To Be NUMBER ONE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 395    Accepted Submission(s): 196
Special Judge

One is an interesting integer. This is also an interesting problem. You are assigned with a simple task.
Find N (3 <= N <= 18) different positive integers Ai (1 <= i <= N), and

Any possible answer will be accepted.

2 3 6
2 4 6 12

算法思想 ：

对分母进行分解：

n可以分解成     1/n  =  1/(n+1)   +  1/(n+1)*n  

例如   从2 3 6 开始

不能有相同的 所以只能从 2分解成 3 6 前面已经含有了，然后只能从3开始改

下一项变成 2 4 6 12

下一项不能分解 2因为会多出来6 所以应当选择 2 5 6 12 20

post code:

#include<stdio.h>
#include<iostream>
#include<algorithm>

using namespace std;

int a[21][21];
int b[21];
int main()
{
    a[3][1]=2;
    a[3][2]=3;
    a[3][3]=6;
    int i,j,k,len,temp,flag=0;
    for( i=3; i<=17; i++ )
      {
          len=i;
          
          for( j=1; j<=len; j++)
          {
            b[j] = a[i][j];                   
          }
          for( j=1; j<=len; j++)
          {
             flag=0;
             if(b[j]+1!=b[j+1]){     temp=b[j]*(b[j]+1);
                                     for(k=1;k<=len;k++)    
                                        {
                                            if(temp==b[k]){flag=1;break;}
                                        }
                                     if(flag==1)continue;
                                     b[j]=b[j]+1;
                                     b[len+1]=temp;
                                     break;
                               }
          }
          sort(b+1,b+len+1+1);
          for( j=1; j<=len+1; j++)
          {
               
              a[i+1][j]=b[j];     
          }
                           
      }     
    for( i=3; i<=18; i++ )
     {
     for( j=1; j<=i; j++ )
       {
         printf("%d",a[i][j]);
         if(j!=i)printf(" ");
       }
        printf("\n");
     }
    
}