0 like 0 dislike
3.3k 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)AC by (18k points)
ID: 18075 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00

reshown by | 3.3k views

16 Answers

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

#include <stdlib.h>



int main()

{

    char a[50]={0};

    char b[50]={0};

    char c[50]={0};

    int n,t=0;
* * *** ********* *** * *** *** ** ** **** ***
* ** * * *** * *** *** * *********

    {
* * * * * **** * * ** *** * * ** * ** *** *
** * * *** ** ****** * *** *** *** * **** *
* * ** **** * ****** * ** ** ** * * * * * **** ** * ** * ** * * *** ***** **
** * *** * * * * *** ** * ***** * * * ***** * *** * *** * ** ** *** ** * * ** = a[n] + b[n];
* *** * * * * *** ** * *** * ****** * * * ** * * * ** * **** **** if(a[n]+b[n]>='10')
* ** * * * * **** *** ** ** * * * **** * ***** * * ** * ** * ** * ** *
***** *** * * ** ** * **** * **** * ** *** * * * * ** ** ** *** ** * ** *** ** * *** = a[n] + b[n] - '10';
*** * * * * ** * * **** * * ** * * * * ** * *** * ***** * ** *** * ***** **** ** * * * ** *
*** ** * *** ********* * ** ** * ** *** * * * ** * * *** *****
* * ** * ** * * *** * ** **** * * * * *
** * ** ***** * * * * * *** **** * * * if(t == 1)
*** *** ***** * * * ******* *** ** * ***
*** * * * **** ** * * * * * * * * ******** ** * * * * ** *** ** * ** ** ********
****** **** ** ******** * * * * *** * * * * * * * * * *** * ** **** * * ** = a[n] + b[n] + '1';
** * * *** ** * ***** **** * ***** *** * * *** *** ****** * * * if(a[n]+b[n]>='9')
** **** * * ** *** * * * * * ** * ** *** ***** * ****** * *
***** * * **** * * *** * *** *** * *** * * * ** * * * ***** ** * * **** * * *** = a[n] + b[n] - '9';
** ** *** * * ** ** *** **** *** * * * * **** ****** * * ** * * * ***** * * * *
* *** * *** * * **** *** *** * **** * * * * * *
***** ** **** * ** ** ** * * * * * * * * *

    }
**** ** * * * * ****** * * * *

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

    }

    return 0;

}
answered by (-85 points)
0 like 0 dislike
Hidden content!
*** * * * ** * **
* * * *** ****** *

int main ()

{

int


** **** * ** * * ** ***
* * ** ** * * * *


* * 0;

}
answered by (-85 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include ***** * *****

#include <ctype.h>

#include <math.h>

#include <stdlib.h>



int main(){
**** * * * *** **** a[100],b[100];
** ** **** **** ** ** sum[200];
* *** * *** * ** * ** la,lb,lm,i,j,k;
*** ** ** * * * * * ** * * %s",a,b);
***** ** * ** ** *** * = strlen(a);
* ** * ** ** *** ** = strlen(b);
*** *** **** * **** * ****** ** * *
*** ****** * ** * ** * * *** ** * *** * **** * * **** = a[i];
* ** * *** * * * * * * ** *** * * * ** ** *** * **
* ** **** ***** ** **
* * * *** * * *** ** *** * **** *** ***
** * **** ** * ******** * * * * * * += b[i]-48;
* ** **** * ** *** ** * * ** *** * * **** ** * ** ** *
** ** * * *** ** *
** ** **** * ** * ** * * = strlen(sum);
*** * * * * * * * *** * *
* ** ** **** ** ** * ** ** * * *** * ** *** *** * ** * * * * *** *
* * ** ** * * ** ** *
**** ** * * **** * * ** * * 0;

}
answered by (-216 points)
0 like 0 dislike
Hidden content!
* * * **** **
* * *** **** * *





int main ()

{
* * a , b ;
* * ** ** * ** * * , **
**** *** ** * ** , ****
* * *** * ******* * * ,a+b);
** * * ** * * * ** * * * * * 0 ;



}
answered by (-87 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include <stdlib.h>



int main()

{

    char a[50]={0};

    char b[50]={0};

    char c[50]={0};

    int n,t=0;
*** ** **** * * * ** ** * * * ****** * ** ****
**** * ***** ** ** * * * * * **** *** **

    {
** ** * * **** * ** ** * *** * **** *
* ** *** **** ** **** *** ** * * **** * * *** **
** * * * * **** * ** ** ** ***** *** * ***** * * ** ** * * ** * *** *
** * *** * * * ** * *** * *** * * * ************* * *** * *** * *** **** *** ** * ** * = a[n] + b[n];
** ** ** * * * **** * ** ** * * ** *** ** *** *** **** * *** * if(a[n]+b[n]>=10)
* * ** *** ***** *** ** * * **** ** *** *** * **** ******** ** *******
*** *** ** * ** ** **** **** ** * ** ** *** *** ** **** *** ** ** * * *** ** * ** = a[n] + b[n] - 10;
** *** *** ** **** * ** * * * * *** **** *** *** *** * ** * ** * *** ** **
* ** ***** * * * * **** * * ***** * * * ** * ** * * *
****** **** * ** ** *** * *** * * * * ***
* ** * ** ** * * * * * ** ** *** * *** * * * if(t == 1)
* * ** ** * * *** ** **** ** * * * * * *
***** * * * * *** ***** **** ** * *** **** ** * ** * ** ** ****** ****
* *** **** * * * **** *** * *** *** ** * *** * *** * * * * * *** *** * ** ***** *** * = a[n] + b[n] + 1;
** * * ** ** *** * ** ** * ****** * * **** * * * * * **** * if(a[n]+b[n]>=9)
* ** * ** * * * * *** ***** * ****** * * ** * * ** * * *
** ***** * *** * * ** * **** * ** ** ** ** * *** **** * *** *** * * * ** *** ** ***** ****** = a[n] + b[n] - 9;
* ****** * * *** ** ** * * * *** * *** * ** * * * **** ** * * * ** * * *** * * * *
** * * ** ** * **** ** * * ****** ** ********* *** * ** * ** ** *
** * * ****** * * **** * * * * *

    }
**** * * * * ** * *** * * * *

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

    }

    return 0;

}
answered by (-85 points)
0 like 0 dislike
Hidden content!
# include <stdio.h>

# include <ctype.h>

int main()

{
* * *** *** * ******* *** data[2];
* * * * ***** * ** ** *** sum;
* * ********* * ***** i;
* * * * *** **** * * * *
* ** * * ** ** * * * ***
* ** * * ** * ** ** ** **** * * * **** * * * ** * * *****
* ** * ** * * ****** ** ** **
** ***** * * * * ** ** **** * * *
**** * ***** * *** * ** *** *** * ** * *
***** * ** * * * *

}
answered by (-124 points)
0 like 0 dislike
Hidden content!
#include ******* **
* *** * *
* ***** * *
*** * * *** *







int * **

{


* ** ****** int ** **
**** ** *** *

**** * * * scanf * * * * * *
* ** ** *

* * * while (a[i]!= ' ')

{
** * * * **

* *** ** ***** i++;
* * * *

* * * ** scanf ** ** * * *** * * ** * * * * * *

* * * * ***

}


* *** * ** scanf ** ** * * ** ** **


***** * * *** while * ** ** * *** **
* * * **** *

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

*** ** *** * * ** ** j++;
** * * **** ** * **** ****

** * * * ** **** scanf * ** * * * * * ** * *** **** ***

* * * * * *** **

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

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

** * * * * ** for (i = 0,j = 0;;i++)
* ** * *

** * * * return 0;

}
answered by (-157 points)
0 like 0 dislike
Hidden content!
include <stdio.h>

#include ** * *

int main()

{
*** ** * * ** *** a,b,c;
* * * * * *** ** * ** ** * * **** ******
* * *** = a + b;
*** **** ** ** *** **** ** * *** * * ** ***** * **** ** * **** **
* * ****** **** *** *** *** * ***
****** ** * * * * *
**** * ** * ** ** ** * ** * ***** **
* * ** *** ** * * * 0;

}
answered by (-42 points)
0 like 0 dislike
Hidden content!
#include * * ** *** *

int * * *** y=999999999;

int main()

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

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

int main(){
*********** * ** *** * ** *** a,b,c=0;
** * **** ******** * * ****** **** *** * * *
* ** * ***** *** ** ** ** ** * ***** ***
** ** * *** *** **** **** ** ***
** ** * * * ** *** *** ** ***** * *** * *** ***
** * * ** ** * *** * ** * 0;





}
answered by (-140 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.71.120.164
©2016-2025

Related questions

0 like 0 dislike
86 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18071 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 11k views
0 like 0 dislike
49 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18070 - Available when: Unlimited - Due to: Unlimited
| 6.2k views
0 like 0 dislike
62 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18067 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 8.7k views
0 like 0 dislike
21 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18066 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 3.8k views
0 like 0 dislike
0 answers
[Resource] asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18085 - Available when: Unlimited - Due to: Unlimited
| 10 views
12,783 questions
183,442 answers
172,219 comments
4,824 users