无符号的乘法C++代码完善

Question

目前的代码只实现了结果，代码想体现具体过程，如图所示样子

有专家帮忙完善一下吗

#include<iostream>
#include <stdio.h>
#define N 100
using namespace std;
int Lengh(char a[])
{
    int i=0,j=0;
    while(a[i]!='\0')
    {
        if(a[i]!=' ')
        j++;
        i++;
    }
    return j;
}
void mul(char int1[],char int2[],int n)
{
    char *sum = new char[2*(n+1)];
    char *temp = new char[2*(n+1)];
    for(int i = 0; i < 2*(n+1); i++) {
        sum[i] = '0';
        temp[i] = '0';
    }
    for(int i = n; i > -1; i--) {
        if(int2[i]=='1')
        {
            for(int k = 0; k < 2*(n+1); k++)
            {
                temp[k] = '0';
            }
            for(int j = n; j > -1; j--)
            {
                temp[j+i+1] = int1[j];
            }
            bool carry = false;
            for(int i = (2*n + 1); i > -1; i--)
            {
                if(!carry)
                {
                    if((sum[i]^temp[i])==0)
                    {
                        if((sum[i]=='1')&&(temp[i]=='1'))
                        {
                            carry = true;
                        }
                        sum[i] = '0';
                    }
                    else
                    {
                        sum[i] = '1';
                    }
                } else
                {
                    if((sum[i]^temp[i])==0)
                    {
                        carry = ((sum[i]=='1')&&(temp[i]=='1')) ? true : false;
                        sum[i] = '1';
                    }
                    else
                    {
                        sum[i] = '0';
                        carry = true;
                    }
                }
            }
        }
    }
    for(int i = 0; i < (n+1); i++)
    {
        int1[i] = sum[i];
    }
    for(int i = (n+1); i < 2*(n+1); i++)
    {
        int2[i-n-1] = sum[i];
    }
}





int main()
{
    LOOP:
    int a=0,b=0;
    char *c=new char[N];
    char *d=new char[N];
    cout<<"请输入两个二进制乘数，可以加空格，输入quit退出程序"<<endl;
    cout<<"A=";
    cin>>c;

    cout<<"B=";
    cin>>d;
    
    a=Lengh(c);
    b=Lengh(d);
    if(a!=b)
    {
        cout<<"输入的二进制数不正确，两数长度不匹配"<<endl;
        goto LOOP;
    }
    a=a-1;
    mul(c,d,a);
    cout<<c<<d;
    goto LOOP;
}

Answer 1

你题目的解答代码如下：

#include<iostream>
#include <cstring>
#include <stdio.h>
#define N 100
using namespace std;
int Lengh(char a[])
{
    int i=0,j=0;
    while(a[i]!='\0')
    {
        if(a[i]!=' ')
            j++;
        i++;
    }
    return j;
}
void mul(char int1[],char int2[],int n)
{
    char *sum = new char[2*(n+1)];
    char *temp = new char[2*(n+1)];
    printf("%*s\n",2*(n+1),int1);
    printf("%*s\n",2*(n+1),int2);
    printf("%*s",n+1,"X ");
    for(int i = 0; i < n+1; i++)
        printf("-");
    printf("\n");

    for(int i = 0; i < 2*(n+1); i++) {
        sum[i] = '0';
        temp[i] = '0';
    }
    for(int i = n; i > -1; i--) {
        if(int2[i]=='1')
        {
            for(int k = 0; k < 2*(n+1); k++)
            {
                temp[k] = '0';
            }
            for(int j = n; j > -1; j--)
            {
                temp[j+i+1] = int1[j];
            }
            bool carry = false;
            for(int i = (2*n + 1); i > -1; i--)
            {
                if(!carry)
                {
                    if((sum[i]^temp[i])==0)
                    {
                        if((sum[i]=='1')&&(temp[i]=='1'))
                        {
                            carry = true;
                        }
                        sum[i] = '0';
                    }
                    else
                    {
                        sum[i] = '1';
                    }
                } else
                {
                    if((sum[i]^temp[i])==0)
                    {
                        carry = ((sum[i]=='1')&&(temp[i]=='1')) ? true : false;
                        sum[i] = '1';
                    }
                    else
                    {
                        sum[i] = '0';
                        carry = true;
                    }
                }
            }
            printf("%*s\n",n+2+i,int1);
        }
        else
        {
            printf("%*s",i+1,"");
            for(int k = 0; k < n+1; k++)
                printf("0");
            printf("\n");

        }


    }
    for(int i = 0; i < (n+1); i++)
    {
        int1[i] = sum[i];
    }
    for(int i = (n+1); i < 2*(n+1); i++)
    {
        int2[i-n-1] = sum[i];
    }
    for(int i = 0; i < 2*(n+1); i++)
        printf("-");
    printf("\n");
    printf("%s\n",sum);

}





int main()
{
    LOOP:
    int a=0,b=0;
    char *c=new char[N];
    char *d=new char[N];
    cout<<"请输入两个二进制乘数，可以加空格，输入quit退出程序"<<endl;
    cout<<"A=";
    cin>>c;
    if(strcmp(c, "quit") == 0)
        return 0;

    cout<<"B=";
    cin>>d;

    a=Lengh(c);
    b=Lengh(d);
    if(a!=b)
    {
        cout<<"输入的二进制数不正确，两数长度不匹配"<<endl;
        goto LOOP;
    }
    a=a-1;
    mul(c,d,a);
    goto LOOP;
}

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