标签：building room apartments windows Apartments Number three five CF1430A

Recently a new building with a new layout was constructed in Monocarp’s hometown. According to this new layout, the building consists of three types of apartments: three-room, five-room, and seven-room apartments. It’s also known that each room of each apartment has exactly one window. In other words, a three-room apartment has three windows, a five-room — five windows, and a seven-room — seven windows.

Monocarp went around the building and counted n windows. Now he is wondering, how many apartments of each type the building may have.

Unfortunately, Monocarp only recently has learned to count, so he is asking you to help him to calculate the possible quantities of three-room, five-room, and seven-room apartments in the building that has n windows. If there are multiple answers, you can print any of them.

Here are some examples:

if Monocarp has counted 30 windows, there could have been 2 three-room apartments, 2 five-room apartments and 2 seven-room apartments, since 2⋅3+2⋅5+2⋅7=30;
if Monocarp has counted 67 windows, there could have been 7 three-room apartments, 5 five-room apartments and 3 seven-room apartments, since 7⋅3+5⋅5+3⋅7=67;
if Monocarp has counted 4 windows, he should have mistaken since no building with the aforementioned layout can have 4 windows.
Input
Th first line contains one integer t (1≤t≤1000) — the number of test cases.

The only line of each test case contains one integer n (1≤n≤1000) — the number of windows in the building.

Output
For each test case, if a building with the new layout and the given number of windows just can’t exist, print −1.

Otherwise, print three non-negative integers — the possible number of three-room, five-room, and seven-room apartments. If there are multiple answers, print any of them.

Example
input
4
30
67
4
14
output
2 2 2
7 5 3
-1
0 0 2

#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
typedef long long ll;

int main()
{
	int t; cin >> t;
	while (t--)
	{
		int n; cin >> n;

		if (n == 1 || n == 2 || n == 4)
		{
			cout << -1 << endl;
			continue;
		}

		if (n % 3 == 0) cout << n / 3 << ' ' << 0 << ' ' << 0 << endl;
		else if (n % 3 == 1) cout << (n - 7) / 3 << ' ' << 0 << ' ' << 1 << endl;
		else cout << (n - 5) / 3 << ' ' << 1 << ' ' << 0 << endl;
	}


	return 0;
}

结论： 3 ， 5 ， 7 3，5，7 3，5，7, 可以组成任何数，除了 1 ， 2 ， 4 1 ，2， 4 1，2，4 。

标签：building,room,apartments,windows,Apartments,Number,three,five,CF1430A
来源： https://blog.csdn.net/j_x323u/article/details/120401338

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。