i 服了 u。居然用3 个字节的 int，太bt了。
可是服务器如果是 big endian 的你怎么办？

> #include <iostream>
> #include <string>
> #include <vector>
> #include <algorithm>
> #include <functional>
> using namespace std;
> 
> const int lut[] =
> {
> 	2, 2, 2,
> 	3, 3, 3,
> 	4, 4, 4,
> 	5, 5, 5,
> 	6, 6, 6,
> 	7, 0, 7, 7,
> 	8, 8, 8,
> 	9, 9, 9, 0,
> };
> 
> int main()
> {
> 	int count;
> 	cin >> count;
> 	vector<string> vec_num;
> 	vec_num.reserve(count);
> 	string str, num_str;
> 	char ch;
> 	char* bucket = new char[10000001*3];
> 	memset(bucket, 0, 10000001 * 3 * sizeof(char));
> 
> 	while ( count-- && cin >> str )
> 	{
> 		int i = 0;
> 		num_str = "";
> 		while ( (ch=str[i++])!='\0')
> 		{
> 			if ( ch == '-')
> 				continue;
> 			else if ( ch > '9')
> 				num_str.append(1, lut[ch-'A'] +'0');
> 			else
> 				num_str.append(1, ch);
> 		}
> 		i = atoi(num_str.c_str()) * 3;
> 		(*((int*)&(bucket[i])))++;
> 	}
> 
> 	bool bNoDuplicate = true;
> 	char* p = bucket;
> 	char des[20];
> 	for ( int i = 0; i < 10000000; i++, p+=3 )
> 	{
> 		if ((count = (*((int*)p) & 0x00FFFFFF)) > 1)
> 		{
> 			bNoDuplicate = false;
> 			sprintf(des, "%07d %d", i, count);
> 			str = des;
> 			str.insert(3, "-");
> 			cout << str << endl;
> 		}
> 	}
> 
> 	if ( bNoDuplicate )
> 		cout << "No duplicates." << endl;
> 
> 	delete[] bucket;
> 	return 0;
> }