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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/lipching/p/3904061.html
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