Gerald has n younger brothers and their number happens to be even. One day he bought n2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer k from 1 to n2 he has exactly one bag with k candies.
Help him give n bags of candies to each brother so that all brothers got the same number of candies.
Input
The single line contains a single integer n (n is even, 2 ≤ n ≤ 100) — the number of Gerald's brothers.
Output
Let's assume that Gerald indexes his brothers with numbers from 1 to n. You need to print n lines, on the i-th line print n integers — the numbers of candies in the bags for the i-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to n2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Examples
Input
2
Output
1 4 2 3
Note
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
分析：题目要求给出一个偶数n，然后有n^2个袋子，里面的糖果数量依次从一到n^2;让分成n伙，每一伙的糖果数量总和相等。看了别人的思路，用二维数组的相对位置来做挺简单的。
比如输入4，这个时候糖果的分布情况，我们可以用二维数组表示
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
在每一伙里面按相对位置里面取一个，例如取1就不能取5,9,13,然后取了6之后10,14就不能取了.依此规律便找到了这n份。这样每一伙里面的总和肯定相等。
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#include<iostream>
-
using
namespace
std;
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int main()
-
{
-
int n,i,j,flag=
0;
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int a[
102][
102];
-
while(
cin>>n)
-
{
-
for(i=
0;i<n;i++)
-
for(j=
1;j<=n;j++)
-
a[i][j
-1]=i*n+j;
//从1开始到n^2赋值给二维数组
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for(i=
0;i<n;i++)
-
{
-
flag=
0;
-
for(j=
0;j<n;j++)
-
{
-
if(flag==
0)
-
{
-
flag=
1;
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cout<<a[j][(j+i)%n];
//按照相对位置的不同输出
-
}
-
else
-
cout<<
" "<<a[j][(j+i)%n];
-
}
-
cout<<
endl;
-
}
-
}
-
return
0;
-
}
Gerald has n younger brothers and their number happens to be even. One day he bought n2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer k from 1 to n2 he has exactly one bag with k candies.