计算都没啥问题

#include<iostream>
#include<ctype.h>
using namespace std;
int a[3000],b[3000],c[3000];
void reverse(int *num,int len) {
	for(int i = 0; i < len/2; i++) {
		int temp = num[i];
		num[i] = num[len - 1 - i];
		num[len - 1 - i] = temp;
	}
	return;
}
int main()
{
	int temp,len1 = 0,len2 = 0;
	char ch;
	while(isdigit(ch = getchar())){
		a[len1++] = ch - '0';
	}
	while(isdigit(ch = getchar())){
		b[len2++] = ch - '0';
	}
	reverse(a,len1),reverse(b,len2);
	for(int i = 0;i < len2;i++){
		int carry = 0;//保存低位的进位 
		for(int j = 0;j < len1;j++){
			temp = a[j] * b[i] + carry;//使用temp，保证每一位为个位数 
			carry = temp / 10;
			a[j] = temp % 10;
		}//乘完之后，j现在在目前最高位的下一位，通过carry来处理进位 
		while(carry){
			a[len1++] = carry % 10;
			carry /= 10;
		}
	}
	for(int i = len1 - 1;i >= 0;i--) cout << a[i];
	return 0;
}