Problem Description

Give you a number on base ten,you should output it on base two.(0 < n < 1000)

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

 

1.普通方法

代码如下：

1 #include "stdafx.h" 2 #include
       
         3 #include
        
          4 using namespace std; 5 int main() 6 { 7     int a[100]; 8     int n; 9     int m;10     int i;11     while(cin>>n)12     {13         i=0;14         while(n)15         {16             i=i+1;17             a[i]=n%2;18             n/=2;19         }20         for(i;i>0;i--)21            cout<
        
       

 

运用桟解决:

代码如下：

 

1 #include "stdafx.h" 2 #include
       
         3 #include
        
          4 using namespace std; 5 #define OK 1 6 #define ERROR 0 7 #define OVERFLOW -2 8 #define MAXSIZE 100 9 typedef int Status;10 typedef int SElemType;11 12 typedef struct13 {14     SElemType *base;15     SElemType *top;16     int stacksize;17 }SqStack;18 19 Status InitStack(SqStack &S)           //桟的初始化20 {21     S.base = new SElemType[MAXSIZE];22     if (!S.base)23         exit(OVERFLOW);24     S.top = S.base;25     S.stacksize = MAXSIZE;26     return OK;27 }28 29 Status Push(SqStack &S, SElemType e)   //进栈30 {31     if (S.top - S.base == S.stacksize)32         return ERROR;33     *S.top++ = e;34     return OK;35 }36 37 Status Pop(SqStack &S, SElemType &e)   //出桟38 {39     if (S.top == S.base)40         return ERROR;41     e = *--S.top;42     return OK;43 }44 int main()45 {46     int N;47     SqStack S;48     SElemType e;49     InitStack(S);50     while (cin >> N)51     {52         while (N != 0)53         {54             e = N % 2;55             Push(S, e);56             N = N / 2;57         }58         while (!(S.top == S.base))59         {60             Pop(S, e);61             cout << e;62         }63         cout << endl;64     }65     return 0;66 }