CodeForces 509B Painting Pebbles

最新推荐文章于 2022-07-22 13:33:59 发布

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本文探讨了如何在给定数量的鹅卵石堆中，使用指定颜色进行着色，使得任何两个堆中同一颜色的鹅卵石数量差不超过一。通过输入堆的数量和颜色种类，以及每堆的鹅卵石数量，本文提供了满足条件的着色方案，或者指出不存在这样的方案。详细阐述了解决问题的方法和示例。

There are n piles of pebbles on the table, the i-th pile contains ai 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 iand 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 bi, 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 |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, 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, ..., an (1 ≤ ai ≤ 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 ai space-separated integers. j-th (1 ≤ j ≤ ai) 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
YES
1
1 4
1 2 4
1 2 3 4

input
5 2
3 2 4 1 3

output
NO

input
5 4
3 2 4 3 5

output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4

题目大意：有n行和k个数字，已知每行数字个数，每行数字由1-k的k个数字组成，可重复，要求任意两列相同数字的出现次数的差值不大于1，若满足条件输出“YES”以及n行数字，不满足输出“NO”。

解题思路：不满足情况的条件很简单，任意两行的差值大于k就不满足。然后每行按1 2 3...k 1 2..的顺序输出即可。

代码如下：

#include <cstdio>
int a[105];
int main()
{
    int n,m,i,j,max=0,min=105;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
		if(max<a[i])
			max=a[i];
		if(min>a[i])
			min=a[i];
    }
	if(max-min>m)
	{
        printf("NO\n");
        return 0;
    }	
	printf("YES\n");
    for(i=0;i<n;i++)
    {
        for(j=1;j<=a[i];j++)
		{
			if(j!=1)
				printf(" ");
			if(j%m)
				printf("%d",j%m);
			else
				printf("%d",m);
		};
        printf("\n");
    }
    return 0;
}