0 like 0 dislike
1.6k 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 | 1.6k 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:172.70.80.197
©2016-2025

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
| 11.4k 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
| 2.5k 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
| 8.3k 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
| 2.6k 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