POJ 3673 Cow Multiplication

Cow Multiplication

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13312		Accepted: 9307

Description

Bessie is tired of multiplying pairs of numbers the usual way, so she invented her own style of multiplication. In her style, A*B is equal to the sum of all possible pairwise products between the digits of A and B. For example, the product 123*45 is equal to 1*4 + 1*5 + 2*4 + 2*5 + 3*4 + 3*5 = 54. Given two integers A and B (1 ≤ A, B ≤ 1,000,000,000), determine A*B in Bessie's style of multiplication.

Input

* Line 1: Two space-separated integers: A and B.

Output

* Line 1: A single line that is the A*B in Bessie's style of multiplication.

Sample Input

123 45

Sample Output

54

Source

USACO 2008 February Bronze

题目链接：http://poj.org/problem?id=3673

分析：算是处理字符串吧！用两个数组去做，用一个数去计算它们的和即可！

下面给出AC代码：

 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <iostream>
 using namespace std;
 int main()
 {
     char a[],b[];
     int c[];
     while(scanf("%s%s",&a,&b)!=EOF)
     {
         memset(c,,sizeof(c));
         int s=;
         int lena=strlen(a);
         int lenb=strlen(b);
         for(int i=;i<lena;i++)
         {
             for(int j=;j<lenb;j++)
             {
                 c[i]=(int)(a[i]-'')*(int)(b[j]-'');
                 s+=c[i];
             }
         }
         cout<<s<<endl;
     }
     return ;
 }