題干：

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

題目大意：

一個橘子分成n*k塊（對應編號1~n*k），k個人分，每個人都有一塊最想得到的編號，現在要你每個人都分到n塊，應該怎么分。

解題報告：

特殊塊特殊給，剩下的平均分。直接用整除，，算是個小技巧了吧、、反正數據量也小，，怎么搞都行。。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; int n,k; int a[MAX]; int opa[MAX]; int main() {cin>>n>>k;memset(opa,-1,sizeof opa);for(int i = 1; i<=k; i++) {scanf("%d",a+i);opa[a[i]] = i;}if(n == 1) {for(int i = 1; i<=k; i++) {printf("%d\n",a[i]);}return 0 ;}int cur = 0;for(int i = 1; i<=n*k; i++) {if(opa[i] != -1) continue;opa[i] = cur/(n-1)+1;cur++;}int times = 0;while(1) {times++;if(times == k+1) break;for(int i = 1; i<=n*k; i++) {if(opa[i] == times) {printf("%d ",i);}}printf("\n");}return 0 ;}

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