JOJ 1006解题

妖狐艹你老母 2022-06-08 07:09 128阅读 0赞

## ![art.gif][] [1006][]: All your base ##

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 <tbody> 
  <tr> 
   <th style="background-color:rgb(255,238,79); padding-right:5px"> Result</th> 
   <th style="background-color:rgb(255,238,79); padding-right:5px">TIME Limit</th> 
   <th style="background-color:rgb(255,238,79); padding-right:5px">MEMORY Limit</th> 
   <th style="background-color:rgb(255,238,79); padding-right:5px">Run Times</th> 
   <th style="background-color:rgb(255,238,79); padding-right:5px">AC Times</th> 
   <th style="background-color:rgb(255,238,79); padding-right:5px">JUDGE</th> 
  </tr> 
  <tr> 
   <td align="center" style="font-family:Verdana,Arial,Helvetica,sans-serif; font-size:14px"> <span><img src="http://acm.jlu.edu.cn/joj/images/ok1.gif" width="20" height="20" alt="" style="border:0px"></span></td> 
   <td align="center" style="font-family:Verdana,Arial,Helvetica,sans-serif; font-size:14px"> 3s</td> 
   <td align="center" style="font-family:Verdana,Arial,Helvetica,sans-serif; font-size:14px"> 8192K</td> 
   <td align="center" style="font-family:Verdana,Arial,Helvetica,sans-serif; font-size:14px"> 6195</td> 
   <td align="center" style="font-family:Verdana,Arial,Helvetica,sans-serif; font-size:14px"> 2080</td> 
   <td align="center" style="font-family:Verdana,Arial,Helvetica,sans-serif; font-size:14px"> Standard</td> 
  </tr> 
 </tbody> 
</table>

Given a base, and two positive integers in decimal (base 10), convert the two numbers to the new base, add them, and display their sum in the new base.

### Input ###

Three positive integers denoting the base and the two numbers, respectively. Input numbers will be integers between 0 and 65535. Bases will be between 2 and 10 inclusive. Each case will be on a separate line. The end of input will be denoted by three zeros.

### Output ###

An equation for the sum of the two numbers, in the new base.

### Example ###

In this example, we add 10 and 3 in base 2, and we add 15 and 4 in base 3.

In base 2, 10 = 8 + 2 = 1\*23 \+ 0\*22 \+ 1\*21 \+ 0\*20 and 3 = 2 + 1 = 1\*21 \+ 1\*20, so their base 2 equivalents are 1010 and 11, respectively. 10 + 3 = 13 = 1\*23\+1\*22\+0\*21\+1\*20, so the base 2 equivalent of 13 is 1101.

In base 3, 15 = 9 + 6 = 1\*32\+2\*31\+0\*30 and 4 = 3 + 1 = 1\*31 \+ 1\*30, so their base 2 equivalents are 120 and 11, respectively. 15 + 4 = 19 = 2\*32 \+ 0\*31 \+ 1\*30, so the base 3 equivalent of 19 is 201.

### Input ###

2 10 3
    3 15 4
    0 0 0

### Output ###

1010 + 11 = 1101
    120 + 11 = 201

解题代码如下：

#include <iostream>
    #include <stack>
    #include <queue>
    #include <vector>
    using namespace std;
    
    
    typedef struct{
        int a ;
        int b ;
        int c ;
    }inputNum;
    
    void converse(int base, int num, stack<char> *result)
    {
        int a = num;
        int b = base;
        while (a != 0)
        {
            result->push(a%b + '0');
            a = a/b;
        }
    }
    
    int main ()
    {
        queue<inputNum> input;
        int a,b,c;
        while(cin>>a>>b>>c && !(a == 0 && b == 0 && c == 0))
        {
            inputNum temp;
            temp.a = a;
            temp.b = b;
            temp.c = c;
            input.push(temp);
        }
        while (!input.empty())
        {
            inputNum temp ;
            temp = input.front();
            input.pop();
            stack<char> result;
            converse(temp.a, temp.b, &result);
            stack<char> result2;
            converse(temp.a, temp.c, &result2);
            stack<char> result3;
            converse(temp.a, temp.b+temp.c , &result3);
            while (!result.empty()) {
                cout<<result.top();
                result.pop();
            }
            cout << " + ";
            while (!result2.empty()) {
                cout<<result2.top();
                result2.pop();
            }
    
            cout <<" = ";
            while (!result3.empty()) {
                cout<<result3.top();
                result3.pop();
            }
            cout<<endl;
            
        }
        
        return 0;
    }

[art.gif]: http://acm.jlu.edu.cn/joj/images/art.gif
[1006]: http://acm.jlu.edu.cn/joj/problemstatus.php?pid=1006