用C语言编程实现十进制转化为二进制 c语言编程，如何将十进制转化为2进制
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#include void main(){ int c[32],i=0,n; scanf("%d", &n) ; //\u8f93\u5165\u5f85\u8f6c\u6362\u6574\u6570n do { c[i++]=n%2; n/=2; } while(n!=0) ; for(i--;i>=0;i--) printf("%d",c[i] ); printf("\n");}
#include
"stdio.h"
#define
StackSize
100
typedef
int
ElemType;
typedef
struct
{
ElemType
elem[StackSize];
int
top;
}SqStack;
InitStack(SqStack
*pS)
{
pS->top=0;
/*
top\u6307\u5411\u6808\u9876\u7684\u4e0a\u4e00\u4e2a\u5143\u7d20
*/
}
int
Push(SqStack
*pS,ElemType
e)
{
if
(pS->top==StackSize-1)
/*
\u6808\u6ee1
*/
return
0;
pS->elem[pS->top]=e;
pS->top=pS->top+1;
return
1;
}
int
Pop(SqStack
*pS,ElemType
*pe)
{
if
(pS->top==0)
/*
\u6808\u7a7a
*/
return
0;
pS->top
=
pS->top
-
1;
*pe
=
pS->elem[pS->top];
return
1;
}
main()
{
SqStack
S;
ElemType
e;
int
N;
InitStack(&S);
printf("\u8bf7\u8f93\u5165\u4e00\u4e2a\u5341\u8fdb\u5236\u6574\u6570\uff1a\n");
scanf("%d",&N);
do
{
e
=
N
%
2;
Push(&S,e);
N
=
N/2;
}
while(N);
while(Pop(&S,&e))
{
printf("%d",e);
}
printf("\n");
return
0;
}
有两种计算方法，运行的结果是完全相同的：
一是采用位运算的方法：
#include
int main()
{ int n,i;
scanf("%d",&n);
for(i=31;i>=0;i--)
printf("%d",(n&(1<<i))!=0);
return 0;
}
二是采用楼主指定的方法：
#include
int a[32];
int main()
{ int n,i,k=0;
scanf("%d",&n);
if(n<0)k=1;
for(i=31;n;i--)
{a[i]=n%2;
n/=2;
}
if(k==1)
{for(i=31;i>=0;i--)
a[i]=1+a[i];
for(i=31;a[i];i--)
a[i]=0;
a[i]=1;
}
for(i=0;i<32;i++)
printf("%d",a[i]);
return 0;
}
C语言_010_二进制转十进制_上
有一种骚操作可以很快实现这个
发不了代码就发图吧emmm
得到的结果确实是没问题的emmm
如果你是c语言不是c++的话，可以了解一下这样搞
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