E - Dividing Orange

标签：else   each   必须   using   nts   paper   present   cout   AC   

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 a_i (1?≤?a_i?≤?n·k). All numbers a_i 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 kspace-separated integers a₁,?a₂,?...,?a_k (1?≤?a_i?≤?n·k), where a_i 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 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

2 2
4 1

2 4 
1 3 

3 1
2

3 2 1 
解题思路：这道题看辣么就才看懂，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个橙子块的编号即可，编号不能有重复输出，即[1,n*k]中每个编号只输出1次。
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 }

E - Dividing Orange

标签：else   each   必须   using   nts   paper   present   cout   AC   

原文地址：https://www.cnblogs.com/acgoto/p/9157969.html