Usaco contact, puzzle, operator overloading, QuickSort,

I feel like the hardest question so far
reference to the vanishing Person's code: Http://www.nocow.cn/index.php/Code:USACO/contact/C%2B%2B
Reference ideas: http:// www.nocow.cn/index.php/USACO/contact#. E7. ae.80.e6.98.93.e4.bd.86.e9.9d.9e.e5.b8.b8.e9.ab.98.e6.95.88.e7.9a.84.e6.96.b9.e6.b3.95
A simple but very efficient way The Vanishing Person

/* Id:wangxin12 prog:contact lang:c++/#include <iostream> #include <vector> #include <fstream>
#include <string> #include <cstring> using namespace std;
#define MAX 999999999 int A, B, N;
int frequency[12][4100];
int transform, length;

int const_num[13] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096};

Class Unit {Public:int frequency, length, num;};
Vector<unit> Unit_pool;

String original_string;
	void Create_unit (int length, int num) {unit * u = new unit ();
	U->frequency = Frequency[length-1][num];
	u->length = length;
	U->num = num;
Unit_pool.push_back (*u);
		BOOL Operator> (unit U1, unit U2) {if (U1.frequency < u2.frequency) | |
		(u1.frequency = = u2.frequency && u1.length > u2.length) | |
		(u1.frequency = u2.frequency && u1.length = u2.length && u1.num > U2.num))
	return true;
else return false; } template<typename t> int partition (vector<t> &f, int start, int end) {T key = F[start];
	int left = start + 1, right = end;
		while (true) {while (left < right && F[right] > key) right--;
		while (left < right && key > F[left]) left++;
		if (left = right) break;
		T tmp = F[left];
		F[left] = F[right];
	F[right] = tmp;
	} if (F[left] > key) return start;
	F[start] = F[left];
	F[left] = key;
return to left;
	} template<typename t> void Quick_sort (vector<t> &f, int start, int end) {if (start >=) return;
	int boundary = partition (f, start, end);
	Quick_sort (f, start, boundary-1);
Quick_sort (F, boundary + 1, end);
	string transform_to_binary (int num, int length) {string S1 = "", S2 = "";
	int t = num;
	char tmp;
		while (t!= 0) {S1 + = (t% 2 + ' 0 ');
	T/= 2;
	for (int j = s1.length ()-1; J >= 0; j--) s2 + = S1[j];
	S1 = "";
	for (int i = 1; I <= length-s2.length (); i++) S1 + = ' 0 ';
	S1 = s1 + s2;
return s1;
int main () {ifstream fin ("contact.in");	Ofstream fout ("Contact.out");
	fin>>a>>b>>n;
	Original_string want a char to read the char, otherwise the char ch is stopped by reading the newline character;
	while (Fin.get (CH)) {if (ch >= ' 0 ' && ch <= ' 1 ') original_string + = ch; //for (length = A; (length <= B) && (length <= original_string.size ());
			length++) {for (int start = 0; start <= original_string.length ()-length; start++) {transform = 0; for (int j = start; J <= start + length-1 j + +)//for Transform chronicle string size if (original_string[j] = = ' 1 ') transfor
			M + + Const_num[start + length-j-1];
		frequency[length-1][transform]++; for (int i = A; I <= B. i++) {for (int j = 0; J < Const_num[i]; J +) {if (frequency[i-1][j)!= 0) C
		Reate_unit (i, j);

	} quick_sort (Unit_pool, 0, (int) unit_pool.size ()-1);
	int last_frequency = 99999999, sum = 0, count = 0;
			for (int k = 0; k < unit_pool.size (); k++) {if (Unit_pool[k].frequency < last_frequency) {sum++; if (sum = =N + 1) break;
			if (k!= 0) fout<<endl;
			Count = 1;
			fout<<unit_pool[k].frequency<<endl;
			Fout<<transform_to_binary (Unit_pool[k].num, unit_pool[k].length);
		Last_frequency = unit_pool[k].frequency;
			else {count++;
				if (count = = 7) {fout<<endl;
			Count = 1;
			else fout<< "";
		Fout<<transform_to_binary (unit_pool[k].num,unit_pool[k].length);

	}} fout<<endl;
	Fin.close ();

	Fout.close ();
return 0;
 }