B. Painting Pebbles
time limit per test
memory limit per test
input
output
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
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
Input
n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
n positive integers a1, a2, ..., an (1 ≤ ai) denoting number of pebbles in each of the piles.
Output
NO" (without quotes) .
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
代码：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
int main()
{
int n,m;
int a[110];
while(scanf("%d%d",&n,&m)!=EOF)
{
int minn = 999999;
int maxx = -1;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(minn > a[i])
{
minn = a[i];
}
if(maxx < a[i])
{
maxx = a[i];
}
}
int num3 = (maxx - minn);
if(num3>m)
{
printf("NO\n");
continue;
}
printf("YES\n");
for(int i=0;i<n;i++)
{
for(int j=0;j<a[i];j++)
{
if(j == 0)
{
printf("%d",(j%m)+1);
}
else
{
printf(" %d",(j%m)+1);
}
}
printf("\n");
}
}
return 0;
}
