Codeforces Round#194(Div.2) A Candy Bag

最新推荐文章于 2025-04-01 13:53:25 发布

lin375691011

最新推荐文章于 2025-04-01 13:53:25 发布

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本文介绍了一个解决兄弟分糖果问题的算法，通过合理分配糖果数量，确保每个兄弟获得相同数量的糖果。该算法适用于偶数个兄弟的情况，并在特定输入下提供输出实例，演示了解决方案的有效性。

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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 kcandies.

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.

Sample test(s)

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.

这个题想不进去的觉得难，想进去了就好水啊。。

直接贴图，看了图你就明白了：

六个人的时候：

八个人的时候：

所以说我就不多解释了，直接上代码：

#include <stdio.h>
int main()
{
    int n,a[10005],i;
    for(i=1;i<=10000;i++)
    {
        a[i]=i;
    }
    while(scanf("%d",&n)!=EOF)
    {
        int m=n*n,x,falt=0,sum,j;
        for(i=1;i<=n;i++)
        {
            x=i,falt=0,sum=0;
            for(j=1;j<=i;j++)
            {
                if(falt==0)
                {
                    printf("%d",x);
                    falt=1;
                    sum+=x;
                }
                else
                {
                    printf(" %d",x);
                    sum+=x;
                }
                x+=n-1;
            }
            x=n*(i+1);
            for(j=i+1;j<=n;j++)
            {
                if(x>n*n)
                {
                    break;
                }
                printf(" %d",x);
                sum+=x;
                x+=n-1;
            }
            printf(" \n",sum);
        }
    }
    return 0;
}