CF1081E Missing Numbers

思路：

贪心乱搞。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;

 const ll m = 1e13;

 ll a[], b[];

 bool check(ll x)
 {
     ll s = sqrt(x);
     return s * s == x;
 }

 int main()
 {
     int n;
     while (cin >> n)
     {
         for (int i = ; i <= n / ; i++) cin >> a[i];
         ll p = ;
         while (p * p <= m && !check(p * p + a[])) p++;
         if (p * p > m) { cout << "No" << endl; continue; }
         b[] = p * p;
         ll sum = b[] + a[];
         int i = ;
         for ( ; i <= n / ; i++)
         {
             while (p * p <= sum + m && (p * p <= sum || !check(p * p + a[i]))) p++;
             if (p * p > sum + m) break;
             b[i] = p * p - sum; sum = p * p + a[i];
         }
         if (i != n /  + ) { cout << "No" << endl; continue; }
         cout << "Yes" << endl;
         for (int i = ; i <= n / ; i++) cout << b[i] << " " << a[i] << " ";
         cout << endl;
     }
 }