Problem F: 二进制类(1)__运算符重载 hzauoj

Problem F: 二进制类(1)__运算符重载

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 7  Solved: 7
[Submit][Status][Web Board]

Description

将一个16 位二进制数表示成0 和1 的字符序列，即用一个字符数组来存放这个

二进制数。在这个类中设置两个构造函数，一个是传递整数参数的，另一个是传递字符串参数的。因为用户在创建对象时传递的二进制数，可能是以整数形式给出，也可能是以数字串形式给出，系统应该都能接受。另外有一个类型转换函数int()，用来将类类型向整型转换。两个重载运算符“+”，“-”，用来完成两个二进制数之间的加减运算。

class binary { //定义二进制类

char bits[16]; //二进制字模数组

public:

binary(char *); //字符串参数构造函数

binary(int); //整型参数构造函数

friend binary operator +(binary,binary); //重载“+”

friend binary operator -(binary,binary); //重载“-”

operator int(); //类类型转换函数

void print();

};

主函数设计如下：

int main()

{

binary n1="1011";

binary n2=int(n1)+15;

binary n3=n1-binary(7);

n1.print();

n2.print();

n3.print();

cout<<int(n2)+5<<endl;

cout<<n2-binary(5)<<endl;

cout<<n3+binary(5)<<endl;

cout<<int(n3)-5<<endl;

return 0;

}

Input

Output

Sample Output

00000000000010110000000000011010000000000000010031219-1

HINT

[Submit][Status]

#include<iostream>#include<string.h>using namespace std;class binary{public:    binary(){ }    binary(char *);    binary(int n);    friend binary operator + (const binary &b1, const binary &b2);    friend binary operator - (const binary &b1, const binary &b2);    operator int()    {        int temp = 0;        for(int i = 0; i < 16; i++)            temp = temp * 2 + bits[i] - '0';        return temp;    }    void print()    {        for(int i=0; i < 16; i++) { cout << bits[i]; }        cout << endl;    }private:    char bits[16]; //二进制字模数组};binary :: binary(char *p){    int d = strlen(p)-1;    int i, j;    j = d;    for(i = 15; i >= 15 - d; i--,j--)        bits[i] = p[j];    for(; i >= 0; i--)        bits[i] = '0'; } binary :: binary(int n){  int i = 15;  while(n!=0)  {        int t;        t = n % 2;        bits[i] = t + '0';        i--;        n=n / 2;  }  for( int j = 0; j <= i; j++)   bits[j] = '0';}binary operator +(const binary &b1, const binary &b2){     binary tb1 = b1;    binary tb2 = b2;    int temp1 = int(tb1);    int temp2 = int(tb2);    return binary(temp1 + temp2);//    char temp[16];//    for(int i=0; i<16; i++)//       temp[i] = '0';//    for(int i = 15; i >=0; i--)//        if( temp[i] + b1.bits[i] + b2.bits[i] - '0' - '0' - '0' == 0 )//         temp[i] = '0';//        else if (temp[i] + b1.bits[i] + b2.bits[i] - '0' - '0' - '0' == 1)//            temp[i] = '1';//        else if (temp[i] + b1.bits[i] + b2.bits[i] - '0' - '0' - '0' == 2)//        {//            temp[i] = '0';//            temp[i - 1] = '1';//        }//        else if (temp[i] + b1.bits[i] + b2.bits[i] - '0' - '0' - '0' == 3)//        {//            temp[i] = '1';//            temp[i - 1] = '1';//        }//           return binary(temp);}binary operator-(const binary &b1, const binary &b2){   binary tb1 = b1;   binary tb2 = b2;    int temp1 = int(tb1);    int temp2 = int(tb2);    return binary(temp1 - temp2);//    char temp[16];//    char b1[16];//    char b2[16];//    for(int i = 0; i < 16; i++)//    {//         temp[i] = '0';//         b1[i] = b1_.bits[i];//         b2[i] = b2_.bits[i];//    }//     for(int i = 15; i >=0; i--)//        if( b1[i] - b2[i] - '0' - '0'  == 0 )//            temp[i] = '0';//        else if ( b1[i] - b2[i] - '0' - '0' == 1)//            temp[i] = '1';//        else if ( b1[i] - b2[i] - '0' - '0' == -1)//        {//            int j = i ;//            while(b1[j] == '0' && j >= 0) j--;//            b1[j] = '0';//            temp[i] = '1';//            for(int t = j - 1; t < i; t++)//                b1[t] = '1';//        }//    return binary(temp);} int main(){    binary n1="1011";    binary n2=int(n1)+15;    binary n3=n1-binary(7);    n1.print();    n2.print();    n3.print();    cout<<int(n2)+5<<endl;    cout<<n2-binary(5)<<endl;    cout<<n3+binary(5)<<endl;    cout<<int(n3)-5<<endl;    return 0;}

之前一直想用注释部分，利用字符串来解决问题，结果没有出现想要的输出值，然后突然灵机一动想到可以想把二进制转化为十进制，然后在把十进制转化为二进制，结果问题很简单的解决了。