Codeforces Round #289 (Div. 2, ACM ICPC Rules) B. Painting Pebbles

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#codeforces

本文探讨了在给定数量的鹅卵石堆和颜色种类下，如何将每堆鹅卵石染色使得任意两堆间同一颜色的鹅卵石数之差不超过一。文章通过示例展示了算法的具体实现，并提供了完整的C++代码。

There are n piles of pebbles on the table, the i-th pile contains a_i pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that b_i, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |b_i, c - b_j, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then b_i, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain a_i space-separated integers. j-th (1 ≤ j ≤ a_i) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Sample test(s)

Input

4 4
1 2 3 4

Output

Input

5 2
3 2 4 1 3

Output

NO

Input

5 4
3 2 4 3 5

Output

贪心, 水题

/*************************************************************************
    > File Name: cf289b.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月01日 星期日 13时00分21秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
const int N = 120;

int a[N];
int b[N][N];

int main ()
{
	int n, k;
	while (~scanf("%d%d", &n, &k))
	{
		int minc = inf, maxc = -inf;
		memset (b, 0, sizeof(b));
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &a[i]);
			minc = min(minc, a[i]);
		}
		for (int i = 1; i <= n; ++i)
		{
			if (a[i] == minc)
			{
				a[i] -= minc;
				b[i][1] = minc;
			}
			else
			{
				a[i] -= (minc + 1);
				b[i][1] = minc + 1;
			}
			maxc = max (a[i], maxc);
		}
		bool flag = true;
		if (maxc > (k - 1))
		{
			printf("NO\n");
			continue;
		}
		printf("YES\n");
		for (int i = 1;i <= n; ++i)
		{
			int t = 2;
			for (int j = 1; j <= a[i]; ++j)
			{
				b[i][t++] = 1;
			}
		}
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= k; ++j)
			{
				if (b[i][j])
				{
					for (int x = 1; x <= b[i][j]; ++x)
					{
						printf("%d ", j);
					}
				}
			}
			printf("\n");
		}
	}
	return 0;
}