0 like 0 dislike
1.9k 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 | 1.9k 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:104.23.197.65
©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
| 13.3k 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
| 2.8k 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
| 9.6k 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
| 3k 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