To Be NUMBER ONE + 数论

To Be NUMBER ONE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 408    Accepted Submission(s): 207
Special Judge

Problem Description

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.

 

Input

No input file.

 

Output

Your program’s output should contain 16 lines:
The first line contain 3 integers which are the answer for N = 3, separated by a single blank.
The following 15 lines are the answer for N = 4 to 18, as the same format.
The sample output is the first two lines of a possible output.

 

Sample Output

2 3 62 4 6 12
 
/*思路：一个倒数可以分解问2个数的倒数之和1/n= 1/(a*b)=1/a*(a+b) + 1/b*(a+b)*/#include<iostream>#include<cmath>using namespace std ;int main(void){int a[100] ,cnt = 0 ;a[0] = 2 ;a[1] = 3 ;a[2] = 6 ;int k = 3 ;while(cnt < 16){int i ;for( i = 0 ; i < cnt + 2 ; i ++)cout<<a[i]<<" ";cout<<a[cnt+2]<<endl;int j ,flag = 0 ;for( i = 0 ; i < cnt + 3 ; i ++){for( j = 2 ; j <= a[i] ; j ++){if(a[i] % j == 0 ){int b , c , d , tag = 0 , tag1 = 0 ;d = a[i] / j ;b = j*( j + d) ;c = d * ( j + d ) ;if(!(b<=(k+1)*(k+1) && c <=(k+1)*(k+1)))continue ;for(int l = 0 ; l < k ; l++)if(a[l] == b){tag = 1;break;}for( l = 0 ; l < k ; l ++)if(a[l] == j){tag1 = 1;break;}if(tag == 0 || tag == 0){a[i] = b ;a[k ++] =c ;flag = 1;break;}}}if(flag)break;}cnt ++ ;}return 0;}