0 like 0 dislike
3k views

You are requested to help in writing a program to perform addition of two large number up to 50-digits. In C there is no builtin datatype for large number, but we can use characters and array to solve the problem.

Input: The input will contain two positive integers are separated by a blank, the two positive integers do not exceed 50-digits.

Output: Sum.

寫一個程式輸入兩個大數(最大50個位數),計算其總合。C語言不提供大數資料型態,但我們可以用字元和陣列來解決這個問題。

Example input:

999999999999999999999999999999 999999999999999999999999999999

Example output:

1999999999999999999999999999998

 

[Exam] asked in Final Exam
ID: 43460 - Available when: 2018-01-20 09:00 - Due to: Unlimited
| 3k views

4 Answers

0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char input[500],s1[500],s2[500],temp;

int i,space,sum[500],carry,temp2,flag=0;

int main(){
** *** **** ** *** ******* ** * * %s",&s1,&s2);
* ******* * * * ** ** * * ** * *** *

///get space
*** ** *** * * * * * ** * ** ** * *
** ** * ** *** *** ** *** * ** * ** * ** *** * '){space=i;}
    }

///get s1
** * * ** ** * * ****** *** * ** ';i++){
******* * ** * * *** ** * * * **** *
    }

* * ** * * * **** *** *
///reverse input
**** *** * *** ** * ** * * * *** * *******
* * ** * * ** * ** * * * * ** ** ** **** *** * * *
** * ** ******** * * ** ******* * ** **** * **** **
* * * ** ** ** ** * * * * * *** * * * ** ** * ** * **** **
    }

///get reversed s2
* * * ** * * * * * ';i++){
* * *** ****** ** ** ** **** * *** * * ** *** ** * *
    }

///reverse s1

*** * ** * * ** ** * * ** * **** **
*** ** * * * ** * *** *** ** *** ** *** ** * * * **

*** * * * * **** ** * * * *** ** ** * ** ***** * ** * * * *
* * ****** * * ** * * ** * ***** * ** * ***
    }

///sum s1 and s2
******* * * *** *** * *
** ** * ***** * *** ****** ** * ** *** *
** *** * * * *** * ** * *** ** * ** * ***** * * *** * *
* *** * * ******* ** ******** * * ** *** ** * *** ** * * * **** ***
** ** * * * * **** * ****** ***** * *** *** *
*** ** * ** ** **** * * * *** * * ****
** * * *** *** ** * * * * * * * * * ** * * **

    }

* * *** * * **** * *** ** *** * * ** * rev:%s\n",input);
*** ** * * * * * * ** * ** ** * * * rev:%s\n",s1);
** * ** * * * * *** * * ** * rev:%s\n",s2);
* * ** ** * * ** *** *** * * * ** * ** ** ****** * * * ** * *** * *

///reverse output
*** ** ** * *** ** ** **
* * * * ** * ** *** ** * * * *** ******

**** * *** * * ** *** * **** *** ** * * *
**** * ** * *** ** ** ****** * ** * * ***** * *
    }

///print
** ** *** * * ** *** **** *** ******
* ** * * * ** * * ***** * *** * * ** *** * ***** * ** * * **
*** * ** ** ** * ** * ** * *** * * * * * ** * *** * * *** * * * ** ** ** * ** * * * *
* **** * *** ** * * ** **** * ** * * * * * **** *** * *** ** * *** ******* * * ** ****** * ** * ** ***
**** * * * * * *** * * *** * * **** * ** ***** ** ** **** ** * ** ** * *** ****
* * ** * * * * * *** ** * * *** * **** *** ** ** * *****

* ** * *** * *** *** * * * ***** ** * *

    }
** * **** ** ** *** * ** * ** * * * ** ****
    return 0;
}
answered by (-286 points)
edited by
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
0 0
prog.c: In function 'main':
prog.c:80:12: warning: multi-character character constant [-Wmultichar]
     printf('10');
            ^~~~
prog.c:80:12: warning: passing argument 1 of 'printf' makes pointer from integer without a cast [-Wint-conversion]
In file included from prog.c:1:0:
/usr/include/stdio.h:364:12: note: expected 'const char * restrict' but argument is of type 'int'
 extern int printf (const char *__restrict __format, ...);
            ^~~~~~
prog.c:80:5: warning: format not a string literal and no format arguments [-Wformat-security]
     printf('10');
     ^~~~~~
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
0 like 0 dislike
Hidden content!
#include<stdio.h>
#include<string.h>
int main()
{
*** ** ** *** * * * * string1[50];
* ***** * * ** * string2[50];

* * * * ** * * ** ** * ** ** * * ** * string1);
** ** * * * **** ** ******* * ** ** string2);
* **** * * * **** arr[51]={0},i;

* * ** * ** ****** * ** i<strlen(string1); i++)
* * * ****** * ***
* * * *** * *** ** ** * ***** * * * * += string1[strlen(string1) - 1 - i] - 48;
    }

*** ** * **** * * * ***** i<strlen(string2); i++)
* * ** * * *** *
**** ** ***** * * ****** * * ** * ***** += string2[strlen(string2) - 1 - i] - 48;
**** ** * *** ** * *

*** ** ** * ** * **** i<50; i++)
** ** ** * * ** ****
** ** * *** * ** *** ** * * * ** *** *** > 9)
** *** * * * * ** **** * * ** ** * * **
** ****** *** *** * ***** * ** *** ** ** ** * *** ******* * -= 10;
* *** ** ** * * * ** ***** * * * * ** * ** *** * * * *
* ** * ***** ** * ** * * ** * *** * *
** **** ********* * * *

* * * * * * * * *** ** * * i>=0; i--)
* ******* * * * * *** * *
* *** *** * * *** **** **** * * * ****** * * * * ** ** * != 0)
** * ***** * ***** *** ****** ** * *** * *** *** * *** ** *
* * ** ******* * *

***** *** ** * ** **** * *** i--)
* **** ** ***** ** * ** ** *** ****** * ** ** *** ** **** ** arr[i]);
}
answered by (-168 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
0 like 0 dislike
Hidden content!
#include <stdio.h>
int main (){
int s1,s2,l1,l2;
int num1[50],num2[50],sum[50];
printf("enter input 1: ");
***** ** ** * ** * *
printf("enter input 2: ");
**** ** * * * ***
for(l1=s1[l1]!='\0')
**** * ** * * **
for(l2=s2[l2]!='\0')
* * * * * ** * * *** **** * **
int carry=0;
int k=0;
int i=l1-1;
int j=l2-1;
* * * * * * * ***
{
** * ** * * * ** * * **** * * **
* * * * ** * ** *** *
*** * ** * ** *** * * * *** * * * * **** * *** *
* * * *** **** ** *** ** ** * * ***** **** * * ***** * * * * * * * ** ** ***
** *** * * * ** * ** * ** * *** ** ** *** * * * **** ****** ** * * **** *** *
* * * * *** ** * if (j>=0)
* * *** ** * * ** * ** * *
*** ** ** * * * * * * * * *** ****** * * ** * * * * *** **** * *** * **** *** * ** *
* * * * * *** * ********* * *** * ** * * *** * * ** ** * * ** * * *** ****
* **** ******* ** * ** * * * * **** * * **** if(carr>0)
* *** ** * * * *** ***** *** ** * * *
* * ** * * ***** **** *** ***** **** * ** **** * * ** ** * ");
* * * * *** *** * *** * ***** * **** ** * * *
}
for(k--;k>=0;k--)
** ****** ** * * ********
* ** *** * ** ** * *** *** * ***
** * * * * ******
  return 0;
}
answered by (16 points)
0 0
prog.c: In function 'main':
prog.c:9:10: error: subscripted value is neither array nor pointer nor vector
 for(l1=s1[l1]!='\0')
          ^
prog.c:9:20: error: expected ';' before ')' token
 for(l1=s1[l1]!='\0')
                    ^
prog.c:9:20: error: expected expression before ')' token
prog.c:10:10: error: subscripted value is neither array nor pointer nor vector
   num1=s1[l1]-'0';
          ^
prog.c:11:10: error: subscripted value is neither array nor pointer nor vector
 for(l2=s2[l2]!='\0')
          ^
prog.c:11:20: error: expected ';' before ')' token
 for(l2=s2[l2]!='\0')
                    ^
prog.c:11:20: error: expected expression before ')' token
prog.c:12:12: error: subscripted value is neither array nor pointer nor vector
     num2=s2[l2]-'0';
            ^
prog.c:17:15: error: expected ')' before ';' token
 for(;i>=0;j>=0;i--;j--;k++)
               ^
prog.c:23:13: error: expected ';' before 'carry'
             carry=(num1[i--]+num2[j]+carry)/10;
             ^~~~~
prog.c:27:13: error: expected ';' before 'carry'
             carry=(num1[j--]+num2[j]+carry)/10;
             ^~~~~
prog.c:28:18: error: 'carr' undeclared (first use in this function)
         }else if(carr>0)
                  ^~~~
prog.c:28:18: note: each undeclared identifier is reported only once for each function it appears in
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


int main()
{
***** * *** **** **** * a[50],b[50];
**** * ** * ** * ** ** * c[51]={0};
* ** *** *** ** ** * * i;
* *** ** * ** ***** ** ** x=0;

**** ** **** ** * *** * * %s",a,b);

* ** * ** ** * *** ** * ****** ** ** *
    {
* * * * * * ** * * * *** * * * * ** * **** * * * ***** *
    }

**** * ** **** * *** * ** * * ** * **** ***
** * * ** *** ** ** **
* * * * * ** * * * * ** * * * ***** ** ** * ** ** *
* *** ** * **** ** *


** ** * * **** ** ** *
** * ** ** * **** *** *
* ** *** *** * * * * * * * ** ** **** * *** * * * * *
* * * ** * ** * * * ** * * ** *** * * *
**** ** * ** * ** * ** * **** **** * *** * *** * * ** * ****
* ** ** ** * * ** ** * *** *** ****** *** ** *** ** * * ** ** *****
********* *** **** *** * * ** ****** * **** *
    }

** ** * * * *** *** * * ** *
* * * * * * ** ** * * *
* * **** ** ***** ** * ** * != 0)
***** * ** *** * **** ** ** * *** * * ** ** ** * ***
*** * * * ** * * * *** ** ** ** * * ** * ******** * *
* * * * **** ** ** ** *** **** **** *** ** *** ** * ******* *
**** **** ***** * * **** * *** *** * * * * *
    }

*** * * ***** * *** ** ** **
    {
** * * *** * * * ** * * * ** * **** * * * *** * * **** *
    }


*** ** * * * ***** ** ** 0;
}
answered by (-193 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:104.23.243.164
©2016-2026

Related questions

0 like 0 dislike
13 answers
[Exam] asked Jan 19, 2018 in Final Exam
ID: 43463 - Available when: 2018-01-20 09:00 - Due to: Unlimited
| 4.7k views
0 like 0 dislike
6 answers
[Exam] asked Jan 19, 2018 in Final Exam
ID: 43462 - Available when: 2018-01-20 09:00 - Due to: Unlimited
| 2.6k views
0 like 0 dislike
8 answers
[Exam] asked Jan 19, 2018 in Final Exam
ID: 43461 - Available when: 2018-01-20 09:00 - Due to: Unlimited
| 3k views
0 like 0 dislike
12 answers
[Exam] asked Jan 19, 2018 in Final Exam
ID: 43459 - Available when: 2018-01-20 09:00 - Due to: Unlimited
| 4.9k views
1 like 0 dislike
37 answers
[Exam] asked Jan 16, 2018 in Final Exam
ID: 42299 - Available when: 2018-01-17 14:00 - Due to: Unlimited
| 15.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users