思路：
- 题意：判断1到n的平方和是否是一个可开平方的数，若是输出 “Fake news!”，不然输出 “Nobody knows it better than me!”.
- 因为1到n的平方和有公式 n*(n+1)*(2n+1)/6 ，刚开始一直讨论觉得可能需要统计下质因数的个数书否都为偶数。后面一气之下就特判了下 1 和 24 两组数据竟然过了！！！
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, n;
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> n;
if(n == 1 || n == 24) cout << "Fake news!" << endl;
else cout << "Nobody knows it better than me!" << endl;
}
return 0;
}