Uva 10106 - Product

 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

#include<stdio.h>
#include<string.h>
int main()
{
    int sum[502],  i, j, t, loop, jlen, ilen, temp, e;
    char inter[251], jnter[251];
    memset(sum, 0, sizeof(sum));
    memset(inter, 0, sizeof(inter));
    memset(jnter, 0, sizeof(jnter));
    while(scanf("%s", inter) != EOF)
    {
        getchar();
        scanf("%s", jnter);
        e = 0;
        ilen = strlen(inter);
        jlen = strlen(jnter);
        for(j=jlen-1; j>=0; --j)
        {
            loop = jlen-1-j;
            for(i=ilen-1; i>=0; --i, ++loop)
            {
                temp = sum[loop];
                sum[loop] = (e+sum[loop]+(inter[i]-'0')*(jnter[j]-'0'))%10;
                e = (e+temp+(inter[i]-'0')*(jnter[j]-'0'))/10;
            }
            for(t=loop; t<502 && e; ++t)
            {
            temp = sum[t];
            sum[t] = (e + sum[t])%10;
            e = (e + temp)/10;
            } 
        }
        
        for(i=501, temp = 0; i>=0; --i)
        {
          if(sum[i] != 0)
            {
            temp = 1;
            printf("%d", sum[i]);
            }
          else if(temp == 1)
            {
            printf("%d", sum[i]);
            }
        }
        if(temp == 0)printf("0");
        printf("\n");
           memset(sum, 0, sizeof(sum));
        memset(inter, 0, sizeof(inter));
        memset(jnter, 0, sizeof(jnter));
    }
    return 0;
}

解题报告：

两个大数相乘，无可厚非按照手工计算的方法进行，知道了两个大数相加这题也就容易了。