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[10005],b[10005];
    int c[10005];
    while(scanf("%s%s",&a,&b)!=EOF)
    {
        memset(c,0,sizeof(c));
        int s=0;
        int lena=strlen(a);
        int lenb=strlen(b);
        for(int i=0;i<lena;i++)
        {
            for(int j=0;j<lenb;j++)
            {
                c[i]=(int)(a[i]-'0')*(int)(b[j]-'0');
                s+=c[i];
            }
        }
        cout<<s<<endl;
    }
    return 0;
}