zoukankan      html  css  js  c++  java
  • Candy Bags

    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 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("
    ");
        }
        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("
    ");
            }
        }
    }


    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
    ",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("
    ");
                }
            }
    
        return 0;
    }
  • 相关阅读:
    TeXLive安装过程
    js 如何获取class的元素 以及创建方法getElementsByClassName
    点击返回
    MVC框架实现文件的上传(支持多文件上传)
    团购倒计时抢购功能
    获取取当前页 地址分割字符串
    网页内容截取部分打印
    HttpFileCollection 实现多文件上传
    JS如何获取URL
    正则表达式 占位符 替换
  • 原文地址:https://www.cnblogs.com/lipching/p/3904061.html
Copyright © 2011-2022 走看看