CF1081E Missing Numbers

思路：
贪心乱搞。
实现：
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 
 5 const ll m = 1e13;
 6 
 7 ll a[100005], b[100005];
 8 
 9 bool check(ll x)
10 {
11     ll s = sqrt(x);
12     return s * s == x;
13 }
14 
15 int main()
16 {
17     int n;
18     while (cin >> n)
19     {
20         for (int i = 1; i <= n / 2; i++) cin >> a[i];
21         ll p = 1;
22         while (p * p <= m && !check(p * p + a[1])) p++;
23         if (p * p > m) { cout << "No" << endl; continue; }
24         b[1] = p * p;
25         ll sum = b[1] + a[1];
26         int i = 2;
27         for ( ; i <= n / 2; i++)
28         {
29             while (p * p <= sum + m && (p * p <= sum || !check(p * p + a[i]))) p++;
30             if (p * p > sum + m) break;
31             b[i] = p * p - sum; sum = p * p + a[i];
32         }
33         if (i != n / 2 + 1) { cout << "No" << endl; continue; }
34         cout << "Yes" << endl;
35         for (int i = 1; i <= n / 2; i++) cout << b[i] << " " << a[i] << " ";
36         cout << endl;
37     }
38 }