题目：
One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 to n·k.
There were k children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the i-th (1 ≤ i ≤ k) child wrote the number ai (1 ≤ ai ≤ n·k). All numbers ai accidentally turned out to be different.
Now the children wonder, how to divide the orange so as to meet these conditions:
- each child gets exactly n orange segments;
- the i-th child gets the segment with number ai for sure;
- no segment goes to two children simultaneously.
Help the children, divide the orange and fulfill the requirements, described above.
Input
The first line contains two integers n, k (1 ≤ n, k ≤ 30). The second line contains kspace-separated integers a1, a2, ..., ak (1 ≤ ai ≤ n·k), where ai is the number of the orange segment that the i-th child would like to get.
It is guaranteed that all numbers ai are distinct.
Output
Print exactly n·k distinct integers. The first n integers represent the indexes of the segments the first child will get, the second n integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.
You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.
Examples
Input
2 2 4 1
Output
2 4 1 3
Input
3 1 2
Output
3 2 1
解题报告：存在一个橘子，可以分成n*k瓣，有k个孩子，每个人都有了自己一定要有的瓣数（标记）的，问每个的分配情况。就是在一定的区间里面，成功的将所有的瓣数分配出去，利用mark标记就可以实现。
ac代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1005;
int num[35];
int flag[maxn];
int main()
{
int n,k;
scanf("%d%d",&n,&k);
int l=0;
memset(num,0,sizeof(num));
memset(flag,0,sizeof(flag));
for(int i=0;i<k;i++)
{
scanf("%d",&num[i]);
flag[num[i]-1]=1;
}
for(int i=0;i<k;i++)
{
printf("%d ",num[i]);
for(int j=0;j<n-1;)
{
if(!flag[l])
{
printf("%d ",1+l);
j++;
flag[l]=1;
}
l++;
}
printf("\n");
}
return 0;
}