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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 Input
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
/*
把个数从少到多进行排序，用来判断前后是否可能，只要统计循环的次数以及循环到哪里了就可以轻松判断，如果循环次数差的大于1并且差的位数也大于1那么肯定是不行的，for example:
m = 2 a[1] = 2, a[2] = 5
1 2 b[1] = 1 c[1] = 0;
1 2 1 2 1 b[1] = 2 c[2] = 1
如果循环次数差的大于2也肯定不行这个显然
所以先判是否可以如果可以直接就根据循环和位输出当前值
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct edge{
int a, b, c;
}G[110];
bool cmp(edge i, edge j){
return i.a < j.a;
}
int main()
{
int n, m;
int a[110];
int b[110];
int c[110];
while(~scanf("%d%d", &n, &m)){
for(int i = 1; i <= n ; i++)
scanf("%d", &a[i]);
for(int i = 1; i <= n; i++){
int k = a[i] / m;
int t = a[i] % m ;
b[i] = k;
c[i] = t;
}
for(int i = 1; i <= n ;i++){
G[i].a = a[i];
G[i].b = b[i];
G[i].c = c[i];
}
sort(G + 1, G + n + 1, cmp);
/* for(int i = 1; i <= n ;i++){
printf("%d %d %d", G[i].a, G[i].b, G[i].c);
puts("");
}
*/
int flag = 0;
for(int i = 1; i <= n ; i++){
for(int j = i + 1; j <= n ; j++){
if(G[j].b > G[i].b && G[j].c > G[i].c || G[j].b - G[i].b >= 2 ){
flag = 1 ;
break;
}
}
if(flag == 1)
break;
}
if(flag == 1) {printf("NO\n"); }
else {
printf("YES\n");
for(int i = 1; i <= n ; i++){
for(int j = 1; j <= b[i]; j++){
for(int k = 1; k <= m; k++)
printf("%d ",k);
}
for(int j = 1; j <= c[i]; j++)
printf("%d ", j);
puts("");
}
}
}
return 0;
}
Codeforces Round #289 (Div. 2, ACM ICPC Rules)——B贪心——Painting Pebbles
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原文地址：http://www.cnblogs.com/zero-begin/p/4652404.html
