程序解答高数方程

Fake News 传送门

题目大意

给定一个数字 n，问是否存在 x 满足 x ² = $\sum_{k=1}^n$ K²

输入描述

There are multiple test cases. The first line of the input contains an integer T (1 ≤ T ≤ 1e6) indicating the number of test cases.
For each test case, the first line contains one integers n (1 ≤ n≤ 1e15 ).

输出描述

If the sum is a square number, please output ‘Fake news!’ in one line. Otherwise, please output ‘Nobody knows it better than me!’ instead.

输入

5
1
2
3
4
5

输出

Fake news!
Nobody knows it better than me!
Nobody knows it better than me!
Nobody knows it better than me!
Nobody knows it better than me!

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解题思路

我们可以先在本地打表测试一下哪些数字满足条件，打完表后发现只有两个数满足条件，故我们直接特判这两个数就好了
cin 和 cout 加速之后还是会被 T 掉的，直接上 scanf 和 printf 就可以 ac
至于为什么是 1 和 24 ，其实是可以证明出来的 知乎的证明帖子 - 传送门

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AC代码

#include<iostream> #include<algorithm> #include<string.h> #include<string> #include<cmath> #include<cstdio> #include<cstdlib> using namespace std; typedef long long ll; //const int maxn = 1e8; //double sum[maxn]; //void table() //{ //	sum[0] = 0; //	for(int i = 1 ;  i <= maxn ; i++) //	{ //		sum[i] += i*i + sum[i-1]; //		int tmp = sqrt(sum[i]); //		if(tmp * tmp == sum[i]) //		{ //			cout<<"sum["<<i<<"] = "<<sum[i]<<endl; //		} //	} //} int main() { //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); //table(); int t; scanf("%d",&t); while(t--) { char n[20]; scanf("%s",n); if(strlen(n) == 1 && n[0] == '1') { printf("Fake news!\n"); }else if(strlen(n) == 2 && n[0] == '2' && n[1] == '4') { printf("Fake news!\n"); }else { printf("Nobody knows it better than me!\n"); } } return 0; } 

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