bzoj4173 数学

[φ(n)*φ(m)*sum_{kin S(n,m)}φ(k)~mod~p,p=998244353且S(n,m)={n~mod~k+m~mod~kge~k}\ 打表都找出规律了555,写龟速乘写挂了,抱灵了,要不然就rank1了100pts->0\ 因为n~\%~k=n-lfloor frac nk floor*k\ S(n,m)等价于lfloorfrac{n+m}{k} floor-lfloorfrac nk floor-lfloorfrac mk floor=1\ sumφ(k)[n\%k+m\%kge k]=sumφ(k)[lfloorfrac{n+m}{k} floor-lfloorfrac nk floor-lfloorfrac mk floor=1]\ 因为0lelfloorfrac{n+m}{k} floor-(lfloorfrac nk floor+lfloorfrac mk floor)<2\ 所以上式可以化成sum_{k=1}^{n+m}φ(k)(lfloorfrac{n+m}{k} floor-lfloorfrac nk floor-lfloorfrac mk floor)\ =sum_{i=1}^{n+m}φ(k)lfloorfrac{n+m}{k} floor-sum_{k=1}^nlfloorfrac nk floorφ(k)-sum_{k=1}^mlfloorfrac mk floorφ(k),这里开始没搞懂,因为枚举超过n和m时就取0了\ sum_{k=1}^nφ(k)*lfloorfrac nk floor咋搞?通过sum_{k|n}φ(k)=n \ sum_{k=1}^nφ(k)*lfloorfrac nk floor=sum_{i=1}^nsum_{k|i}φ(k)=sum_{i=1}^ni\ [1,n]中约数有k的个数是lfloorfrac nk floor\ 然后等差求和了原式就为φ(m)*φ(n)*mn ]

#include<bits/stdc++.h>
#define mod 998244353
using namespace std;
typedef long long ll;
const int inf =0x3f3f3f3f;
const double eps=1e-8;
ll read()
{
    ll res=0,f=1;
    char ch=getchar();
    while(ch<'1'||ch>'9'){f=ch=='-'?-1:1;ch=getchar();}
    while(ch>'1'&&ch<'9'){res=res*10+ch-'0';ch=getchar();}
    return res*f;
}
ll n,m;
ll phi(ll n)
{
    ll i,re=n;
    for(i=2;i*i<=n;i++){
        if(n%i==0){
            re=re/i*(i-1);
            while(n%i==0){n/=i;}
        }
    }
    if(n!=1){re=re/n*(n-1);}
    return re;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    printf("%lld
",phi(n)%mod*(phi(m)%mod)%mod*(n%mod)%mod*(m%mod)%mod);
    return 0;
}