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一个文件中有40亿个整数，每个整数为四个字节，内存为1GB，写出一个算法：求出这个文件里的整数里不包含的一个整数
下面的代码片段仅仅是一个样例。
4个字节的整数最大可表示为2^32=4294967296, 一个数一个数的读入内存，建立一个bit map，共需要4294967296个bits(也就是0.5G字节的内存，并没有超过1G内存的限制)，读入每一个数，置相应的bit为1。

 1     int N = 20; // # of number
 2     int M = 1000;   // number range
 3     std::vector<int> a(N);  // can be imported from external file number by number
 4     for (int i = 0; i < N; i++)
 5         a[i] = (int)rand()%M;
 6     std::copy(a.begin(), a.end(), std::ostream_iterator<int>(std::cout, " "));
 7     std::cout << "\n";
 8     // bit map setup for existence of each number
 9     unsigned int nbytes = M%8 ? (M/8+1) : (M/8);
10     std::cout << "nbytes = " << nbytes << "\n";
11 
12     char* p = new char [nbytes];
13     memset(p, 0, sizeof(char)*nbytes);
14 
15     for (int i = 0; i < N; i++) {
16         unsigned int index = a[i]/8;
17         unsigned int bitpos = a[i]%8;
18         char* tmp = p+index;
19         *tmp |= 1 << bitpos;
20         //std::cout << "bit pos set to 1 : " << 8*index+bitpos << "\n";
21     }
22     for (int i = nbytes-1; i >= 0; i--) {
23         printf("%02X ", (char)*(p+i)&0xFF);
24     }
25     std::cout << "\n";
26     delete [] p;
27