POJ 2389 Bull Math 高精度（大数）乘法模板-优快云博客

本文提供了一个解决POJ2389 BullMath问题的具体方法，该问题要求进行大数乘法运算并输出结果，避免使用特殊库函数。文章中包含了一个C语言实现的大数乘法算法，采用逐位相乘和进位处理的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Bull Math

Time Limit:1000MS		Memory Limit:65536K

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros).

FJ asks that you do this yourself; don't use a special library function for the multiplication.

Input

* Lines 1..2: Each line contains a single decimal number.

Output

* Line 1: The exact product of the two input lines

Sample Input

11111111111111
1111111111

Sample Output

12345679011110987654321

/* Author : yan
 * Question : POJ 2389 Bull Math
 * Date && Time : Sunday, January 30 2011 12:54 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
#define MAX 45
char mul1[MAX],mul2[MAX];
char ans[2*MAX];

void multiply(char* a,char* b,char* c)
{
    int i,j,ca,cb,* s;
    ca=strlen(a);
    cb=strlen(b);
    s=(int*)malloc(sizeof(int)*(ca+cb));
    for (i=0;i<ca+cb;i++)
        s[i]=0;
    for (i=0;i<ca;i++)
        for (j=0;j<cb;j++)
            s[i+j+1]+=(a[i]-'0')*(b[j]-'0');//i+j+1的目的就是为了防止最高位进位而产生错误
    for (i=ca+cb-1;i>=0;i--)
        if (s[i]>=10)
        {
            s[i-1]+=s[i]/10;
            s[i]%=10;
        }
    i=0;
    while (s[i]==0)
        i++;//去除前导0
       for (j=0;i<ca+cb;i++,j++)
           c[j]=s[i]+'0';
    c[j]='/0';//将结果存储到字符数组
    free(s);
}
int main()
{
	//freopen("input","r",stdin);
	scanf("%s",mul1);
	scanf("%s",mul2);
	multiply(mul1,mul2,ans);
	printf("%s",ans);
	return 0;
}