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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

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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;

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    return 0;
}
answered by (-286 points)
edited by
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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
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Hidden content!
#include<stdio.h>
#include<string.h>
int main()
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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: ");
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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
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Hidden content!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


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* * * * * ** * ** * * 0;
}
answered by (-193 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
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