Fake NewsSilence的博客-fake news csdn

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 =
$sum_{k=1}^n$

∑k=1n​ 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!

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

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; }