Solution of ZOJ 1205 Martian Addition

In the 22nd Century, scientists have discovered intelligent residents live on the Mars. Martians are very fond of mathematics. Every year, they would hold an Arithmetic Contest on Mars (ACM). The task of the contest is to calculate the sum of two 100-digit numbers, and the winner is the one who uses least time. This year they also invite people on Earth to join the contest.

As the only delegate of Earth, you're sent to Mars to demonstrate the power of mankind. Fortunately you have taken your laptop computer with you which can help you do the job quickly. Now the remaining problem is only to write a short program to calculate the sum of 2 given numbers. However, before you begin to program, you remember that the Martians use a 20-based number system as they usually have 20 fingers.

Input:
You're given several pairs of Martian numbers, each number on a line.
Martian number consists of digits from 0 to 9, and lower case letters from a to j (lower case letters starting from a to present 10, 11, ..., 19).
The length of the given number is never greater than 100.

Output:
For each pair of numbers, write the sum of the 2 numbers in a single line.

Sample Input:

1234567890
abcdefghij
99999jjjjj
9999900001

Sample Output:

bdfi02467j
iiiij00000

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Source: Zhejiang University Local Contest 2002, Preliminary

 

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思路分析：由于参与运算的数字位数较长，因此选用字符串来表示数字是一种恰当的策略。为简化计算过程，可首先将两字符串反转，这样可从下标0开始运算。各位完成加法运算后，进位需要保留下来。另外此题容易出错的一个地方在于参与运算的数字长度可能不一致，因此可先将长度较短的数字翻转后追加若干0，使两个数字的长度相同。

 

下面给出我的实现代码，为了加快速度，我使用了查表的方法，即事先建立火星数字与字母的一一对应关系。

 

模块一：字符串长度调整

void adjust_str( string & str1, string & str2 ) { if( str1.size() > str2.size() ) { str2.append( str1.size() - str2.size(), '0' ); } else if( str1.size() < str2.size() ) { str1.append( str2.size() - str1.size(), '0' ); } }

模块二：字符串加法

void martian_addition( string & str1, string & str2 ) { reverse( str1.begin(), str1.end() ); reverse( str2.begin(), str2.end() ); adjust_str( str1, str2 ); string str_sum; int val = 0, incr = 0; int sum = 0; for( int i = 0; i < str1.size(); ++ i ) { val = char2num[str1[i]] + char2num[str2[i]] + incr; if( val >= 20 ) { incr = val / 20; val %= 20; } else { incr = 0; } str_sum += num2char[val]; } if( incr > 0 ) str_sum += num2char[incr]; reverse( str_sum.begin(), str_sum.end() ); printf( "%s/n", str_sum.c_str() ); //<< str_sum << endl; }

 

查表所用的两个全局变量

// 20-based maritian numbers const char num2char[] = "0123456789abcdefghij"; // char to number mapping map<char, int> char2num;  

 

主函数

 int main() { for( int c = '0'; c <= '9'; ++c ) char2num[c] = static_cast<int>( c - '0' ); for( char c = 'a'; c <= 'j'; ++c ) char2num[c] = static_cast<int>( c - 'a' + 10 ); string str1, str2; while( cin >> str1 >> str2 ) { martian_addition( str1, str2 ); } return 0; }