Timus 1796. Amusement Park 聪明题

On a sunny Sunday, a group of children headed by their teacher came to an amusement park. Aunt Frosya, who was a very kind and quiet person, worked at the ticket window on that day. The teacher gave
her the money but didn't say how many tickets she wanted to buy. Could Aunt Frosya determine it knowing only the numbers of different notes the teacher gave? It is assumed that the teacher didn't give extra notes, which means that there would not be enough
money for the tickets if any of the notes was taken away.

Input

The first line contains six nonnegative integers separated with a space; these are the numbers of 10, 50, 100, 500, 1000, and 5000 rouble notes the teacher gave to Aunt Frosya. In the second line you
are given the price of one ticket; it is a positive integer. All the integers in the input data do not exceed 1000.

Output

Find the number of tickets the teacher wanted to buy. Output the number of possible answers in the first line. The variants in ascending order separated with a space must be given in the second line.
It is guaranteed that there is at least one variant of the answer.

Samples

input	output
0 2 0 0 0 0 10	5 6 7 8 9 10
1 2 0 0 0 0 10	1 11

这是一道考人是否聪明的题目，没有现成的算法。

所以须要模拟人计算的过程。用计算机的程序思维去思考。

过程这种：

1 先算出总钱数能购买多少张票

2 总钱数减去一张最小面值的钱，然后模票价，然后加上最小面值的钱，在减去一张票价。最后就得到灵活度的钱

3 灵活度的钱除以票价，就得到灵活度了，灵活度的钱除以票价得到零。那么就仅仅有一种可能了，得到1就有两种可能

难以理解的话，就细心想想人是怎样计算的就能够攻克了。

#include <iostream>
using namespace std;

static const int AmusePartRoubles[6] = {10, 50, 100, 500, 1000, 5000};

void AmusementPark1796()
{
	int A[6] = {0};
	int money = 0;
	for (int i = 0; i < 6; i++)
	{
		cin>>A[i];
		money += A[i] * AmusePartRoubles[i];
	}
	int ticket = 0;
	cin>>ticket;
	int total = money / ticket;

	int i = 0;
	for ( ; i < 6 && A[i] == 0; i++);

	int leftMoney = (money - AmusePartRoubles[i]) % ticket;
	leftMoney += AmusePartRoubles[i] - ticket;
	int flex = leftMoney / ticket;

	cout<<flex+1<<endl;
	for (int j = flex; j >= 0 ; j--)
	{
		cout<<total - j<<' ';
	}
}

int main()
{
	AmusementPark1796();
	return 0;
}