题目链接
将给出的已经有了的排序， 在前面加上0， 后面加上1e9+1。
然后对相邻的两项判断。 如果相邻两项之间的数的和小于m， 那么全都选上， m减去相应的值。
如果大于m， 那么二分判断最多能选多少个。
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 1e5+5;
int ans[maxn], a[maxn], cnt;
int check(int l, int r, int sum) {
int len = r-l+1;
ll tmp = 1LL*(l+r)*len/2;
if(tmp<=sum)
return 1;
return 0;
}
void bin(int l, int r, int sum) {
int pos = l;
int tmpl = l;
while(l<=r) {
int m = l+r>>1;
if(check(tmpl, m, sum)) {
pos = m;
l = m+1;
} else
r = m-1;
}
for(int i = tmpl; i <= pos; i++)
ans[cnt++] = i;
}
int main()
{
int n, m;
cin>>n>>m;
a[0] = 0;
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
a[++n] = 1e9+1;
sort(a, a+n+1);
for(int i = 0; i < n; i++) {
int tmp = a[i+1]-a[i];
if(tmp==1)
continue;
if(tmp==2) {
if(m>=a[i]+1) {
m -= (a[i]+1);
ans[cnt++] = a[i]+1;
continue;
} else
break;
}
int a1 = a[i]+1, an = a[i+1]-1;
tmp--;
ll sum = 1LL*(a1+an)*tmp/2;
if(sum<=m) {
m -= sum;
for(int j = a[i]+1; j<=a[i+1]-1; j++) {
ans[cnt++] = j;
}
} else {
if(a1<=m) {
bin(a1, an, m);
}
break;
}
}
cout<<cnt<<endl;
for(int i = 0; i < cnt; i++)
printf("%d ", ans[i]);
return 0;
}