1 second
256 megabytes
standard input
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.
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.
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.
4 4 1 2 3 4
YES 1 1 4 1 2 4 1 2 3 4
5 2 3 2 4 1 3
NO
5 4 3 2 4 3 5
YES 1 2 3 1 3 1 2 3 4 1 3 4 1 1 2 3 4
贪心, 水题
/*************************************************************************
> File Name: cf289b.cpp
> Author: ALex
> Mail: zchao1995@gmail.com
> Created Time: 2015年02月01日 星期日 13时00分21秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
const int N = 120;
int a[N];
int b[N][N];
int main ()
{
int n, k;
while (~scanf("%d%d", &n, &k))
{
int minc = inf, maxc = -inf;
memset (b, 0, sizeof(b));
for (int i = 1; i <= n; ++i)
{
scanf("%d", &a[i]);
minc = min(minc, a[i]);
}
for (int i = 1; i <= n; ++i)
{
if (a[i] == minc)
{
a[i] -= minc;
b[i][1] = minc;
}
else
{
a[i] -= (minc + 1);
b[i][1] = minc + 1;
}
maxc = max (a[i], maxc);
}
bool flag = true;
if (maxc > (k - 1))
{
printf("NO\n");
continue;
}
printf("YES\n");
for (int i = 1;i <= n; ++i)
{
int t = 2;
for (int j = 1; j <= a[i]; ++j)
{
b[i][t++] = 1;
}
}
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= k; ++j)
{
if (b[i][j])
{
for (int x = 1; x <= b[i][j]; ++x)
{
printf("%d ", j);
}
}
}
printf("\n");
}
}
return 0;
}