0 like 0 dislike
1.6k 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 | 1.6k 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:172.70.80.121
©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
| 1.4k 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
| 5k 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
| 2.5k 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
| 2.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users