0 like 0 dislike
1.1k 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 | 1.1k 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:172.70.179.148
©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
| 993 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
| 3.3k 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
| 1.8k 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
| 1.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users