CodeForces244A Dividing Orange（STL+思维）
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 k space-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
- 题目大意：
有n*k瓣橘子，编号分别是1–n * k，有k个孩子，每个孩子都有一个想要的橘子瓣的编号，然后 要求每个人的要求都要满足。每个人都要分得n瓣橘子，让你输出一种分配的方式。 - 解题思路：
每个人都有一个想要的，就把这些个被人预定的橘子都给放进 set 里面。然后从第一个人开始，先把他想要的那个给他安排上，然后在从第一个开始，往后给他安排上n-1 个 ，每次判断一下，这个是不是在set里，在俩面就跳过去，然后下一个人继续往下安排。 - 代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
using namespace std;
set<int>s;
int a[100];
int main()
{
int n,k;
cin>>n>>k;
for(int i=1;i<=k;i++)
{
cin>>a[i];
s.insert(a[i]);
}
int cnt=1;
for(int i=1;i<=k;i++)
{
cout<<a[i];
int num=1;
for(int j=cnt;j<=n*k;j++)
{
if(s.count(j)==0)
{
cout<<" "<<j;
num++;
}
if(num==n)
{
cout<<endl;
cnt=j+1;
break;
}
}
}
return 0;
}