码迷,mamicode.com
首页 > 其他好文 > 详细

Candy Bags

时间:2014-08-11 11:25:02      阅读:250      评论:0      收藏:0      [点我收藏+]

标签:des   style   blog   http   color   java   os   io   

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

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 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;
}

 

Candy Bags,布布扣,bubuko.com

Candy Bags

标签:des   style   blog   http   color   java   os   io   

原文地址:http://www.cnblogs.com/lipching/p/3904061.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!