Candy Bags

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 334A

Description

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.

Sample Input

2

1 4
2 3

 

分析：对角线问题，根据对角线进行运算。

代码：

 

#include<stdio.h>
#include<string.h>
int N;
int g[125][125];
int main()
{
    int i,j,k;
    scanf("%d",&N);
    for (i=1;i<=N;i++) g[1][i]=i;
    for (i=2;i<=N;i++)
    {
        for (j=1;j<=N;j++)
        {
            if (j==1) g[i][j]=g[i-1][N];
            else g[i][j]=g[i-1][j-1];
        }
    }
    for (i=1;i<=N;i++)
    {
        int tmp=0;
        for (j=1;j<=N;j++)
        {
            printf("%d ",g[i][j]+tmp);
            tmp+=N;
        }
        printf("\n");
    }
    return 0;
}

 

 youcuowu

 

#include<stdio.h>
#include<string.h>
int main()
{
    int n,a[10001],i,j,x,f;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
            a[i]=i;
        f=0;
        for(i=1;i<=n;i++)
        {
            x=i;
            f=0;
            for(j=1;j<=i;j++)
            {
                if(f==0)
                {
                    printf("%d",x);
                    f=1;
                }
                else
                {
                    printf(" %d",x);
                }
                x+=n-1;
            }
            x=n*(i+1);
            for(j=i+1;j<=n;j++)
            {
                if(x>n*n)
                   break;
                printf(" %d",x);
                x+=n-1;
            }
            printf("\n");
        }
    }
}

另一种方法:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

#define MAXSIZE 100100
#define eps 1e-8
#define LL __int4
#define N 110

int a[N][N];

int main()
{
    int n;
    int i,j;
    while(~scanf("%d",&n)){
            int cnt = 1;
        for(i = 0;i<n;i++)
            for( j = 0;j<n;j++){
                a[i][j] = cnt;
                cnt+=1;
            }

           //  for(i = 0;i<n;i++)
          //  for( j = 0;j<n;j++)
               // printf("%d\n",a[i][j]);
            int flag ;
            int sum = 0;
            for(j = 0;j<n;j++){
                    flag = 1;
                for( i = 0;i<n;i++){
                        if(i+j>=n)
                          sum = i+j-n;
                        else
                          sum = i+j;
                    if(flag == 1){
                       printf("%d",a[i][sum]);
                       // printf("%d %d",i,sum);
                        flag = 0;
                    }
                    else{
                      printf(" %d",a[i][sum]);
                      //  printf(" %d %d",i,sum);
                    }
                }
                printf("\n");
            }
        }

    return 0;
}