Coloring

题目链接

题目描述：

Cirno_9baka has a paper tape with $n$ cells in a row on it. As he thinks that the blank paper tape is too dull, he wants to paint these cells with $m$ kinds of colors. For some aesthetic reasons, he thinks that the $i$-th color must be used exactly $a_i$ times, and for every $k$ consecutive cells, their colors have to be distinct.

Help Cirno_9baka to figure out if there is such a way to paint the cells.

输入描述：

The first line contains a single integer $t(1\le t\le 10000)$ — the number of test cases. The description of test cases follows.

The first line of each test case contains three integers $n$, $m$, $k(1\le k\le n\le {10}^9,1\le m\le {10}^5,m\le n)$. Here $n$ denotes the number of cells, $m$ denotes the number of colors, and $k$ means that for every $k$ consecutive cells, their colors have to be distinct.

The second line of each test case contains $m$ integers $a_1,a_2,\cdots ,a_m(1\le a_i\le n)$ — the numbers of times that colors have to be used. It's guaranteed that $a_1+a_2+\dots +a_m=n$.

It is guaranteed that the sum of $m$ over all test cases does not exceed ${10}^5$.

输出描述：

For each test case, output "YES" if there is at least one possible coloring scheme; otherwise, output "NO".

You may print each letter in any case (for example, "YES", "Yes", "yes", and "yEs" will all be recognized as positive answers).

样例：

input:

2
12 6 2
1 1 1 1 1 7
12 6 2
2 2 2 2 2 2

output:

NO
YES

Note:

In the first test case, there is no way to color the cells satisfying all the conditions.

In the second test case, we can color the cells as follows: $(1,2,1,2,3,4,3,4,5,6,5,6)$. For any $2$ consecutive cells, their colors are distinct.

AC代码：

#include <bits/stdc++.h>
 
using namespace std;
 
void solve()
{
	int n, m, k;
	scanf("%d%d%d", &n, &m, &k);
 
	// x 代表将 n 分为 k 段， y 代表最后不足 k 的长度
	int x = n / k, y = n % k;
 
	bool f = 1;
 
	while (m--)
	{
		int a;
		scanf("%d", &a);
 
		// 颜色需要涂的次数如果小于 x 就代表可以把这个颜色涂在 a 个段里
		if (a <= x)
			continue;
		// 如果大于x就代表需要用到最后那一段
		// 如果最后一段有剩余并且只会占用一格的话就说明可以涂完
		// 否则则要占用大于一格或者最后一段已经被涂满，不满足没有相同颜色
		if (a == x + 1 && y)
			y--;
		else
			f = 0;
	}
 
	if (f)
		cout << "YES\n";
	else
		cout << "NO\n";
}
 
int main()
{
	int T;
	scanf("%d", &T);
 
	while (T--)
		solve();
 
	return 0;
}