Codeforces 479D - Long Jumps

479D - Long Jumps, 480B - Long Jumps
It is easy to see that the answer is always 0, 1 or 2. If we can already measure both x and y, output 0. Then try to measure both x and y by adding one more mark. If it was not successful, print two marks: one at x, other at y.

So, how to check if the answer is 1? Consider all existing marks. Let some mark be at r. Try to add the new mark in each of the following positions: r - x, r + x, r - y, r + y. If it become possible to measure both x and y, you have found the answer. It is easy to check this: if, for example, we are trying to add the mark at r + x, we just check if there is a mark at r + x + y or r + x - y (by a binary search, since the marks are sorted). Make sure that the adde marks are in [0, L].

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define ll long long
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
using namespace std;
const int INF = 0x3f3f3f3f;

int a[100001];
int n, l;

bool Ok(int num){
    if(num > 0 && num <= l) return true;
    return false;
}

int main(){
    int i, j, k, t, x, y;
    while(EOF != scanf("%d%d%d%d",&n,&l,&x,&y)){
        for(i = 1; i <= n; ++i) scanf("%d",&a[i]);
        bool xx = false;
        bool yy = false;
        for(i = 1; i <= n; ++i){
            if(binary_search(a, a + n + 1, a[i] + x)){
                xx = true;
            }
            if(binary_search(a, a + n + 1, a[i] + y)){
                yy = true;
            }
        }
        if(xx && yy){
            printf("0\n");
            continue;
        }
        bool status = false;
        for(i = 1; i <= n; ++i){
            if(Ok(a[i] - x)){
                if(yy){
                    printf("1\n%d\n",a[i] - x);
                    status = true;
                    break;
                }
                if(binary_search(a, a + n + 1, a[i] - x - y) || binary_search(a, a + n + 1, a[i] - x + y)){
                    printf("1\n%d\n",a[i] - x);
                    status = true;
                    break;
                }
            }
            if(Ok(a[i] + x)){
                if(yy){
                    printf("1\n%d\n",a[i] + x);
                    status = true;
                    break;
                }
                if(binary_search(a, a + n + 1, a[i] + x - y) || binary_search(a, a + n + 1, a[i] + x + y)){
                    printf("1\n%d\n",a[i] + x);
                    status = true;
                    break;
                }
            }
            if(Ok(a[i] - y)){
                if(xx){
                    printf("1\n%d\n",a[i] - y);
                    status = true;
                    break;
                }
                if(binary_search(a, a + n + 1, a[i] - y - x) || binary_search(a, a + n + 1, a[i] - y - x)){
                    printf("1\n%d\n",a[i] - y);
                    status = true;
                    break;
                }
            }
            if(Ok(a[i] + y)){
                if(xx){
                    printf("1\n%d\n",a[i] + y);
                    status = true;
                    break;
                }
                if(binary_search(a, a + n + 1, a[i] + y - x) || binary_search(a, a + n + 1, a[i] + y + x)){
                    printf("1\n%d\n",a[i] + y);
                    status = true;
                    break;
                }
            }
        }
        if(!status){
            printf("2\n%d %d\n",x,y);
        }
    }
    return 0;
}