題面
已知
\[\large{S(n,m)=\{k_{1},k_{2},\cdots k_{i}\}} \]
且每個 \(k\) 滿足
\[\large{n \%k+m\%k\geq k} \]
求
\[\large{\phi(n)\times \phi(m)\times\sum_{k\in S(n,m) }\phi(k)\%998244353} \]
Part 1
令
\[\large{n=a_{1} \times k +b_{1} ,m=a_{2} \times k +b_{2}} \]
所以有
\[\large{b_{1}+b_{2} \geq k} \]
\[\large{(a_{1} \times k +b_{1})+(a_{2} \times k +b_{2}) \geq (a_{1}+a_{2}+1)\times k} \]
所以
\[\large{n+m \geq (a_{1}+a_{2}+1)\times k} \]
兩邊同時除以 \(k\) 并向下取整得
\[\large{\lfloor \frac{n+m}{k} \rfloor \geq a_{1}+a_{2}+1} \]
因為
\[\large{a_{1}=\lfloor \frac{n}{k} \rfloor ,a_{2}=\lfloor \frac{m}{k} \rfloor} \]
所以
\[\large{\lfloor \frac{n+m}{k} \rfloor \geq \lfloor \frac{n}{k} \rfloor+\lfloor \frac{m}{k} \rfloor+1} \]
\[\large{\lfloor \frac{n+m}{k} \rfloor - \lfloor \frac{n}{k} \rfloor - \lfloor \frac{m}{k} \rfloor\geq 1} \]
已知
\[\large{\lfloor\frac{x}{y}\rfloor=\frac{x}{y}-\{\frac{x}{y}\}} \]
所以式子可化為
\[\large{\frac{n+m}{k}-\{\frac{n+m}{k}\}-(\frac{n}{k}-\{\frac{n}{k}\}+\frac{m}{k}-\{\frac{m}{k}\})} \geq 1 \]
化簡得
\[\large{\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}}\geq 1 \]
因為
\[\large{0\leq\{\frac{n}{k}\}},\{\frac{m}{k}\},\{\frac{n+m}{k}\}<1 \]
所以
\[\large{1<\{\frac{n}{k}\}}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}<2 \]
又因為
\[\large{\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}}\geq 1,\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}\in N^{+} \]
所以
\[\large{\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}}= 1 \]
即
\[\large{\lfloor \frac{n+m}{k} \rfloor - \lfloor \frac{n}{k} \rfloor - \lfloor \frac{m}{k} \rfloor= 1} \]
Part2
先忽視要求式子的部分， 得
\[\large{\sum_{k\in S(n,m)}\phi(k)} \]
即
\[\large{\sum_{n \%k+m\%k\geq k }\phi(k)} \]
即
\[\large{\sum_{k=1}^{n+m}\phi(k)\times\lfloor \frac{n+m}{k} \rfloor}-\sum_{k=1}^{n}\phi(k)\times\lfloor \frac{n}{k} \rfloor-\sum_{k=1}^{m}\phi(k)\times\lfloor \frac{m}{k} \rfloor \]
因為
\[\large{n=\sum_{d|n}\phi(d)} \]
所以
\[\large{\sum_{i=1}^{n+m}i-\sum_{i=1}^{n}i-\sum_{i=1}^{m}i=\frac{(n+m)\times(n+m-1)}{2}-\frac{n\times(n-1)}{2}-\frac{m\times(m-1)}{2}-} \]
\[\large{=n\times m} \]
結論
\[\large{ans=\large{\phi(n)\times \phi(m)\times n\times m\%998244353}} \]
代碼
#include <bits/stdc++.h>
using namespace std;
const int mod=998244353;
unsigned long long n,m;
unsigned long long phi(unsigned long long x)
{
unsigned long long ans=x;
for (unsigned long long i=2;i*i<=x;i++)
{
if (x%i==0)
{
ans-=ans/i;
while (x%i==0) x/=i;
}
}
if (x>1) ans-=ans/x;
return ans%mod;
}
int main()
{
cin>>n>>m;
cout<<(phi(n)%mod)*(phi(m)%mod)%mod*(n%mod)%mod*(m%mod)%mod;
return 0;
}
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標籤：C++