poj 2389 BullMath(大数乘法)_分治大整数乘 poj-优快云博客

本文介绍了一个解决大数乘法问题的程序，该程序能够处理多达40位数的整数相乘，并准确输出结果。通过逐位相乘并进位的方法，实现了不依赖于特殊数学库的大数乘法运算。

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Bull Math

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14987		Accepted: 7707

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

tips:水题

#include<iostream>
#include<cstring>
#include<string>


using namespace std;

string s1,s2;
char ans[88];
void multi()
{
	int ans[88];
	int a[88],b[88],lena=0,lenb=0;
	memset(a,0,sizeof(a));memset(b,0,sizeof(b));
	memset(ans,0,sizeof(ans));
	for(int i=s1.size()-1;i>=0;i--)a[lena++]=s1[i]-'0';
	for(int i=s2.size()-1;i>=0;i--)b[lenb++]=s2[i]-'0';
	
	for(int j=0;j<lenb;j++)
	for(int i=0;i<lena;i++)
	{
		ans[i+j]+=b[j]*a[i];
		ans[i+j+1]+=ans[i+j]/10;
		ans[i+j]=ans[i+j]%10;
	}
	int flag=0;
	for(int i=80;i>=0;i--)
	{
		if(ans[i]!=0||flag==1){
			flag=1;cout<<ans[i];
		}
	}
	cout<<endl;
}
int main()
{
	
	while(cin>>s1>>s2)
	{
		multi();
	}	
	
	return 0;
 }