Problem description
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
解题思路:这道题看辣么久才看懂,真的是怀疑自己的智商QAQ。输出有k行,每行必须包含一个儿童想要的那块橙子块编号(该行的任意位置输出都可以),再输出n-1个橙子块的编号,编号不能有重复输出,[1,n*k]中每个编号只输出1次即可。
AC代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 int main(){ 4 int n,k,cnt=1,a[1000],b[35]; 5 cin>>n>>k; 6 for(int i=1;i<=n*k;++i)a[i]=i; 7 for(int i=1;i<=k;++i){cin>>b[i];a[b[i]]=0;} 8 for(int i=1;i<=k;++i){ 9 cout<<b[i];//每一行先输出b[i] 10 int t=1;//t为计数器,一行输出n个数 11 while(t<n){ 12 if(a[cnt]){cout<<' '<<a[cnt++];t++;} 13 else cnt++; 14 } 15 cout<<endl; 16 } 17 return 0; 18 }