0 like 0 dislike
1.4k views
請設計一個程式,由使用者輸入一個二進位的數,程式會將他轉成十進位後輸出。

sample input:

10101010

sample output

170
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AC by (18k points)
ID: 15399 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00

reshown by | 1.4k views
0 0
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()

{
    char a[100];
    int n,c,i;
    scanf("%s",a);
    i=1;
    c=0;
    for(n=strlen(a)-1;n>=0;n--)
    {
        c=c+i*(a[n]-'0');
        i=i*2;
    }
    printf("%d",c);
    return 0;
}

7 Answers

0 like 0 dislike
Hidden content!
#include <stdio.h>

main()\

{
* * *** * *** ** a[99], b[99], i, j=0, k;
* * * *** * ** * * h;
* * **** * **** * * * * &i);
* * **** ****** *** * ** *

   (
* ** * * *** * ** * * * * ** a[i];
* **** ** ** * ** ** * * **** * *** *** *** *******

   )
*** **** ***** ** * **

   (
* ** *** **** *** * b[i];
** ** * ** ******** * * * ** *** * *
* *** * * * * * *** *** ** * *** ** * * ** ****
* ***** **** * ** *** * *
* ** * ** * ** ** * * *** ** ****** * * ** ** * *
*** **** * ** * ** *** **** ** **** * ** * * ******
* ** ** * ** * * ** * * * ***** * * * *** *** * ** * = b[j];
* ****** ********** * * * * * ** ** ** **** **** = a[k];
* * ***** * ** * * * * **** ***** * * * ***** * * = temp;
** ** * * * **** * * * * * ** ** * * *** *
* * **** **** * * * * ** * *

   )



   if (i/2!=0)
** ** * * ******* * **** (b[i/2] + b[(i/25)+1)/2
**** ** **** *** * *
* * ** * **** * * *** * ******* * ** = b[i/2];
****** * * * * * * * * * b[j])
* *** *** * * ** ** 0;

}
answered by (-52 points)
0 like 0 dislike
Hidden content!
*** ** * *****
* *** * * * *** * ***



int * *

{
* **** * * * **** * * * *** **


**** * **** * * ***** * * *** * ** * * * *** * * **** *** * * ** ** *** **


* *** * ** *** * = 128 * a + 64 * b + 32 * c + 16 * d + 8 * e + 4 * f + 2 * g + h;


* *** * ** **** * *** *** * ** **


* *** * ** * * ** 0;

}
answered by (180 points)
0 like 0 dislike
Hidden content!
** ** * * ** **
** ** * * * *** ** *



int * *

{
** *** * * ** ** * * ** *


**** * **** ** ** ** **** *** ** * * ********* * ** ** ** * ** ***** *** * *** ** * **


*** * ** *** * * * = 128 * a + 64 * b + 32 * c + 16 * d + 8 * e + 4 * f + 2 * g + h;


*** * * *** * *** * * ** ****** **


** **** * * *** ** * 0;

}
answered by (180 points)
0 like 0 dislike
Hidden content!
#include **** * ***

#include * * ***** **



int main( int argc, char )

{
** ******* **** * * *** iNum;
* **** **** * * * * * *** * *** ** ****** ** * * ** *
* ***** ** * * ** **** ** * ****
***** * ** * * ** ** * * **** * * ** ** * ** * %d\n", iNum, iNum);


** * * ** *** *** * ** * ** * *** ******
*** *** ** * **** * ** ** * * * ** * 0;

}
answered by (-74 points)
0 like 0 dislike
Hidden content!
#include ** ** *** * *

#include **** *** *

#include **** ** *****
*** * * **** ** * **

int()

{
* * *** ** ***** * * x;
*** * * ***** temp = 0;
**** **** * ***** * * * * *** x;
** ** * * * * * (x > 0)
** * ** * * **
** * * * * *** ** = 2 * temp +%10;
* * * ** ** * *
** ***** ** *** *
** ** * * * * * * temp *** endl;


** * * ******** 0;



}
answered by (-74 points)
0 like 0 dislike
Hidden content!
**** * ** **** *
* *** ** *** ***

int main()

{
* * ** ** ** ** *


******* ** ** **



if(n%10!=0)
**** * *** * **** *

a=(n%10)*1;

}



else if(n%100!=0)

{
* *** * *** *

}

else if(n%1000!=0)

{
** ** * * **** * * ***

}

else * * *

{
* ** **** * *** * * * *** **

}

else *** * *

{
* * ** ********* * * * * * ** ***** *

}

else ** * * *

{
** ** * * * * *** * * * *** *** * * * **

}

else * *

{
* * ** * **** * ** **** * * * **** * ** * *** *

}

else * * ** *

{
** * *** * ** ***** * * *** ** * ** ** *** ** ** *

}
** * *****
* ** ** *

return 0;

}
answered by (-248 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>





int main(){
** *** * * ** * * ***** * ** x,y=0,i,j=0,n=1;
* * ** * *** * ** *** * ** * * ***** ** ** ** * * *

    do{
* * * *** ** ******* ** * *** ** **** * * * *
*** ** ** *** ** * * ** **** * * ** * ***
* ** *** * * ** ** ****** ** * *** * * ** *
**** ** **** ** ***** * * * ** * *** * **


* **** * * ********** ** * ** *
* ** * * * * * * ** ** * * * ** ****
* ** * * ** * ***** * * 0;

}
answered by (-126 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.69.130.13
©2016-2025

Related questions

0 like 0 dislike
0 answers
[Resource] asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15475 - Available when: Unlimited - Due to: Unlimited
| 13 views
0 like 0 dislike
5 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15447 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 1.3k views
1 like 0 dislike
37 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15446 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 4.4k views
1 like 0 dislike
17 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15442 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 2.3k views
1 like 0 dislike
18 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15440 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 2.3k views
12,783 questions
183,442 answers
172,219 comments
4,824 users