[COGS 2877]老m凯的疑惑

Posted 2020-10-19 Z-Y-Y-S

Description

Margatroid退役之后沉迷文化课

这天，写完数学作业之后的他脑洞大开，决定出一道比NOIP2017 D1T1《小凯的疑惑math》还要好的题

题面是这样的

 

$$ f(n)=n^2\\\\ g(n)=\\sum_{i=1}^{n^3}[f(i)<n]\\\\\\\\ k(n)=\\sum_{i=1}^{n^3}[g(i)<n] $$

 

试求$k(n)\\ \\text{mod}\\ 998244353$

Input

 

一行一个整数$n$

Output

 

一行一个整数$k(n)$

Sample Input

 

1

Sample Output

 

1

由题： $$g(n) = \\sum_{i=1} [i^2 < n]$$

显然：

$$g(n) =\\begin{cases}
\\sqrt n-1& \\text{ n 是完全平方数}\\\\
\\lfloor \\sqrt n \\rfloor& \\text{otherwise}
\\end{cases}$$

构造等价函数： $$g(n) = \\lfloor \\sqrt {n-1} \\rfloor$$

同理，由题： $$k(n) = \\sum_{i=1} [\\sqrt {i-1} < n]$$

因为 $n$ 是正整数，所以 $k(n)$ 等价于：

\\begin{aligned}    

     k(n) &= \\sum_{i=1} [i-1 < n^2]\\\\
     & = \\sum_{i=1} [i <= n^2]\\\\
     & = n^2
\\end{aligned}

转载自Navi：http://www.cnblogs.com/NaVi-Awson/p/8175894.html

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 long long Mod=998244353;
 7 long long n;
 8 int main()
 9 {
10   cin>>n;
11   cout<<((n%Mod)*(n%Mod))%Mod;
12 }

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