0 like 0 dislike
2.2k 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 | 2.2k 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:104.23.197.95
©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
| 14.9k 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
| 3.2k 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
| 10.8k 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
| 3.4k 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