题目大意:
\(f(x)=\begin{cases} x,1\le x\le9\\ f(x的各数位之和),x>9\\ \end{cases}\)
求\(\sum_{i=1}^{n}f(i)\)。
根据打表找规律,我们会发现\(f(x)=(x-1)\bmod 9+1\)。
所以\(\sum_{i=1}^{n}f(i)\)
\(=\sum_{i=1}^{\lfloor\frac{n}{9}\rfloor\cdot 9}f(x)+\sum_{i=\lfloor\frac{n}{9}\rfloor\cdot 9+1}^{n}f(x)\)
\(=\lfloor\frac{n}{9}\rfloor\cdot\sum_{i=1}^{9}f(i)+\sum_{i=1}^{n\bmod 9}f(i)\)
\(=\lfloor\frac{n}{9}\rfloor\cdot\sum_{i=1}^{9}i+\sum_{i=1}^{n\bmod 9}i\)
\(=\lfloor\frac{n}{9}\rfloor\cdot\frac{(1+9)\times9}{2}+\frac{(1+n\bmod 9)*(n\bmod 9)}{2}\)
\(=\lfloor\frac{n}{9}\rfloor\cdot 45+\frac{(1+n\bmod 9)*(n\bmod 9)}{2}\)
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const ll N=1e6+10,inf=1e18+10,mod=1e9+7;
int main(){
int T;
cin >> T;
while(T--){
ll n;
cin >> n;
cout << n/9*45+(1+n%9)*(n%9)/2 << endl;
}
return 0;
}
标签:Origin,lfloor,frac,数论,sum,rfloor,cdot,bmod,CodeChef
From: https://www.cnblogs.com/alric/p/18195884