0 like 0 dislike
456 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 | 456 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:172.69.6.248
©2016-2024

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
| 1.5k 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
| 911 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
| 1.1k 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
| 480 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
183,443 answers
172,219 comments
4,824 users