【CodeForces】 652B
2017-10-09 22:34
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A student of z-school found a kind of sorting called z-sort.
The array a with n elements
are z-sorted if two conditions hold:
ai ≥ ai - 1 for
all even i,
ai ≤ ai - 1 for
all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted
while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000)
— the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109)
— the elements of the array a.
Output
If it's possible to make the array a z-sorted
print n space separated integers ai —
the elements after z-sort. Otherwise print the only word "Impossible".
Examples
input
output
input
output
题意:偶数位置上的数的大小必须大于或等于它左右位置上的数;
题解:可以先从小到大排序,然后分成前后两部分,按顺序一小一大输出(注意个数的奇偶,分别讨论)
代码:
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int n,a[1010],b[1010],c[1010];
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
for(int i=1;i<=n/2;i++)
b[i]=a[i];
int k=1;
for(int i=n/2+1;i<=n;i++)
c[k++]=a[i];
if(n%2==0) //总个数为偶数
for(int i=1;i<=n/2;i++)
{
printf("%d %d ",b[i],c[i]);
}
if(n%2!=0) //总个数为奇数
{
int j=1;
for(int i=n;i>=n/2+1;i--)
c[j++]=a[i];
for(int i=1;i<=n/2;i++)
{
printf("%d %d ",b[i],c[i]);
}
printf("%d\n",c[(n+1)/2]);
}
return 0;
}
The array a with n elements
are z-sorted if two conditions hold:
ai ≥ ai - 1 for
all even i,
ai ≤ ai - 1 for
all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted
while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000)
— the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109)
— the elements of the array a.
Output
If it's possible to make the array a z-sorted
print n space separated integers ai —
the elements after z-sort. Otherwise print the only word "Impossible".
Examples
input
4 1 2 2 1
output
1 2 1 2
input
5 1 3 2 2 5
output
1 5 2 3 2
题意:偶数位置上的数的大小必须大于或等于它左右位置上的数;
题解:可以先从小到大排序,然后分成前后两部分,按顺序一小一大输出(注意个数的奇偶,分别讨论)
代码:
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int n,a[1010],b[1010],c[1010];
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
for(int i=1;i<=n/2;i++)
b[i]=a[i];
int k=1;
for(int i=n/2+1;i<=n;i++)
c[k++]=a[i];
if(n%2==0) //总个数为偶数
for(int i=1;i<=n/2;i++)
{
printf("%d %d ",b[i],c[i]);
}
if(n%2!=0) //总个数为奇数
{
int j=1;
for(int i=n;i>=n/2+1;i--)
c[j++]=a[i];
for(int i=1;i<=n/2;i++)
{
printf("%d %d ",b[i],c[i]);
}
printf("%d\n",c[(n+1)/2]);
}
return 0;
}
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