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[100];
    int n;
    int m;
    int i;
    while(cin>>n)
    {
        i=0;
        while(n)
        {
            i=i+1;
            a[i]=n%2;
            n/=2;
        }
        for(i;i>0;i--)
           cout<<a[i];
        cout<<endl;
    }
    return 0;
}

 

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 != 0)
        {
            e = N % 2;
            Push(S, e);
            N = N / 2;
        }
        while (!(S.top == S.base))
        {
            Pop(S, e);
            cout << e;
        }
        cout << endl;
    }
    return 0;
}