「BZOJ4173」数学

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题面

已知

\[\large{S(n,m)=\{k_{1},k_{2},\cdots k_{i}\}}
\]

且每个 \(k\) 满足

\[\large{n \%k+m\%k\geq k}
\]

\[\large{\phi(n)\times \phi(m)\times\sum_{k\in S(n,m) }\phi(k)\%998244353}
\]

Part 1

\[\large{n=a_{1} \times k +b_{1} ,m=a_{2} \times k +b_{2}}
\]

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

\[\large{n+m \geq (a_{1}+a_{2}+1)\times k}
\]

两边同时除以 \(k\) 并向下取整得

\[\large{\lfloor \frac{n+m}{k} \rfloor \geq a_{1}+a_{2}+1}
\]

因为

\[\large{a_{1}=\lfloor \frac{n}{k} \rfloor ,a_{2}=\lfloor \frac{m}{k} \rfloor}
\]

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

\[\large{\lfloor\frac{x}{y}\rfloor=\frac{x}{y}-\{\frac{x}{y}\}}
\]

所以式子可化为

\[\large{\frac{n+m}{k}-\{\frac{n+m}{k}\}-(\frac{n}{k}-\{\frac{n}{k}\}+\frac{m}{k}-\{\frac{m}{k}\})} \geq 1
\]

化简得

\[\large{\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}}\geq 1
\]

因为

\[\large{0\leq\{\frac{n}{k}\}},\{\frac{m}{k}\},\{\frac{n+m}{k}\}<1
\]

所以

\[\large{1<\{\frac{n}{k}\}}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}<2
\]

又因为

\[\large{\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}}\geq 1,\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}\in N^{+}
\]

所以

\[\large{\{\frac{n}{k}\}+\{\frac{m}{k}\}-\{\frac{n+m}{k}\}}= 1
\]

\[\large{\lfloor \frac{n+m}{k} \rfloor – \lfloor \frac{n}{k} \rfloor – \lfloor \frac{m}{k} \rfloor= 1}
\]

Part2

先忽视要求式子的部分, 得

\[\large{\sum_{k\in S(n,m)}\phi(k)}
\]

\[\large{\sum_{n \%k+m\%k\geq k }\phi(k)}
\]

\[\large{\sum_{k=1}^{n+m}\phi(k)\times\lfloor \frac{n+m}{k} \rfloor}-\sum_{k=1}^{n}\phi(k)\times\lfloor \frac{n}{k} \rfloor-\sum_{k=1}^{m}\phi(k)\times\lfloor \frac{m}{k} \rfloor
\]

因为

\[\large{n=\sum_{d|n}\phi(d)}
\]

所以

\[\large{\sum_{i=1}^{n+m}i-\sum_{i=1}^{n}i-\sum_{i=1}^{m}i=\frac{(n+m)\times(n+m-1)}{2}-\frac{n\times(n-1)}{2}-\frac{m\times(m-1)}{2}-}
\]

\[\large{=n\times m}
\]

结论

\[\large{ans=\large{\phi(n)\times \phi(m)\times n\times m\%998244353}}
\]

代码

#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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