• One is an interesting integer. This is also an interesting problem. You are assigned with a simple task. Find N (3 ) different positive integers Ai (1 ), and Any possible answer will be accepted. ... ai as lines
• 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 On
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): 207Special JudgeProblem 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 6
2 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 ,cnt = 0 ;
a = 2 ;
a = 3 ;
a = 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;
}


展开全文 • 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 Problem Description


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): 196Special 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 6
2 4 6 12

Author

iSea@WHU

Source

The
6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Preliminary

Recommend

lcy

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;
int b;
int main()
{
a=2;
a=3;
a=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");
}

}


展开全文  output input 算法
• One is an interesting integer. This is also an interesting problem. You are assigned with a simple task. Find N (3 ) different positive integers Ai (1 ), and Any possible answer will ...
这是一道典型的数学题，只要套公式即可。
1/n=1/(n+1)+1/(1+n)*(n)
2分解成3和6，3分解成4和12，4分解成5和20。。。。注意每次从最小的分解。
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 6 Accepted Submission(s) : 4
Special Judge

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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 6
2 4 6 12

代码：
#include<iostream>
using namespace std;
int f;
int main()
{
int i,j,flag;
memset(f,0,sizeof(f));
f=1;
f=1;
f=1;
printf("2 3 6\n");
i=4;
while(i<=18)
{
i++;
for(j=2;j<=i*i;j++)
{
if(f[j]&&!f[j+1]&&!f[(j+1)*j])
{
f[j]=0;
f[j+1]=1;
f[(j+1)*j]=1;
break;
}
}
flag=1;
for(j=2;j<=(i+1)*(i+1);j++)
{
if(f[j])
{
if(flag){ printf("%d",j);flag=0;}
else printf(" %d",j);
}
}
printf("%\n");
}
//system("pause");
return 0;
}




展开全文  output input
• To Be NUMBER ONE Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 383 Accepted Submission(s): 190 Special Judge Problem Description On
To Be NUMBER ONE
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 383    Accepted Submission(s): 190Special JudgeProblem 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 6
2 4 6 12

题意
分别用3到18个数 的倒数相加 等于1
并且每个数要求 小于等于 个数+1的平方
比如 用3个数
那么要找3个数满足 这三个数小于等于（3+1）*（3+1) 且倒数之和为1

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)

故 在已知 上层的情况下 可以直接将上一层的数分解为2个数  以此类推即可

手算  我没有写程序
#include<stdio.h>
int main()
{

printf("2 3 6\n");
printf("2 4 6 12\n");
printf("2 5 6 12 20\n");
printf("2 5 7 12 20 42\n");
printf("3 5 6 7 12 20 42\n");
printf("3 5 6 8 12 20 42 56\n");
printf("3 5 6 8 16 20 42 48 56\n");
printf("3 5 6 12 16 20 24 42 48 56\n");
printf("3 5 6 16 20 21 24 28 42 48 56\n");
printf("3 5 6 16 21 24 28 36 42 45 48 56\n");
printf("3 5 6 16 24 28 30 36 42 45 48 56 70\n");
printf("3 5 6 16 28 30 33 36 42 45 48 56 70 88\n");
printf("3 5 6 16 28 33 36 39 42 45 48 56 70 88 130\n");
printf("3 5 6 16 33 36 39 42 44 45 48 56 70 77 88 130\n");
printf("3 5 6 20 33 36 39 42 44 45 48 56 70 77 80 88 130\n");
printf("3 5 6 24 33 36 39 42 44 45 48 56 70 77 80 88 120 130\n");

return 0;
}


展开全文  output input
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• If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to ... Golang erlang r语言
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