题目传送门

　　传送门I

　　传送门II

　　传送门III

题目大意

　　给定将$\left \{ 0, 1, \dots, m - 1\right \}$分成了不相交的两个非空集合$A$和$B$，给定$A$，问存在多少个整数$x$满足$0\leqslant x < m$且对于任意$a\in A, b \in B$满足$a + b\not \equiv x \pmod{m}$

 /**

  * Codeforces

  * Problem#1045B

  * Accepted

  * Time: 77ms

  * Memory: 7300k

  */

 #include <algorithm>

 #include <iostream>

 #include <cstdlib>

 #include <cstdio>

 #include <vector>

 using namespace std;

 typedef bool boolean;

 #define ull unsigned long long

 const int N = 2e5 + ;

 const ull base = 1e9 + ;

 int n, M;

 int ar[N];

 int cr[N];

 ull ph[N], sh[N], pb[N];

 int add(int a, int b) {

     return ((a += b) >= M) ? (a - M) : (a);

 }

 inline void init() {

     scanf("%d%d", &n, &M);

     for (int i = ; i <= n; i++)

         scanf("%d", ar + i);

 }

 ull fquery(int l, int r) {    // forward

     if (l > r)

         return ;

     return ph[r] - ph[l - ] * pb[r - l + ];

 }

 ull rquery(int l, int r) {    // backward

     if (l > r)

         return ;

     return sh[l] - sh[r + ] * pb[r - l + ];

 }

 boolean ispar(int l, int r) {

     return fquery(l, r) == rquery(l, r);

 }

 vector<int> vec;

 inline void solve() {

     ph[] = , sh[n] = , pb[] = ;

     for (int i = ; i < n; i++)

         ph[i] = ph[i - ] * base + (cr[i] = ar[i + ] - ar[i]);

     for (int i = n - ; i; i--)

         sh[i] = sh[i + ] * base + cr[i];

     for (int i = ; i < n; i++)

         pb[i] = pb[i - ] * base;

     for (int i = , x; i < n; i++) {

         x = ar[] + ar[i];

 //        if (x >= M)

 //            break;

         if (add(ar[i + ], ar[n]) == x && ispar(, i - ) && ispar(i + , n - ))

             vec.push_back(x);

     }

     int x = add(ar[], ar[n]);

     if ((x <= ar[] || x >= ar[n]) && ispar(, n - ))

         vec.push_back(x);

     printf("%d\n", (signed) vec.size());

     sort(vec.begin(), vec.end());

     for (int i = ; i < (signed) vec.size(); i++)

         printf("%d ", vec[i]);

     putchar('\n');

 }

 int main() {

     init();

     solve();

     return ;

 }