大数乘大数 Bull Math

末蓝、 2024-02-19 19:38 35阅读 0赞

Bull Math

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 16580</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 8451</td> 
  </tr> 
 </tbody> 
</table>

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros).  
  
FJ asks that you do this yourself; don't use a special library function for the multiplication.

Input

\* Lines 1..2: Each line contains a single decimal number.

Output

\* Line 1: The exact product of the two input lines

Sample Input

11111111111111
    1111111111

Sample Output

12345679011110987654321

Source

[USACO 2004 November][]

过程不懂的话，看这篇文章

[https://blog.csdn.net/qq\_41325698/article/details/88929961][https_blog.csdn.net_qq_41325698_article_details_88929961]

#include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    #define range 103
    void multi(char a[],char b[],char s[])
    {
    	int i,j,k=0,alen=strlen(a),blen=strlen(b),sum=0,res[range][range]={0},flag=0;
    	char result[range];
    	for(i=0;i<alen;i++)
    	for(j=0;j<blen;j++)
    	res[i][j]=(a[i]-'0')*(b[j]-'0');
    	
    	// 模拟乘法 
    	for(i=alen-1;i>=0;i--)
    	{
    		for(j=blen-1;j>=0;j--)
    		sum+=res[i+blen-1-j][j];
    		
    		result[k]=sum%10;
    		k++;
    		sum/=10;
    	}
    	for(i=blen-2;i>=0;i--)
    	{
    		for(j=0;j<=i;j++)
    		sum+=res[i-j][j];
    		
    		result[k]=sum%10;
    		k++;
    		sum/=10;
    	}
    	
    	// 处理最后的进位 
    	if(sum)
    	{
    		result[k]=sum;
    		k++;
    	}
    	
    	for(i=0;i<k;i++)
    	result[i]+='0';
    	// 字符串翻转 
    	for(i=k-1;i>=0;i--)
    	s[i]=result[k-1-i];
    	s[k]='\0';
    	printf("%s\n",s);
    }
    int main()
    {
    	char a[range],b[range],res[range];
    	cin>>a>>b;
    	multi(a,b,res);
    	return 0;
    }

[USACO 2004 November]: http://poj.org/searchproblem?field=source&key=USACO+2004+November
[https_blog.csdn.net_qq_41325698_article_details_88929961]: https://blog.csdn.net/qq_41325698/article/details/88929961