最大公约数——414AMashmokh and Numbers

C. Mashmokh and Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.

In this game Mashmokh writes sequence of n distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers x and y from the board, he gets gcd(x, y) points. At the beginning of the game Bimokh has zero points.

Mashmokh wants to win in the game. For this reason he wants his boss to get exactly k points in total. But the guy doesn't know how choose the initial sequence in the right way.

Please, help him. Find n distinct integers a1, a2, ..., an such that his boss will score exactly k points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.

Input

The first line of input contains two space-separated integers n, k (1 ≤ n ≤ 105; 0 ≤ k ≤ 108).

Output

If such sequence doesn't exist output -1 otherwise output n distinct space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Examples

input

5 2

output

1 2 3 4 5

input

5 3

output

2 4 3 7 1

input

7 2

output

-1

Note

gcd(x, y) is greatest common divisor of x and y.

题意：输出一个长度为n序列，然后对这个序列进行操作，每次取前两个数，并从序列中去掉，求它们的最大公约数，当余下的不足两个时停止，将这些最大公约数累加起来，要等于k。如果这个序列不存在就输出-1。

题解： 长度为n的序列，那么我们可以取n/2次，当K<n/2时肯定是输出-1 的，因为每对数公约数最下为1。当k==n/2时，只需要输出n个连续的数字就可以了，因为连续的两个数字的公约数为1，当k>n/2时，我们先输出最大公约数为k-n/2+的两个数，余下的同样为连续的数字。

 1 /*C*/
 2 #include<cstdio>
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     int n,k;
 8     while(scanf("%d%d",&n,&k)!=EOF)
 9     {
10         int t=n/2;
11         if(t>k||(k!=0&&t==0))
12         {
13             printf("-1\n");
14             continue;
15         }
16         else if(t==k)
17         {
18             for(int i=1;i<=n;i++)
19                 printf("%d ",i);
20             printf("\n");
21         }
22         else
23         {
24             int temp=k-t+1;
25             printf("%d %d ",temp,temp*2);
26             int d=2;
27             for(int i=temp*2+1;;i++)
28             {
29                 if(d==n)    break;
30                 printf("%d ",i);
31                 d++;
32                 
33             }
34             printf("\n");
35         }
36     }
37     return 0;
38  } 

 

转载于:https://www.cnblogs.com/yepiaoling/p/5316979.html