[C++]백준1041번 주사위

문제는 다음과 같다.

오늘은 백준1041번 주사위 문제를 풀어보도록 하자. 그리디 문제라기보다는 그냥 구현문제였다. 주사위가 탁자 위에 올라와있는 것을 생각하며 하나하나 구하면 된다.

면이 3개, 2개, 1개 보이는 것의 갯수를 각각 찾으면 된다.

                                     [코드]
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	long long unsigned N, three_picks = 987654321, two_picks = 987654321, one_pick = 987654321;
	long long unsigned dice[6];
	long long unsigned answer;

	cin >> N;

	for (int i = 0; i < 6; i++) cin >> dice[i];

	three_picks = min(three_picks, dice[0] + dice[1] + dice[2]);
	three_picks = min(three_picks, dice[0] + dice[1] + dice[3]);
	three_picks = min(three_picks, dice[0] + dice[2] + dice[4]);
	three_picks = min(three_picks, dice[0] + dice[3] + dice[4]);

	three_picks = min(three_picks, dice[1] + dice[2] + dice[5]);
	three_picks = min(three_picks, dice[1] + dice[3] + dice[5]);
	three_picks = min(three_picks, dice[2] + dice[4] + dice[5]);
	three_picks = min(three_picks, dice[3] + dice[4] + dice[5]);


	two_picks = min(two_picks, dice[0] + dice[1]);
	two_picks = min(two_picks, dice[0] + dice[2]);
	two_picks = min(two_picks, dice[0] + dice[3]);
	two_picks = min(two_picks, dice[0] + dice[4]);

	two_picks = min(two_picks, dice[5] + dice[1]);
	two_picks = min(two_picks, dice[5] + dice[2]);
	two_picks = min(two_picks, dice[5] + dice[3]);
	two_picks = min(two_picks, dice[5] + dice[4]);

	two_picks = min(two_picks, dice[1] + dice[2]);
	two_picks = min(two_picks, dice[1] + dice[3]);
	two_picks = min(two_picks, dice[2] + dice[4]);
	two_picks = min(two_picks, dice[3] + dice[4]);

	for (int i = 0; i < 6; i++) one_pick = min(one_pick, dice[i]);

	if (N == 1) {
		sort(dice, dice + 6);
		answer = dice[0] + dice[1] + dice[2] + dice[3] + dice[4];
		cout << answer << '\n';
	}
	else if (N == 2) {
		answer = three_picks * 4 + two_picks * 4;
		cout << answer << '\n';
	}
	else {
		answer = three_picks * 4 + two_picks * 4 * (N - 1) + two_picks * (N - 2) * 4 + one_pick * (N - 2) * (N - 2) * 5 + one_pick * (N - 2) * 4;
		cout << answer << '\n';
	}
	return 0;
}