0 like 0 dislike
3.6k 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 | 3.6k 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:162.159.115.31
©2016-2026

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
| 12k 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
| 6.7k 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
| 9.5k 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
| 4.1k 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