CodeForces 509B Painting Pebbles

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 41 2 3 4

output

YES11 41 2 41 2 3 4

input

5 23 2 4 1 3

output

NO

input

5 43 2 4 3 5

output

YES1 2 31 31 2 3 41 3 41 1 2 3 4

大致题意是给定颜色的种类，问是否能用这些颜色涂石子

如果是，输出满足条件的一种涂色方案

使任意两堆石子中涂有任意颜色的石子数的差值至多为一个

有一点很容易发现：当这些石堆中有两堆的石子数差值的绝对值大于给的颜色数时，这样的方案显然是不存在的

                                而对于其它情况，方案似乎是存在的（其实我还没想出严格的证明方法o(╯□╰)o）

那么假定对于其它情况这样的方案是存在的

该怎么输出涂色方案捏

为了让答案更清晰

我们有序地对石子进行涂色

对于任一堆石子，从下至上，我们都从颜色1开始涂，一种颜色涂一块石子

例如总共有4块石子，那么从下到上的颜色分别为1，2，3，4

似不似很简单~(≧▽≦)/~

那么问题来了，如果我们现在只有3种或更少的颜色怎么办o(╯□╰)o

很容易想到再从颜色1开始继续按顺序涂

例如总共有5块石子，而颜色总数只有3种，那么从下到上的颜色分别为1，2，3，1，2

由于一堆石块中各石块的涂色顺序与题目是无关的

因此，当我们按照这样的方法涂色时，给出的必定是符合题意的方案（咦，前面刚说没想出证明方法来着怎么就证出来了╮(╯▽╰)╭）

代码如下：

#include<iostream>using namespace std;const int maxn = 100+5;int main() {  int n, k, min_num = maxn, max_num = -maxn;  cin >> n >> k;  int a[n];  for (int i = 0; i < n; i++) {    cin >> a[i];    if (a[i] > max_num) max_num = a[i];    if (a[i] < min_num) min_num = a[i];  }  if (max_num - min_num > k) { cout << "NO" << endl; return 0; }  cout << "YES" << endl;    for (int i = 0; i < n; i++) {    int first = 1;    for (int j = 0; j < a[i]; j++) {      if (first) first = 0;      else cout << " ";      cout << j%k+1;    }    cout << endl;  }  return 0;}