如何添加多位数？ [英] How to add multi-digit number?

问题描述

如何添加：

12334568757678978709709880988987687576546346543432432531253153213515 + 987328746329887632987216389721634879231648971236498723146238644348364893274

  void  add（ char  * a， char  * b， char  * c）
 {
 
  / /  在此处编写代码 
} 
 
  int  _tmain（ int  argc，_TCHAR * argv []）
 {
  char  a [ 1024 ]; 
  char  b [ 1024 ]; 
  char  c [ 2048 ]; 
 printf（ 输入第一个值）; 
 scanf_s（ ％s，a）; 
 scanf_s（ ％s，b）; 
 
 
 add（a，b，c）; 
 
 printf（  c is％s，c）; 
  return   0 ; 
} 

解决方案

首先要做的事情：c不需要是a和b的两倍，它只是需要是一个更大的字符，因为两个整数之和总是小于或等于最大值的两倍：

  9  +  9  =  18  
  99  +  99  =  198  
  999  +  999  =  1998  

其次，这并不困难，只是看起来就是这样！

首先想一想如何手动完成：

1）你把它们写下来，所以每个的最低位数都在同一个柱子中

 12334568757678978709709880988987687576546346543432432531253153213515 
 + 98732874 6329887632987216389721634879231648971236498723146238644348364893274 

2）然后，您将添加最低有效数字，并使用结果模10作为结果数字，任何多余数字都是进位：

 12334568757678978709709880988987687576546346543432432531253153213515 
 + 987328746329887632987216389721634879231648971236498723146238644348364893274 
 ___________________________________________________________________________ 
 ... 9 
 ___________________________________________________________________________ 
 Carry：0 

3）然后你会为下一个数字对重复这个，包括每次进位，直到你用完数字。



所以试试：从字符串的右端并添加数字对。

这并不复杂 - 但它是你的功课，所以我给你无码！

how to add:
12334568757678978709709880988987687576546346543432432531253153213515 + 987328746329887632987216389721634879231648971236498723146238644348364893274

void add(char *a, char *b, char *c)
{

    // Write code here
}
 
int _tmain(int argc, _TCHAR* argv[])
{
    char a[1024];
    char b[1024];
    char c[2048];
    printf("enter First values");
    scanf_s("%s",a);
    scanf_s("%s",b);
   
 
    add(a,b,c);
 
    printf("c is %s",c);
    return 0;
}

解决方案

First things first: c doesn't need to be twice the size of a and b, it only needs to be one character larger as the sum of two integers is always less than or equal to double the largest value:

  9 +   9 =   18
 99 +  99 =  198
999 + 999 = 1998

Second, this isn't difficult, it just seems that way!
Start by thinking how you would do it manually:
1) Your write them down so the least significant digit of each was in the same columne

         12334568757678978709709880988987687576546346543432432531253153213515
+ 987328746329887632987216389721634879231648971236498723146238644348364893274

2) Then, you would add the least significant digits, and use the result modulo 10 as the result digit, with any excess being the carry:

         12334568757678978709709880988987687576546346543432432531253153213515
+ 987328746329887632987216389721634879231648971236498723146238644348364893274
  ___________________________________________________________________________
 ...                                                                        9
  ___________________________________________________________________________
Carry:                                                                     0 

3) You would then repeat this for the next digit pair, including the carry each time, until you ran out of digits.

So try that: start from the right end of the strings and add digit pairs.
It's not complex - but it's your homework, so I'll give you no code!

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