POJ 3673 Cow Multiplication

Cow Multiplication

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：

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <iostream>
 5 using namespace std;
 6 int main()
 7 {
 8     char a[10005],b[10005];
 9     int c[10005];
10     while(scanf("%s%s",&a,&b)!=EOF)
11     {
12         memset(c,0,sizeof(c));
13         int s=0;
14         int lena=strlen(a);
15         int lenb=strlen(b);
16         for(int i=0;i<lena;i++)
17         {
18             for(int j=0;j<lenb;j++)
19             {
20                 c[i]=(int)(a[i]-'0')*(int)(b[j]-'0');
21                 s+=c[i];
22             }
23         }
24         cout<<s<<endl;
25     }
26     return 0;
27 }