10106 - Product

题目：10106 - Product

题目大意：高精度数乘法

注意：每次输入的那两个结构体内的数组要初始化为零，数组不能开太小；

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
const int N = 16255;
	

class begin {

public:
	int len , s[N];
	begin() {

		len = 0;
		memset(s, 0, sizeof (s));
	}
	begin operator = (const char * s1);
	begin operator += (const begin s1);
	void  multiply (const begin & n);
} b[2], sum;

	

begin begin::operator += (const begin s1) {
		
	int i , a = 0;
	int l = (len > s1.len) ? len: s1.len;
	for (i = 0; a || i < l; i++) {

			a = a + s1.s[i] + s[i] ;
			s[i] = a % 10;
			a = a / 10;
	}
	len = i;
		return *this;
}

void  begin::multiply (const begin& n) {
	
	memset(sum.s,0,sizeof(sum.s));
	sum.len = 0;
	begin c;
	c.len = 0;
	int i , j , k,  a = 0;

	for(i = 0; i < len; i++) {
			
			for ( k = i, j = 0;  a || j < n.len; j++, k++) {

				a = a + s[i] * n.s[j];
				c.s[k] = a % 10;
				a = a / 10;
			}
			c.len = j + i;
			sum += c;
			memset(c.s, 0 , sizeof(c.s)) ;
		}

}

begin begin ::operator = (const char * s1) {

		len = strlen(s1);
		for (int i = 0 ; i < len ; i++) {
			
			s[i] = s1[len - 1 - i] - '0';

		}
		return *this;

}
int main () {

	char a[2][N];
	while ( scanf("%s", &a[0]) != EOF && scanf("%s", &a[1]) != EOF) {
			
			memset(b[0].s, 0, sizeof(b[0].s));
			memset(b[1].s, 0, sizeof(b[1].s));
			if(strcmp(a[0],"0") == 0 || strcmp(a[1],"0") == 0)
			{
				printf("0\n");
				continue;
			}
			b[0] = a[0];
			b[1] = a[1];
		
			if(b[0].len < b[1].len)
				b[0].multiply(b[1]);
			else
				b[1].multiply(b[0]);
			
			for (int i = sum.len - 1; i >= 0; i--) 
				printf("%d", sum.s[i]);
			printf("\n");
			
			
	}
	return 0;

}