大数乘法（C++）

POJ2398大数乘法实现

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本文提供了一个解决POJ2398问题的C++代码示例，该问题要求不使用特殊库函数手动实现两个大整数的乘法运算，并输出结果。代码中采用字符串逆序处理，逐位相乘并进位的方法来计算乘积。

题目：POJ 2398

Bull Math

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13410		Accepted: 6903

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;

const int maxn = 100;
void reverse(char a[])
{
    int len = strlen(a);
    for(int i = 0 ; i < len / 2; i++)
    {
        int temp = a[i];
        a[i] = a[len - i - 1];
        a[len - i -1] = temp;
    }
}
int main()
{
    char a[maxn],b[maxn];
    int t[100] = {0};
    //printf("Please enter 2 numbers: ");
    scanf("%s%s",a,b);
    reverse(a);
    reverse(b);
    if(strcmp(a,"0")==0||strcmp(b,"0")==0)
        cout<<"0"<<endl;
    else
    {
        int i,j;
        for(i = 0; i <strlen(b); i++)
        {
            int cnt = 0;
            for(j = 0; j < strlen(a); j++)
            {
                int temp = (b[i] - '0') * (a[j] - '0');
                int tt= t[i+j] + temp + cnt;
                t[j+i] = tt % 10;
                cnt = tt / 10;
            }
            while(cnt != 0)
            {
                t[j+i] = cnt % 10;
                cnt = cnt / 10;
                j++;
            }
        }
        for(int k = i + j - 2; k >= 0; k--)
        {
            cout<<t[k];
        }
    }
    return 0;
}