Uva 10106 - Product

最新推荐文章于 2021-07-29 15:39:30 发布

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最新推荐文章于 2021-07-29 15:39:30 发布

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原文链接：http://www.cnblogs.com/liaoguifa/archive/2012/11/07/2758092.html

本文介绍了一种解决大数乘法问题的算法实现方法，通过字符串存储大数并模拟手动计算过程完成乘法运算。文章提供了一个完整的C语言代码示例，并详细解释了解题思路。

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;
}