A. Dividing Orange

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 ton·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 a_{i (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 a_{i 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 k space-separated integers a_{1, 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 a_{i 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 secondn 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.

Sample test(s)input

2 24 1

output

2 4 1 3 

input

3 12

output

3 2 1 

解題說明：此題考慮到n k的值不是很大，為了判斷哪個值已經被選中，可以考慮用數組儲存每個位置的資訊，如果被選中則標記為1。最後輸出時只需要判斷位置是否為0，連續找n-1個為0的位置，加上指定的位置即可

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int main(){int n,k,i,j;int a[31];int pos[901];int count;scanf("%d %d",&n,&k);memset(pos,0,sizeof(pos));for(i=0;i<k;i++){scanf("%d",&a[i]);pos[a[i]]=1;}j=1;for(i=0;i<k;i++){count=1;while(1){if(pos[j]==0){pos[j]=1;count++;printf("%d ",j);j++;}if(pos[j]==1){j++;}if(count==n){printf("%d\n",a[i]);break;}}}return 0;}