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  • Poj OpenJudge 百练 2389 Bull Math

    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 lines

    Sample Input

    11111111111111
    1111111111

    Sample Output

    12345679011110987654321

    Source

    3.Method:

    直接套大数相乘模板

    http://www.cnblogs.com/mobileliker/p/3516920.html

    4.Code:

     1 #include <iostream>
     2 #include <string>
     3 #include <vector>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 string mul(string str1,string str2)
     9 {
    10     vector<int> v_res(str1.size()+str2.size(),0);
    11     string::size_type i,j;
    12     vector<int>::size_type k,p;
    13 
    14     reverse(str1.begin(),str1.end());
    15     reverse(str2.begin(),str2.end());
    16     for(i = 0; i != str1.size(); ++i)
    17     {
    18         for(j = 0; j != str2.size(); ++j)
    19         {
    20             v_res[i+j] += (str1[i]-'0') * (str2[j] - '0');
    21         }
    22     }
    23     for(k = 0; k != v_res.size() - 1; ++k)
    24     {
    25         v_res[k+1] += v_res[k] / 10;
    26         v_res[k] = v_res[k] % 10;
    27     }
    28 
    29     for(p = v_res.size() - 1; p != -1; --p)
    30     {
    31         if(v_res[p] != 0) break;
    32     }
    33     if(p == -1) p = 0;
    34 
    35     string s_res(p+1,'0');
    36     for(k = p; k != -1; --k) s_res[p-k] = char(v_res[k] + '0');
    37     
    38 
    39     return s_res;
    40 
    41 }
    42 
    43 int main()
    44 {
    45     //freopen("D://input.txt","r",stdin);
    46     
    47     string str1,str2;
    48     cin >> str1 >> str2;
    49 
    50     cout << mul(str1,str2) << endl;
    51 
    52     return 0;
    53 }

    5.Reference:

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  • 原文地址:https://www.cnblogs.com/mobileliker/p/4095425.html
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