백준 6318번 Box of Bricks // C++

문제

 

Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. “Look, I’ve built a wall!”, he tells his older sister Alice. “Nah, you should make all stacks the same height. Then you would have a real wall.”, she retorts. After a little consideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one, such that all stacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimum number of bricks moved. Can you help?

입력

The input consists of several data sets. Each set begins with a line containing the number n of stacks Bob has built. The next line contains n numbers, the heights hi of the n stacks. You may assume 1 ≤ n ≤ 50 and 1 ≤ hi ≤ 100.

The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange the bricks such that all stacks have the same height.

The input is terminated by a set starting with n 0. This set should not be processed.

출력

For each set, first print the number of the set, as shown in the sample output. Then print the line “The minimum number of moves is k.”, where k is the minimum number of bricks that have to be moved in order to make all the stacks the same height.

Output a blank line after each set.

 

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풀이

 

블럭의 개수와, 각 블럭의 높이가 주어진다.

 

이 블럭들의 높이를 최대한 같게 만들려고 할 때, 최소 블럭의 움직임을 출력하는 문제이다.

 

처음에는 오름차순으로 블럭들을 정렬한 뒤, 제일 높은 블럭에서 제일 낮은 블럭으로 하나씩 옮겨주면서 횟수를 세려고 했다.

 

그런데 그렇게 하면 시간 초과가 발생할 것이 뻔했기 때문에 다른 방법을 생각해냈다.

 

어차피 높이를 최대한 같게 만들게 되면 블럭들의 높이의 합의 평균이 될 것이고, 평균과 차이가 나는 블럭의 개수만 세주면 실제로 옮기지 않아도 정답을 구할 수 있다.

 

주의할 점은, 각 세트마다 한 줄씩 띄워서 출력을 해야한다.

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코드

#include <iostream>
#include <vector>

using namespace std;

int main() {
	int n;
	int t = 0;

	while (true) {
		cin >> n;
		t++;

		if (n == 0)
			break;

		vector<int> v;
		int sum = 0;

		for (int i = 0; i < n; i++) {
			int num;
			cin >> num;

			sum += num;
			v.push_back(num);
		}

		int avg = sum / n;
		int count = 0;

		for (int i = 0; i < n; i++) {
			if (v[i] < avg)
				count += avg - v[i];
		}

		printf("Set #%d\n", t);
		printf("The minimum number of moves is %d.\n\n", count);
	}
}