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[100005], b[100005];

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

int main()
{
    int n;
    while (cin >> n)
    {
        for (int i = 1; i <= n / 2; i++) cin >> a[i];
        ll p = 1;
        while (p * p <= m && !check(p * p + a[1])) p++;
        if (p * p > m) { cout << "No" << endl; continue; }
        b[1] = p * p;
        ll sum = b[1] + a[1];
        int i = 2;
        for ( ; i <= n / 2; 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 / 2 + 1) { cout << "No" << endl; continue; }
        cout << "Yes" << endl;
        for (int i = 1; i <= n / 2; i++) cout << b[i] << " " << a[i] << " ";
        cout << endl;
    }
}