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Description

Product

The Problem

The problem is to multiply two integers X, Y. (0<=X,Y<10²⁵⁰)

The Input

The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.

The Output

For each input pair of lines the output line should consist one integer the product.

Sample Input

12
12
2
222222222222222222222222

Sample Output

144
444444444444444444444444

注意当为0的情况，一开始忽略了这种情况。当任一个为0的时候别忘了输出0

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char str1[1010],str2[1010];
char a[1010],b[1010],sum[1010];
int main()
{
    int len1,len2;
    int i,j;
    while (~scanf("%s",str1))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(sum,0,sizeof(sum));
        len1= strlen(str1);
        for (i=len1-1;i>=0;i--)
            a[len1-i-1]=str1[i]-'0';
        scanf("%s",str2);
        len2=strlen(str2);
        for (i=len2-1;i>=0;i--)
            b[len2-i-1]=str2[i]-'0';
        for (i=0;i<len1;i++)
        {
            for(j=0;j<len2;j++)
                sum[i+j]+=a[i]*b[j];
            for (j=0;j<1010;j++)
                if (sum[j]>=10)
                {
                    sum[j+1]+=sum[j]/10;
                    sum[j]%=10;
                }
        }
        int flag=0;
        for(i=1010;i>=0;i--)
        {
            if(flag||sum[i])
            {
                flag=1;
                printf("%d",sum[i]);
            }
        }
        if(!flag)
            printf("0");
        printf("\n");
    }
    return 0;
}

UVA 10106-Product(大数乘法)

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原文地址：http://blog.csdn.net/u013486414/article/details/41794631