H - H
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Description
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 Input
Input
3 250 185 2300 185 250
Output
1230
Input
4 250 185 2300 20 185 250
Output
0
Input
2 300 185 2300 300
Output
2185 230
Hint
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.
题意是说有一把尺子，本身有一些刻度，然后需要测量x和y，问最少需要添加多少个刻度，如果需要，这些刻度分别添加在什么位置。
一开始没有看清题目，以为答案最多为4，但是发现，a[1]=0 a[n]=l这两个是确定的，所以答案最多为2，
而不会出现中间有两个刻度，无论是往前刻还是往后都会越界的情况。
先去看看已知的刻度里有没有直接满足的。
除此之外，如果在已知的刻度下无法测量x也无法测量y，我们还可以找找是否能有公用点使得只刻一个刻度就可以满足题意。公用点分两种情况，一种是在两个刻度之间，另一种是在两个刻度的同侧。
前一种没什么坑，后一种要判断是否越界！！！
如果在已知的刻度下能测量x或者能测量出y但不能同时测量的话，就没有必要找公用点。
还有就是查找的话如果线性找复杂度是O(n2)，会TLE在第27个点。
我们用二分...
话说STL里竟然连二分查找也有。。。。简直orz。。。。真是省时省力。
代码：
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int n,l,x,y;
const int N=1E5+7;
int a[N];
int ans,p,q;
bool flag1,flag2,flag3,flag4;
int main()
{
scanf("%d %d %d %d",&n,&l,&x,&y);
flag1 = false;
flag2 = false;
flag3 = false;
flag4 = false;
for ( int i = 1 ; i <= n ; i++ )
scanf("%d",&a[i]);
ans = 2;
p = -1;
q = 0;
for ( int i = 1 ; i <= n ; i++ )
{
if(binary_search(a,a+n+1,a[i]+x))
{
flag1 = true;
}
if (binary_search(a,a+n+1,a[i]+y))
{
flag2 = true;
}
if (binary_search(a,a+n+1,a[i]+x+y))
{
flag3 = true;
p = a[i];
}
if (binary_search(a,a+n+1,a[i]+y-x)&&((a[i]+y<=l)||(a[i]-x>=0)))
{
flag4 = true;
p = a[i];
}
}
if ( flag1) {ans--;q = 1 ;}
if ( flag2 ){ ans--;q = 2 ;}
if ( ans==2&&(flag3||flag4))
{
ans--;
}
cout<<ans<<endl;
if ( ans==2 )
{
cout<<a[1]+x<<" "<<a[1]+y<<endl;
}
if ( ans==1 )
{
if ( p==-1 )
{
if ( q==1 )
cout<<a[1]+y<<endl;
else cout<<a[1]+x<<endl;
}
else
{
if ( p+y<=l )
cout<<p+y<<endl;
else cout<<p-x<<endl;
}
}
return 0;
}