
 

的问题是，总和阵列不是等于它应该是什么。对于同样的例子，和= [1,8] 什么是在code中的问题？

请注意： VARRAY 和 nArray 是否正确生成

编辑：有几个例子和预期的结果。

  V = | N = | VARRAY = | nArray = |总和=
    25 | 9 | [2,5] | [0,9] | [1,6]
    105 | 10 | [1,0,5] | [0,1,0] | [0,9,5]
   1956年| 132 | [1,9,5,6] | [0,1,3,2] | [1,8,2,4]
  369375 | 6593 | [3,6,9,3,7,5] | [0,0,6,5,9,3] | [3,6,2,7,8,2]
 

解决方案

我相信我理解数据结构，如您正在使用的大整数的再presentation。

给出的数字：1234

您V阵列是：[1，2，3，4]

要添加的所有数字（又名总和），我不明白为什么要做到这一点，是：

  INT digit_sum = 0;
的for（int i = 0;我4;;我++）
{
    digit_sum + = V [I]
}
 

要重新presentation转换为正常，试试这个：

  int值= 0;
对（INT I = 0; I＆4; ++ⅰ）
{
  值=（值* 10）+ V [I]
}
 

要执行一个减法，你将不得不执行的步骤，如果你的手这样做。此外，您还需要第二个号码了。

修改1：链接大数量减去结果
这可能帮助：搜索
大数量减去用C 结果
C ++大数算术

I want to calculate the difference between two numbers (let's say v and n, so v-n) using arrays (don't ask why I have to do so). The arrays for each number are made in the following way:

• Their capacity is the number of digits of the greatest number between v and n (=q in the code)
• vArray[i] = ith digit of v except leading zeros to fill the whole array
• nArray[i] = - ith digit of n except leading zeros to fill the whole array

For example, choose v = 10 and n = 2 then,

vArray = [1,0]
nArray = [0,-2]

So I wrote this code to calculate the sum array that will be equal to the digits of the difference (sum = [0,9] for the example above):

long r = 0;
for (int i = q-1 ; i > -1; i--){
    sum[i] = vArray[i] + nArray[i];
    if (sum[i] < 0){
        r = floor(sum[i]/10);
        sum[i-1] -= r;
        sum[i] = sum[i]+10;
    }else{
        r = 0;
    }

    NSLog(@"%li",sum[i]);
}

The problem is that sum array isn't equal to what it should be. For the same example, sum = [1,8] What is the problem in the code?

note : vArray and nArray are properly generated.

EDIT : A few examples and expected results

    v =  |    n =   |  vArray =   |     nArray=    |    sum=
    25   |    9     |    [2,5]    |      [0,9]     |    [1,6]
    105  |    10    |   [1,0,5]   |     [0,1,0]    |   [0,9,5]
   1956  |   132    |  [1,9,5,6]  |    [0,1,3,2]   |  [1,8,2,4]
  369375 |   6593   |[3,6,9,3,7,5]|  [0,0,6,5,9,3] |[3,6,2,7,8,2]

解决方案

I believe I understand the data structure, as you are using a Big Integer representation.

Given the number: 1234

Your V array is: [1, 2, 3, 4].

To add all the digits (a.k.a. sum), which I don't see why you want to do this, is:

int digit_sum = 0;
for (int i = 0;  i < 4; i++)
{
    digit_sum += v[i];
} 

To convert the representation into "normal", try this:

int value = 0;
for (int i = 0; i < 4; ++i)
{
  value = (value * 10) + v[i];
}

To perform a subtraction, you will have to perform the steps as if you doing this by hand. Also, you would need a second number too.

Edit 1: link to big number subtraction
This might help:
Big Number Subtraction in C
C++ Large Number Arithmetic

这篇关于减去的数字使用数组 - C ++的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！