数据结构之(HDU2051 Bitset)

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：

 #include "stdafx.h"
 #include<iostream>
 #include<cstring>
 using namespace std;
 int main()
 {
     int a[];
     int n;
     int m;
     int i;
     while(cin>>n)
     {
         i=;
         while(n)
         {
             i=i+;
             a[i]=n%;
             n/=;
         }
         for(i;i>;i--)
            cout<<a[i];
         cout<<endl;
     }
     return ;
 }

运用桟解决:

代码如下：

 #include "stdafx.h"
 #include<iostream>
 #include<cstdlib>
 using namespace std;
 #define OK 1
 #define ERROR 0
 #define OVERFLOW -2
 #define MAXSIZE 100
 typedef int Status;
 typedef int SElemType;

 typedef struct
 {
     SElemType *base;
     SElemType *top;
     int stacksize;
 }SqStack;

 Status InitStack(SqStack &S)           //桟的初始化
 {
     S.base = new SElemType[MAXSIZE];
     if (!S.base)
         exit(OVERFLOW);
     S.top = S.base;
     S.stacksize = MAXSIZE;
     return OK;
 }

 Status Push(SqStack &S, SElemType e)   //进栈
 {
     if (S.top - S.base == S.stacksize)
         return ERROR;
     *S.top++ = e;
     return OK;
 }

 Status Pop(SqStack &S, SElemType &e)   //出桟
 {
     if (S.top == S.base)
         return ERROR;
     e = *--S.top;
     return OK;
 }
 int main()
 {
     int N;
     SqStack S;
     SElemType e;
     InitStack(S);
     while (cin >> N)
     {
         while (N != )
         {
             e = N % ;
             Push(S, e);
             N = N / ;
         }
         while (!(S.top == S.base))
         {
             Pop(S, e);
             cout << e;
         }
         cout << endl;
     }
     return ;
 }