Volume 1. Elementary Problem Solving->Volume 2 Big Number 10106 - Product

本文链接：https://blog.youkuaiyun.com/yibing2011/article/details/8238597

本文介绍了一种用于实现高精度整数乘法的方法，通过输入一对整数并输出它们的乘积，展示了如何在限制条件下高效进行大整数运算。

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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

题意：高精度乘法

#include<stdio.h>
#include<string.h>
#define m 252
char a[m],b[m];
int ta[m],tb[m],tc[2*m+10];
int main()
{
    int i,j,l1,l2;
    while(scanf("%s%s",a,b)!=EOF)
    {
        memset(ta,0,sizeof(ta));
        memset(tb,0,sizeof(tb));
        memset(tc,0,sizeof(tc));
        l1=strlen(a);
        l2=strlen(b);
        for(i=l1-1;i>=0;i--)
            ta[l1-i-1]=a[i]-'0';
        for(i=l2-1;i>=0;i--)
            tb[l2-i-1]=b[i]-'0';
        for(i=0;i<l1;i++)
        for(j=0;j<l2;j++)
            tc[i+j]+=ta[i]*tb[j];
        for(i=0;i<l1+l2;i++)
        {
            if(tc[i]>=10)
            {
                tc[i+1]+=tc[i]/10;
                tc[i]=tc[i]%10;
            }
        }
        for(i=l1+l2;i>=0;i--)
          if(tc[i]!=0)
            break;
        if(i<0)
          printf("0");
        for(;i>=0;i--)
          printf("%d",tc[i]);
        printf("\n");
    }
    return 0;
}