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

$n\quad mod\quad k+m\quad mod \quad k>=k$

等价于 $n-\lfloor\frac{n}{k}\rfloor*k+m-\lfloor\frac{m}{k}\rfloor*k>=k$

$\lfloor\frac{n+m}{k}\rfloor-\lfloor\frac{n}{k}\rfloor-\lfloor\frac{m}{k}\rfloor=1$

$\sum_{\lfloor\frac{n+m}{k}\rfloor-\lfloor\frac{n}{k}\rfloor-\lfloor\frac{m}{k}\rfloor=1}\phi(k)$

$=\sum_{k=1}^{n+m}\phi(k)\lfloor\frac{m+n}{k}\rfloor-\sum_{k=1}^{n}\phi(k)\lfloor\frac{n}{k}\rfloor-\sum_{k=1}^{m}\phi(k)\lfloor\frac{m}{k}\rfloor$

$\sum_{k=1}^{n}\phi(k)\lfloor\frac{n}{k}\rfloor=\sum_{k=1}^{n}\phi(k)\sum_{k|i}^{n}1=\sum_{i=1}^{n}\sum_{k|i}\phi(k)$

$\because\sum_{k|i}\phi(k)=i$

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

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**