codeforces #479D# Long Jumps(二分upper_bound)

D. Long Jumps

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!

However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has nmarks,
with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end.
The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from
the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an,
where ai denotes
the distance of the i-th mark from the origin (a1 = 0, an = l).

Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n),
such that the distance between the i-th and the j-th
mark is exactly equal to d (in other words, aj - ai = d).

Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y)
centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.

Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y.
Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.

Input

The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l)
— the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.

The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l),
where ai shows
the distance from the i-th mark to the origin.

Output

In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.

In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l).
Number pi means
that the i-th mark should be at the distance of pi centimeters
from the origin. Print the marks in any order. If there are multiple solutions, print any of them.

Sample test(s)

input

3 250 185 230
0 185 250

output

1
230

input

4 250 185 230
0 20 185 250

output

0

input

2 300 185 2300 300

output

2
185 230

Note

In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter
mark or a 230 centimeter mark.

In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters,
so you don't have to add new marks.

In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.

解题思路：这题意思就是给一把尺子上面给定一些刻度，要你从这把尺子上取出x,y两个长度，问你至少要添加几个点才能够实现。

毫无疑问最多是两个点，我开了三个数组，所以内存比较大了些。我是讲 ai +x 和ai-x存到数组里面，然后到ai数组中看看能不能找到对应的点，有就不表示能取到，y也是同理的。这题比较不容易想到一个点就是可能取到一个点，刚好让原来不能测量x,y的尺子能够测量两个数据了。这个要小心了。

/********************************/
/*Problem:  codeforces #483B#   */
/*User: 	    shinelin        */
/*Memory: 	    3400 KB         */
/*Time: 	    62 ms           */
/*Language: 	GNU C++         */
/********************************/
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <string>
#include <list>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

#define INF 0x7fffffff
#define LL long long
int n, l, x, y, d;
vector<int> xNeed, yNeed;
vector<int> point;
vector<int> ans;
bool getx = false, gety = false;
int cnt = 2;

bool if_Find(int d, vector<int>& v)
{
int k = upper_bound(v.begin(), v.end(), d) - v.begin();
if(k > 0 && v[k-1] == d)
{
return true;
}
return false;
}
int main()
{
scanf("%d%d%d%d", &n, &l, &x, &y);
for(int i = 0; i < n; i ++)
{
scanf("%d", &d);
point.push_back(d);
if(d-x >= 0) xNeed.push_back(d-x);
if(d-y >= 0) yNeed.push_back(d-y);
if(d+x <= l) xNeed.push_back(d+x);
if(d+y <= l) yNeed.push_back(d+y);
}
sort(xNeed.begin(), xNeed.end());
sort(yNeed.begin(), yNeed.end());
sort(point.begin(), point.end());
for(int i = 0; i < n; i ++)
{
if(!getx && if_Find(point[i], xNeed))
{
getx = true;
cnt --;
}
if(!gety && if_Find(point[i], yNeed))
{
gety = true;
cnt --;
}
}
if(cnt == 2)
{
for(int i = 0; i < xNeed.size(); i ++)
{
if(if_Find(xNeed[i], yNeed))
{
printf("1\n%d\n", xNeed[i]);
return 0;
}
}
}
printf("%d\n", cnt);
if(cnt == 0) return 0;
if(!getx) ans.push_back(x);
if(!gety) ans.push_back(y);
int k = 0;
while(k < ans.size() - 1)
{
printf("%d ", ans[k++]);
}
printf("%d\n", ans[k]);

return 0;
}