codeforces-660【思维】

题目链接：​​点击打开链接​​

A. Co-prime Array

time limit per test

memory limit per test

input

output

n

109

An array is co-prime if any two adjacent numbers of it are co-prime.

a and b are said to be co-prime if the only positive integer that divides both of them is 1.

Input

n (1 ≤ n ≤ 1000) — the number of elements in the given array.

n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

Output

k on the first line — the least number of elements needed to add to the array a

n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by addingk

If there are multiple answers you can print any one of them.

Example

input

3 2 7 28

output

1 2 7 9 28

题解：就是如果两个相邻的数不互质，添加数让他们互质。那直接添加 1 不就行了，我个ZZ

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int a[1010];
int GCD(int a,int b)
{
  if(a%b==0)  return b;
  return GCD(b,a%b);
}
bool judge(int x,int y)
{
  if(GCD(x,y)==1)  return 0;
  return 1;
}
int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=0;i<n;i++)
      scanf("%d",&a[i]);
    int cnt=0;
    for(int i=0;i<n-1;i++)
    {
      if(judge(a[i],a[i+1]))
        cnt++;
    }
    printf("%d\n",cnt);
    for(int i=0;i<n-1;i++)
    {
      printf(i==0?"%d":" %d",a[i]);
      if(judge(a[i],a[i+1]))
      {
        printf(" 1");
      }
    }
    printf(" %d",a[n-1]);
  }
  return 0;
}

题目链接：

​​点击打开链接​​

B. Seating On Bus

time limit per test

memory limit per test

input

output

2n rows of the seats in a bus. n rows of the seats on the left and n rows of the seats on the right. Each row can be filled by two people. So the total capacity of the bus is 4n.

m (m ≤ 4n) people occupy the seats in the bus. The passengers entering the bus are numbered from 1 to m

1-st row left window seat, 1-st row right window seat, 2-nd row left window seat, 2-nd row right window seat, ... , n-th row left window seat,n-th row right window seat.

m > 2n) the non-window seats are occupied:

1-st row left non-window seat, 1-st row right non-window seat, ... , n-th row left non-window seat, n-th row right non-window seat.

All the passengers go to a single final destination. In the final destination, the passengers get off in the given order.

1-st row left non-window seat, 1-st row left window seat, 1-st row right non-window seat, 1-st row right window seat, ... , n-th row left non-window seat, n-th row left window seat, n-th row right non-window seat, n-th row right window seat.

The seating for n = 9 and m = 36.

n and m. Output m numbers from 1 to m, the order in which the passengers will get off the bus.

Input

n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 4n) — the number of pairs of rows and the number of passengers.

Output

m distinct integers from 1 to m

Examples

input

2 7

output

5 1 6 2 7 3 4

input

9 36

output

19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18

题意：就是按题意输出下车顺序；

脑残代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;
int main()
{
  while(~scanf("%d %d",&n,&m))
  {
    bool flag=0;
    for(int i=1;i<=n;i++)
    {
      if(2*n+1+2*(i-1)<=m)
      {
        if(flag)  putchar(' ');
        printf("%d",2*n+1+2*(i-1));
        flag=1;
      }
      if(2*i-1<=m)
      {
        if(flag)  putchar(' ');
        printf("%d",2*i-1);
        flag=1;
      }
      if(2*n+2+2*(i-1)<=m)
      {
        if(flag)  putchar(' ');
        printf("%d",2*n+2+2*(i-1));
        flag=1;
      }
      if(2*i<=m)
      {
        if(flag)  putchar(' ');
        printf("%d",2*i);
        flag=1;
      }
    }
    puts("");
  }
  return 0;
}

栈模拟：

#include<cstdio>  
#include<cstring>  
#include<stack>  
using namespace std;  
int main()  
{  
    int n,m;  
    while(~scanf("%d%d",&n,&m))  
    {  
        stack<int> seat[205];  
        int cnt=0;  
        for(int i=0;i<m;++i)  
        {  
            seat[i%(2*n)].push(++cnt);  
        }  
        printf("%d",seat[0].top());  
        seat[0].pop();  
        for(int i=0;i<2*n;++i)  
        {  
            while(!seat[i].empty())  
            {  
                printf(" %d",seat[i].top());  
                seat[i].pop();  
            }  
        }  
        printf("\n");  
    }  
    return 0;  
}