0 like 0 dislike
2.3k views
整數的運算會因為整數所佔的記憶體大小而有限制,無法計算太大以及太小的數字,但是如果利用字元以及陣列的結合,就可以大大的提升整數運算的範圍,請設計一個程式,讓使用者輸入兩個很大的整數,程式會把兩個整數相加後輸出。

輸入說明:輸入會包含兩個正整數,兩個正整數之間會用一個空白隔開,兩個正整數相加的結果不會超過五十位數。'

輸出說明:請將兩數相加的結果輸出。

輸入範例:

999999999999999999999999999999 999999999999999999999999999999

輸出範例:

1999999999999999999999999999998
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AD by (18k points)
ID: 18074 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00

reshown by | 2.3k views

5 Answers

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



int main()



{
*** * * * ** ** *** int x,y;
******* *** ** * *
*** **** * * * ** *** * * * * *** ******** ** ** ***
*** * * * **** *
** ** * ** ** ***** * *** * * * *  


* * ** * ****** * * * ******* * *** * *
** ***** * * **** *
**** *** * * ** *** ***** * 0;  

}
answered by (122 points)
0 like 0 dislike
Hidden content!
* ** *
*** * *

int **

int **

int n=0;
* * **** ** * ** * *****

n=a+b;

int x=0;

if ( * (n)){
** * *** ** *** * * * * ** * * *** * **** **
** **


**** * ** * * *** *
*** 0;



}
answered by (-304 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include <stdlib.h>



int main()

{
* **** ** * ** ******* c ;
** ***** * ** ** * i , j , k , m ,a1[50] , a2[50] ;

    
* *** **** *** * * ***** ; c!=32 ; i++)

    {
** **** * * * * *** *** * * * * *** * * * *** *** ** * *** * * *****
** * ** * ***** * * ** * * ** ******* * * * *** ** && c<58)
*** *** * * * * ** ** * * * * **** * * * * *** ** ** *** ** * * ** ****** * ** = c % 48 ;

    }
** ** * ** * *** **** *** * **** ; c!=10 ; j++)

    {
**** * * *** ** ** *** * *** * * * **** *** * * ***
* ** * ** ** * * ****** * ** * ***** * *** * * && c<58)
** ***** *** ** ** ***** ** * * * ** ** **** ** * *** **** *** ** * = c % 48 ;

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

    {
** ** ** * ** *** * **** * **** * * *** ** *** ** *
* * *** *** ***** * **** * ****** * ** * *** ***** * * * *
* * * *** * * * * * * ** ** ** * * ***
** ** *** ** **
** * ** * * *

    {
* * * ** * * * * ** * * *** *** * * ** *
** ** * * * * * * * *** * ** * ** ** * * * ** * ******** ** *** *
***** * ** ** * **** * * * *** ****

    }
** * ** * ********** * a3[i+1] ;
* *** ** ***** * * * * * ** *****
* ** ** ** * * * * *** * * ** * * * **** *

    
**** * ** ** ** * **** ; k<=(i+1) ; k++)
*** * ** * * * ** * *
* ****** ** ** *** * * * * * * **** * * = (a1[k]+a2[k])%10;
* ***** * **** ***** * ** ** **** * ***
** ** ** * * * ***** *** * ** * * *
** ** * ** ** ** ** * **** ** * * * * * * ** * ***** ** ****** * ** ** *** ** * * *** * *
**** * ** **** * ** * ** ** * * ********* ******** ** * * ** ***** ** *** ** * * **** * * ** ** ** *
* ** * * **** *** ** * * * *** * ** *
** *** * ******* * * * ***

    
** ***** ** * ** ** **** * * ** *
* * *** * ** *** *** ***** * ***** ** **** ** * **** ** ** ** **
* * * * * * ** **** * ***** * * * * ********* ** *
* * ******* * * **** * ** * * ** *
* ***** *** * * * 0;

}
answered by (-162 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>
** ** * ** *** *

#include<ctype.h>



int main()



{
** *** ** ** ** *** ***** i,j;
*** *** ** ** *** *
* ** * ** * * ** * * ** ** * * ** **** * *** ** *
** * * * * * * **
* ******* * ******* * * *** ** *


* ****** * ** * ** ** ** ** ** * * *** *
* * * ** * * * * * *
** * **** **** **** 0;  

}
answered by (122 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include <stdlib.h>

#include <string.h>



int main(int argc, char *argv[]){
* ** * ** ** **** * * ** * * * *** * ** **
* ** ** * * *** * ** i,j,len,na,nb,carry;
** * *** * *** **** ****
* * * *** * ** *** *** *** * %s",a,b);
* * ** * ** *** ***** * = strlen(a);
* * * * *** *** * * ** = strlen(b);
*** * * * *** * * ***
**** * *** ** ** * **** * ** ** >= 0; i++,na--)
*** *** ****** ** * ****** * **** * * * = a[na]-'0';
*** *** ******* * *
* ******** * * ** ** * >= 0; i++,nb--){
* * ** **** *** * * * * ***** * * *****
* *** * *** * * ** * * **** * *** * *** = c[i] + b[nb]-'0';
* * * * ** ** * * *** ****** *** * * ** ** %d\n",j);
* * *** *** ** * ** *** ****** * ** >= 10){
*** * **** *** ** * ** * * ** * ** *** * ** * = j/10;
* ** * * * * ** ******** * ** * * * * ** * ** *** = j-10;
** * * *** ** * ** * ** * ** *** * *** *
***** *** * ***** ** * ******** * ** * * * * = j;
*** **** ***** ** ** ** * ** *** ** ** * ** += carry;
** *** * ** ** * * ** * * **** ** ** *** %d %d\n",i,carry);
* * ** *** ** * * *
* * ** * * * ** * * ***** ** %d\n",carry);
* ***** * ** * * *
**** * ** * * * == 0)
* * * *** **** * * * ** ** ** *** * ** * * * * * * * >= 0; i--)
*** ** * * *** ** *** ***** * * ** * ***** * ** *** ** * ** * * *****
* ** * *** ******
** * * ** ** ** **** * *** * >= 0; i--)
* ** ** * * *** *** ** * * * * ** * ****** * *** **** * ******** * * * *

  
* ***** * *** * *** * * * ** *  
* * ** * 0;

}
answered by (-216 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:104.23.197.64
©2016-2026

Related questions

0 like 0 dislike
96 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18072 - Available when: Unlimited - Due to: Unlimited
| 15.5k views
0 like 0 dislike
13 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18069 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 3.3k views
0 like 0 dislike
67 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18068 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 11.3k views
0 like 0 dislike
11 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18065 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 3.6k views
0 like 0 dislike
0 answers
[Resource] asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18084 - Available when: Unlimited - Due to: Unlimited
| 7 views
12,783 questions
183,442 answers
172,219 comments
4,824 users