Codeforces 406B Toy Sum【思维】

B. Toy Sum

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris.

There are exactly s blocks in Chris's set, each block has a unique number from 1 to
s. Chris's teacher picks a subset of blocks
X and keeps it to himself. He will give them back only if Chris can pick such a non-empty subset
Y from the remaining blocks, that the equality holds:



"Are you kidding me?", asks Chris.
For example, consider a case where s = 8 and Chris's teacher took the blocks with numbers 1, 4 and 5. One way for Chris to choose a set is to pick the blocks with numbers 3 and 6, see figure. Then the required sums
would be equal: (1 - 1) + (4 - 1) + (5 - 1) = (8 - 3) + (8 - 6) = 7.



However, now Chris has exactly s = 106 blocks. Given the set
X of blocks his teacher chooses, help Chris to find the required set
Y!

Input
The first line of input contains a single integer n (1 ≤ n ≤ 5·105), the number of blocks in the set
X. The next line contains
n distinct space-separated integers x1,
x2,
..., xn (1 ≤ xi ≤ 106), the numbers of the blocks in
X.

Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

Output
In the first line of output print a single integer m (1 ≤ m ≤ 106 - n), the number of blocks in the set
Y. In the next line output
m distinct space-separated integers y1,
y2,
..., ym (1 ≤ yi ≤ 106), such that the required equality holds.
The sets X and Y should not intersect, i.e.
xi ≠ yj for all
i, j (1 ≤ i ≤ n;
1 ≤ j ≤ m). It is guaranteed that at least one solution always exists. If there are multiple solutions, output any of them.

Examples

Input

3
1 4 5

Output

2
999993 1000000

Input

1
1

Output

1
1000000

题目大意：

给你集合X,让你找一个集合Y，使得集合Y没有集合X中的数，并且集合X值和集合Y的值相同，输出任意解即可。

集合X和集合Y的计算公式按照题干中所示公式来计算。

思路：

1、观察完样例输入输出，我们不难想到对称数，对应集合X中出现一个数a，那么集合Y中对应就可以有一个数1000000-a+1来和这个数对称，并且能够保证这两个数在两个集合中贡献的价值相同。

2、那么如果a和1000000-a+1都出现在集合X中了怎么办呢？不难发现，如果两个数都出现在了X集合中，那么其在X集合中贡献的值为1000000.那么我们其实可以直接找同样对称的一组数b，和1000000-b+1使其直接出现在集合Y中，那么就要保证b和1000000-b+1都没有出现过在集合X中，我们枚举出来这个解即可。

为了节省枚举的时间，我们可以将输入进来的集合X从小到大排序，那么我们如果需要枚举b的时候，从pre枚举即可。pre初始化为1，每次枚举完pre更新为b.优化时间。

Ac代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[500050];
int b[500050];
int vis[1000006];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
int tmp2=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
vis[a[i]]++;
tmp2+=a[i]-1;
}
int pre=1;
int cont=0;
sort(a+1,a+n);
for(int i=1;i<=n;i++)
{
if(vis[a[i]]==-1)continue;
if(vis[1000000-a[i]+1]==0)
{
b[cont++]=1000000-a[i]+1;
}
else
{
vis[1000000-a[i]+1]=-1;
for(int j=pre;j<=1000000;j++)
{
if(vis[j]==0&&vis[1000000-j+1]==0)
{
vis[j]=1;
b[cont++]=j;
b[cont++]=1000000-j+1;
pre=j;
break;
}
}
}
}
int tmp=0;
printf("%d\n",cont);
for(int i=0;i<cont;i++)
{
printf("%d ",b[i]);
tmp+=b[i]-1;
}
printf("\n");
}
}
/*
5
1 1000000 2 500000 500001
*/