BZOJ 4173 数论

思路：

$(m％k+n％k>=k) *phi(k)$

$我们不妨设n=q_1k+r_1　m=q_2k+r$2

$n+m=(q_1+q_2)k+r1+r2$

${\lfloor}\frac{n+m}{k}{\rfloor}-{\lfloor}\frac{m}{k}{\rfloor}-{\lfloor}\frac{n}{k}{\rfloor}=(m％k+n％k>=k)$

$原式=phi(k)*({\lfloor}\frac{n+m}{k}{\rfloor}-{\lfloor}\frac{m}{k}{\rfloor}-{\lfloor}\frac{n}{k}{\rfloor})$

$id=phi|1$

$n=\Sigma_{d|n}phi(d)$

$原式=\Sigma_{i=1}^{n+m}i-\Sigma_{i=1}^mi-\Sigma_{i=1}^ni$
$　　=(n+m)*(n+m-1)/2+m*(m-1)/2+n*(n-2)/2$
$　　=n*m$

 

//By SiriusRen
#include <cstdio>
using namespace std;
typedef long long ll;
ll n,m,mod=998244353;
ll phi(ll x){
    ll res=1;
    for(int i=2;1LL*i*i<=x;i++){
        if(x%i==0){
            while(x%i==0)x/=i,res=res*i;
            res=res/i*(i-1);
        }
    }if(x!=1)res=res*(x-1);
    return res;
}
int main(){
    scanf("%lld%lld",&n,&m);
    printf("%lld\n",((((phi(n)%mod)*(phi(m)%mod))%mod*(n%mod))%mod*(m%mod))%mod);
}