思路
cf题解中合法的x的处于一段连续区间不太明白。在知道这个前提下，将E1的代码改成二分即可。
有空再补回来。
#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <cstring>
#include <string>
#include <stack>
#include <deque>
#include <cmath>
#include <iomanip>
#include <cctype>
#define endl '
'
#define IOS std::ios::sync_with_stdio(0);
#define FILE freopen("..//data_generator//in.txt","r",stdin),freopen("res.txt","w",stdout)
#define FI freopen("..//data_generator//in.txt","r",stdin)
#define FO freopen("res.txt","w",stdout)
#define pb push_back
#define mp make_pair
#define seteps(N) fixed << setprecision(N)
typedef long long ll;
using namespace std;
/*-----------------------------------------------------------------*/
#define INF 0x3f3f3f3f
const int N = 3e5 + 10;
const double eps = 1e-8;
int arr[N];
vector<int> ans;
bool check(int x, int p, int n){
bool ok = true;
for(int i = 0; i < n; i++) {
int num = min(i + 1, i + 1 + (x - arr[i]));
if(num <= 0 || !(num % p)) {
ok = false;
break;
}
}
return ok;
}
int main() {
IOS;
int n, p;
cin >> n >> p;
int mx = -INF;
int mi = INF;
for(int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr, arr + n);
for(int i = 0; i < n; i++) {
mx = max(arr[i], mx);
mi = min(i + 1 - arr[i], mi);
}
int s = max(0, 0 - mi + 1);
int e = mx;
int l = s, r = e;
while(l <= r) {
int mid = (l + r) / 2;
if(check(mid, p, n)) {
l = mid + 1;
} else {
r = mid - 1;
}
}
e = r;
if(s > e) {
cout << 0 << endl;
} else {
cout << e - s + 1 << endl;
for(int x = s; x <= e; x++) cout << x << " ";
}
}