BZOJ 4173 数学

题面：

4173: 数学

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 602  Solved: 289
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Description

 

Input

 输入文件的第一行输入两个正整数 。 

Output

 如题

Sample Input

5 6

Sample Output

240

HINT

 N,M<=10^15

$nquad modquad k+mquad mod quad k>=k$

等价于 $n-lfloorfrac{n}{k} floor*k+m-lfloorfrac{m}{k} floor*k>=k$

$lfloorfrac{n+m}{k} floor-lfloorfrac{n}{k} floor-lfloorfrac{m}{k} floor=1$

$sum_{lfloorfrac{n+m}{k} floor-lfloorfrac{n}{k} floor-lfloorfrac{m}{k} floor=1}phi(k)$

$=sum_{k=1}^{n+m}phi(k)lfloorfrac{m+n}{k} floor-sum_{k=1}^{n}phi(k)lfloorfrac{n}{k} floor-sum_{k=1}^{m}phi(k)lfloorfrac{m}{k} floor$

$sum_{k=1}^{n}phi(k)lfloorfrac{n}{k} floor=sum_{k=1}^{n}phi(k)sum_{k|i}^{n}1=sum_{i=1}^{n}sum_{k|i}phi(k)$

$ecausesum_{k|i}phi(k)=i$

$ hereforesum_{i=1}^{n}sum_{k|i}phi(k)=sum_{i=1}^{n}i$

答案就是$phi(n)*phi(m)*n*m$

#include<iostream>
#include<cstdio>
using namespace std;
#define LL long long
const LL mod=998244353;
LL p(LL x)
{
    LL ans=x;
    for(LL 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;
}
LL n,m;
int main()
{
    scanf("%lld%lld",&n,&m);
    printf("%lld",(((((p(n)%mod)*(p(m)%mod)))%mod)*(((n%mod)*(m%mod))%mod))%mod);
}

感谢lc的**