1.Link:
http://poj.org/problem?id=2389
http://bailian.openjudge.cn/practice/2389/
2.Content:
Bull Math
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13067 Accepted: 6736 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 linesSample Input
11111111111111
1111111111Sample Output
12345679011110987654321Source
3.Method:
直接套大数相乘模板
http://www.cnblogs.com/mobileliker/p/3516920.html
4.Code:
#include <iostream>
#include <string>
#include <vector>
#include <algorithm> using namespace std; string mul(string str1,string str2)
{
vector<int> v_res(str1.size()+str2.size(),);
string::size_type i,j;
vector<int>::size_type k,p; reverse(str1.begin(),str1.end());
reverse(str2.begin(),str2.end());
for(i = ; i != str1.size(); ++i)
{
for(j = ; j != str2.size(); ++j)
{
v_res[i+j] += (str1[i]-'') * (str2[j] - '');
}
}
for(k = ; k != v_res.size() - ; ++k)
{
v_res[k+] += v_res[k] / ;
v_res[k] = v_res[k] % ;
} for(p = v_res.size() - ; p != -; --p)
{
if(v_res[p] != ) break;
}
if(p == -) p = ; string s_res(p+,'');
for(k = p; k != -; --k) s_res[p-k] = char(v_res[k] + ''); return s_res; } int main()
{
//freopen("D://input.txt","r",stdin); string str1,str2;
cin >> str1 >> str2; cout << mul(str1,str2) << endl; return ;
}
5.Reference: