CF 509B Painting Pebbles

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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

（摘自网上）

题目意思：有 n 个piles，第 i 个 piles有 ai 个pebbles，用 k 种颜色去填充所有存在的pebbles，使得任意两个piles，用颜色c填充的pebbles数量之差 <= 1。如果不填充某种颜色，就默认数量为0。

　　这样说还是比较难理解吧～～～以第三组数据为例：

5 43 2 4 3 5

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

第2个 pile 和第5个 pile 的比较结果如下：

cha[1] = 1 （第2个pile用颜色1填充的数量是1，第5个pile为2）

　 cha[2] = 1 （第2个pile用颜色2填充的数量是0，第5个pile为1）

　 cha[3] = 0; cha[4] = 1

要想差尽可能少，就要使得颜色尽量分散。就是说，假如有个 pile 的 pebble 数目为5，可选颜色有4种，这样填就是尽量分散： 1 2 3 4 1。每个pile都这样处理，然后比较两两pile以颜色c填充的数量是否大于1，这样用vector排序比较第一个和最后一个元素之差即可。最后就输出结果了。

总结：这题就是一个构造题，我们能用下面代码的方法构造出符合条件的情况，如果大于M，则不存在，应为只要有一个不是1的数字重复，那么就不成立。

#include<iostream>#include<algorithm>#include<cstdio>using namespace std;int a[200];int main(){int n,m;cin >> n >> m;int maxn = -100000;int minn = 100000;for(int i=0;i<n;i++){cin >> a[i];maxn = max(a[i],maxn);minn = min(a[i],minn);}if(maxn-minn>m){cout<<"NO"<<endl;}else{cout<<"YES"<<endl;for(int i=0;i<n;i++){for(int j=0;j<minn;j++)printf(j==0?"1":" 1");for(int j=minn;j<a[i];j++){printf(" %d",j-minn+1);}printf("\n");}}return 0;}

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