Bitset Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13627    Accepted Submission(s): 10406


Problem Description Give you a number on base ten,you should output it on base two.(0  
Input For each case there is a postive number n on base ten, end of file.  
Output For each case output a number on base two.  
Sample Input

1
2
3

 
Sample Output

1
10
11
//方法一       用itoa函数
#include              //这函数太神了，进制转换直接秒过
#include
int main( )
{
__int64 n;
while(scanf("%I64d",&n)!=EOF)
{
char a[10000];
printf("%s\n",itoa(n,a,2));
}
return 0;
}
              //方法二    用位运算
#include
int main( )
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i,a[100]={0};
for(i=0;n>>1 || n==1;n>>=1,i++)
{
if(n & 1)
    a[i]=1;
    else
    a[i]=0;
}
while(i>0)
printf("%d",a[--i]);
printf("\n");
}
return 0;
}
                 //方法三   常规进制转换法


#include
using namespace std;
int main()
{
int n,s;
char a[1000];

while(cin>>n)
{
int i=0;
while(n)
{
s=n%2;
n=n/2;
a[i]=s+48;
i++;
}
for(s=0;scout<cout<

}
return(0);
}