CodeForces 509B Painting Pebbles

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本文探讨了给定数量石子堆与颜色种类时，如何为每堆石子分配颜色，确保任意两堆中相同颜色石子的数量之差不超过1。通过分析最大堆与最小堆之间的关系，提供了一种简便的解决方案。

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Painting Pebbles

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 509B

Description

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 Input

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

题意：有n堆石子，每堆有a[i] 个石子，有k 种颜色，给每堆石子涂色，要求任意两堆石子中相同的颜色的差 <= 1; 找到满足条件的涂色方法后，输出其中的任一种；

解题思路：先看一下不满足的条件，可以用最大的一堆和最小的一堆来看一下，它们满足的话，其余的肯定都满足。我们可以用商和余数的方法，

maxn = m1 * k + r1;

minn = m2 * k + r2;

分为三种情况：1：当 m1 == m2 时，可以找到 2：当m1 - m2 == 1 时， { (1) r1 - r2 > 0 找不到 3：当 m1 - m2 > 1 时，找不到

{ (2) 其余：可以找到

然后：可以找到的就输出 YES ，再对于每一堆石子循环输出所给的 k种颜色，找不到就直接输出 NO

代码附上：：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
using namespace std;
int a[120];

int main(){
	int n, k, i;
	while(~scanf("%d%d", &n, &k)){
		int minn = 999, maxn = 0;
		for(int i = 0; i < n; ++i){
			scanf("%d", &a[i]);
			minn = min(minn, a[i]);
			maxn = max(maxn, a[i]);
		}
		int m1, m2, r1, r2;
		m1 = maxn / k;
		r1 = maxn % k;
		m2 = minn / k;
		r2 = minn % k;
		if(m1 - m2 > 1){
			cout << "NO" << endl;
		}
		else if (m1 - m2 == 1 && r1 - r2 >= 1){
			cout << "NO" << endl;
		}
		else {
			cout << "YES" << endl;
			for(int i = 0; i < n; ++i){
				for(int j = 0; j < a[i]; ++j){
					cout << j % k + 1 << " " ;
				}
				cout << endl;
			}
		}
	}
	return 0;
}