0 like 0 dislike
2.7k 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 | 2.7k 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:172.70.100.81
©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
| 8.7k 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
| 5.1k 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
| 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
| 3.1k 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