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

B. Painting Pebbles

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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 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

水贪心，，只要求出最多的颜色和最少的颜色之差，，然后再与k相比较，大于k就NO，否则YES，，然后输出就顺序输出就OK啦（可循环输出）

AC代码：

#include <cstdio> 
#include <cstring>
#include <algorithm>
using namespace std;

int a[105];

int main() {
	int n, k;
	while(scanf("%d %d", &n, &k) != EOF) {
		int max = 0, min = 101;
		for(int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
			if(max < a[i]) max = a[i];
			if(min > a[i]) min = a[i];
		}
		int cen = max - min;
		if(cen > k) printf("NO\n");
		else {
			printf("YES\n");
			for(int i = 0; i < n; i++) {
				int c = 1;
				for(int j = 0; j < a[i] - 1; j++) {
					printf("%d ", c++);
					c %= (k + 1);
					if(!c) c++;
				}
				printf("%d\n", c++);
			}
		}
	}
	return 0;
}