POJ 2389 Bull Math(大数乘法,还是Java好)

Bull Math

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14252		Accepted: 7350

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

Source

USACO 2004 November

原题链接:http://poj.org/problem?id=2389

大数乘法,数据只有一组!

AC代码:

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			BigInteger a = sc.nextBigInteger();
			BigInteger b = sc.nextBigInteger();
			System.out.println(a.multiply(b));
		}
	}

}

AC代码2:

#include <stdio.h>
#include <string.h>
#define MAX 220
int main()
{
    char s1[MAX],s2[MAX];
    int a1[MAX],a2[MAX],p[2*MAX];;
    int i,j,len_1,len_2;
    memset(a1,0,sizeof(a1));
    memset(a2,0,sizeof(a2));
    memset(p,0,sizeof(p));
    gets(s1);
    gets(s2);
    len_1=strlen(s1);
    len_2=strlen(s2);
    for (j=0,i=len_1-1;i>=0;i--)
    {
        a1[j++]=s1[i]-'0';
    }

    for (j=0,i=len_2-1;i>=0;i--)
    {
        a2[j++]=s2[i]-'0';
    }

    for (i=0;i<len_1;i++)
    {
        for (j=0;j<len_2;j++)
        {
            p[i+j]+=a1[i]*a2[j];
        }
    }
    for (i=0;i<MAX*2;i++)
    {
        if(p[i]>9)
        {
            p[i+1]+=p[i]/10;
            p[i]%=10;
        }
    }

    int start=0;
    for (i=MAX*2-1;i>=0;i--)
    {
        if(start)
        {
            printf("%d",p[i]);
        }
        else if(p[i])
        {
            printf("%d",p[i]);
            start=1;
        }
    }
    if(!start)
        printf("0");

    return 0;
}