0 like 0 dislike
2.5k 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 | 2.5k 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:172.70.130.43
©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
| 7.8k 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
| 4.6k 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
| 6.1k 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
| 2.7k 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