0 like 0 dislike
2.6k 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 | 2.6k 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:104.23.190.140
©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
| 8.6k 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
| 5k 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
| 6.9k 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
| 3k 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