大數相乘 1

題目：兩個大數相乘。

大數相乘(2)見此。

方法1：將兩個大數分別用陣列儲存，模擬乘法運算，將其中乙個陣列的每一位分別與令乙個大數相乘，結果相加。

**見後面**部分的string multiply1(bignumber &bignumber2)。

方法2：同樣是上面的思路，不同的是儲存過程直接用乙個二維陣列來訪問每一位相乘的結果，然後處理相加的過程。

**見後面**部分的string multiply2(bignumber &bignumber2)。

#include #include #include using namespace std;

class bignumber
}		return true;
}public:
explicit bignumber(const char tocopy)
explicit bignumber(const bignumber &tocopy)
string showvalue()
bool showcheck()
string multiply1(bignumber &bignumber2)
for(count = len2; count > 0; --count)
int *convertpointer = new int[len1 + len2];
memset(convertpointer, 0, (len1 + len2) * sizeof(int));
for(count = len2; count > 0; --count)
}for(count = len1 + len2; count > 0; --count)
}char *resultchar = new char[len1 + len2];
bool checkresultzero = false;
int countresultzero = 0;
for(count = 0; count < len1 + len2; ++count)
}else
}resultchar[countresultzero] = '\0';
string result = resultchar;
delete no1;
delete no2;
delete convertpointer;
delete resultchar;
return result;
}	string multiply2(bignumber &bignumber2)
}int *convertpointer = new int[len1 + len2];
memset(convertpointer, 0, sizeof(int) * (len1 + len2));
int count = 0;
for(count = len1 + len2; count > 0; --count)
if(convertpointer[count - 1] > 9)
}char *resultchar = new char[len1 + len2];
bool checkresultzero = false;
int countresultzero = 0;
for(count = 0; count < len1 + len2; ++count)
}else
}resultchar[countresultzero] = '\0';
string result = resultchar;
for(countcolumn = 0; countcolumn < len1 + 1; ++countcolumn)
delete convertcolumn;
delete convertpointer;
delete resultchar;
return result;
}};int main()

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