A * B Problem Plus                   

Calculate A * B.

Input
Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output
For each case, output A * B in one line.

Sample Input

1
2
1000
2

Sample Output
     
   2
    
   2000
代码：
      

#include<stdio.h>
#include<string>
#include<string.h>
#include<algorithm>
#include<math.h>
#define KK 50010
using namespace std;
char a[KK],b[KK];
long long A[KK],B[KK],ans[KK];
int main()
{
    int lena,lenb,len_ans,lenA,lenB,sum,m;
    while(gets(a)!=NULL)
    {
        gets(b);
        lena=strlen(a);
        lenb=strlen(b);
        lenA=lenB=0;
        for(int i=lena-1; i>=0;)
        {
            sum=0;
            m=1;
            while(i>=0&&m<=100000)
            {
                sum+=(a[i]-'0')*m;
                i--;                                      // 1.把字符型转换成整数型；
                m*=10;                                //2.在每个A【】里存六位数；
            }
            A[lenA++]=sum;
        }
        for(int i=lenb-1; i>=0;)
        {
            sum=0;
            m=1;
            while(i>=0&&m<=100000)
            {
                sum+=(b[i]-'0')*m;
                i--;
                m*=10;

            }
            B[lenB++]=sum;
        }
        len_ans=lenA+lenB;
        for(int i=0; i<lenA; i++)
            for(int j=0; j<lenB; j++)
                ans[i+j]+=A[i]*B[j];                      //相乘；
        for(int i=0; i<len_ans; i++)
        {
            if(ans[i]>1000000)
            {
                ans[i+1]+=ans[i]/1000000;         //进位；
                ans[i]=ans[i]%1000000;
            }
        }
        while(ans[len_ans-1]==0&&len_ans!=0)               //清除 数前 多余的0；
            len_ans--;
        printf("%lld",ans[len_ans-1]);
        for(int i=len_ans-2; i>=0; i--)
            printf("%06lld",ans[i]);            
        printf("\n");
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(ans,0,sizeof(ans));
    }
}