使用数组 - C ++减去数字 [英] Subtract numbers using arrays - C++

问题描述

我想计算两个数字之间的差异（让我们说 v 和 n ，因此 vn ）使用数组（不要问为什么我必须这样做）。每个数字的数组以下列方式产生：

• 它们的容量是<$ c $代码中的 v 和 n （= q ）


• vArray [i] = i code>除了前导零填充整个数组


• nArray [i] = - i除了前导零填充整个数组之外， n 的数字

例如，选择 v = 10和 n = 2然后，

  vArray = [1,0] 
 nArray = [0，-2] 
  pre> 
 
 
所以我写了这段代码来计算 sum 数组，  sum = [0,9] 上面的示例）：
 长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]）; 
} 
  
问题是sum数组不等于它应该是什么。对于同一个示例， sum = [1,8] 代码中有什么问题？
 
 
 
正确生成 vArray 和 nArray 。
 
 
 
一些示例和预期结果
  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
我相信我理解数据结构，因为你使用的是一个大整数表示。 
 
 
 
给定数字：1234 
 
 
 
您的V数组是：[1，2，3，4]。 
 
 
 
要添加所有数字（aka sum），我不知道为什么要这样做：
  int digit_sum = 0; 
 for（int i = 0; i <4; i ++）
 {
 digit_sum + = v [i]; 
} 
  
要将表示转换为normal，请尝试：
  int value = 0; 
 for（int i = 0; i <4; ++ i）
 {
 value =（value * 10）+ v [i] 
} 
  
要执行减法，您必须执行步骤，这手工。 
 
 
 
 编辑1：链接到大数字减法 
 
这可能有助于：
 
  C中的大数字减法 
 
  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  

                        
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