0 like 0 dislike
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)
Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00

reshown by | 122 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)
Get it on Google Play Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.68.65.240
©2016-2020

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)
Available when: Unlimited - Due to: Unlimited
| 445 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)
Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 122 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)
Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 304 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)
Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 108 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)
Available when: Unlimited - Due to: Unlimited
| 7 views
10,934 questions
124,536 answers
118,820 comments
3,819 users