大数乘法C++

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;
}