「BZOJ4173」数学
题目大意：
求
[varphi(n) cdot varphi(m) cdot sum_{k in S(n,m)} varphi(k)quad mathrm{mod} quad 998244353
]
其中 (S(n,m)) 为满足 $ mquad mathrm{mod} quad k + nquad mathrm{mod} quad k ge k$ 的 (k) 的集合。
我们可以对 $ mquad mathrm{mod} quad k + nquad mathrm{mod} quad k ge k$ 进行化简。
[m-lfloor frac{m}{k}
floor*k + n-lfloor frac{n}{k}
floor*k ge k \
m+n ge k(1+lfloor frac{m}{k}
floor + lfloor frac{n}{k}
floor) \
lfloor frac{m+n}{k}
floor -lfloor frac{m}{k}
floor - lfloor frac{n}{k}
floor= 1\
]
即 $ mquad mathrm{mod} quad k + nquad mathrm{mod} quad k ge k iff lfloor frac{m+n}{k} floor -lfloor frac{m}{k} floor - lfloor frac{n}{k} floor= 1$
令 (t=sum_{k in S(n,m)} varphi(k))
则
[t= sum_{k=1}^{n+m} varphi(k) cdot [ lfloor frac{m+n}{k}
floor -lfloor frac{m}{k}
floor - lfloor frac{n}{k}
floor ] \
=sum_{k=1}^{n+m} varphi(k) cdot lfloor frac{m+n}{k}
floor -sum_{k=1}^{m}varphi(k) cdotlfloor frac{m}{k}
floor - sum_{k=1}^{n}varphi(k) cdotlfloor frac{n}{k}
floor \
]
考虑求解 (sum_{k=1}^{Q}varphi(k)cdotlfloor frac{Q}{k} floor)
则其等于
[sum_{k=1}^{Q}varphi(k)cdotlfloor frac{Q}{k}
floor\
=sum_{k=1}^{Q} sum_{t=1}^{t le lfloor frac{Q}{k}
floor} varphi(k)\
=sum_{t=1}^{Q} sum_{k|t} varphi(k)\
=sum_{t=1}^{Q} t
]
于是有 (ans= varphi(n) cdot varphi(m) cdot n cdot m)
时间复杂度 (O(sqrt{n}))
贴代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p=998244353;
ll eular(ll n){
ll ans=n;
ll i=2;
while(i*i<=n){
if(n%i==0){
while(n%i==0) n/=i;
ans=ans/i*(i-1);
}
++i;
}
if(n!=1) ans=ans/n*(n-1);
return ans%p;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
ll n,m;
cin>>n>>m;
cout<<(n%p*(m%p)%p*eular(n)%p*eular(m)%p)%p;
return 0;
}