codeforces 509 B题 Painting Pebbles

转载地址：http://blog.csdn.net/nike0good/article/details/43449739

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 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 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

贪心，每种颜色尽可能放

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,k;
int a[MAXN];
int f[MAXN][MAXN];
int main()
{
//	freopen("Painting.in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n>>k;
	For(i,n) {cin>>a[i];f[i][0]=a[i];}

	int mi=a[1],ma=a[1];
	For(i,n) ma=max(ma,a[i]),mi=min(mi,a[i]);

	if (ma-mi>k)
	{
		cout<<"NO"<<endl;
		return 0;
	}

	cout<<"YES"<<endl;
	For(i,n)
	{
		printf("1");
		Fork(j,2,a[i])
		{
			if (j<=mi) printf(" 1");
			else printf(" %d",j-mi);
		}
		printf("\n");
	}

	return 0;
}