fjnu 1178 Base 9 Calculator

最新推荐文章于 2021-10-17 18:49:55 发布

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#numbers #input #iostream #output #each #hex

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本文介绍了一种解决九进制数相加的问题，输入为两个九进制数，输出为它们相加后的九进制结果。通过将九进制转换为十进制进行运算，并最终返回九进制形式的答案。

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Description

Normally, you use base 10 to do arithmetic. In computer science, you also deal with binary (base 2), octal (base 8), and hex (base 16). In this problem, we'll worry about base 9, which uses the digits 0..8.

Input

each line of input will contain two numbers separated by a space. Each number is in base 9. If converted to base 10, the numbers would be greater than 0, and smaller than 65000.

Output

For each input, your program should add the two numbers together and print the result (in base 9).

Sample Input

148 765
111 888
8734 8345

Sample Output

1024
1110
18180

KEY:这题就是求和，只是他是9进制的就是了

Source:

#include<iostream>
using namespace std;

void Add(char str1[],char str2[])
...{
    int la=strlen(str1),lb=strlen(str2);
    int a[15],b[15];
    int i,j;
    for(i=0;i<=10;i++)
        a[i]=b[i]=0;
    for(i=0;i<la;i++)
        a[i+1]=str1[i]-'0';
    for(i=0;i<lb;i++)
        b[i+1]=str2[i]-'0';
    if(la>=lb)
    ...{
        i=la;
        j=lb;
        while(j>0)
        ...{
            a[i--]+=b[j--];
        }
        for(i=la;i>0;i--)
        ...{
            int t;
            t=a[i];
            a[i-1]+=t/9;
            a[i]=t%9;
        }
        if(a[0]==0) i=1;
        else i=0;
        for( ;i<=la;i++)
            cout<<a[i];
        cout<<endl;
    }
    else
    ...{
        i=la;
        j=lb;
        while(i>0)
        ...{
            b[j--]+=a[i--];
        }
        for(j=lb;j>0;j--)
        ...{
            int t;
            t=b[j];
            b[j-1]+=t/9;
            b[j]=t%9;
        }
        if(b[0]==0) j=1;
        else j=0;
        for( ;j<=lb;j++)
            cout<<b[j];
        cout<<endl;
    }
}

int main()
...{
//    freopen("fjnu_1178.in","r",stdin);
    char str1[10],str2[10];
    while(cin>>str1>>str2)
    ...{
        Add(str1,str2);
    }
    return 0;
}