BZOJ 4173: 数学

4173: 数学

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 460  Solved: 226
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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

Source

分析：

$m mod k+n mod k>=k$可以转化为以下的式子...

$n-left lfloor frac{n}{k} ight floor*k+m-left lfloor frac{m}{k} ight floor*k>=k$

$n+m-k*( left lfloor frac{n}{k} ight floor+left lfloor frac{m}{k} ight floor )>=k$

$left lfloor frac{n+m}{k} ight floor-left lfloor frac{n}{k} ight floor-left lfloor frac{m}{k} ight floor>=1$

也就是$left lfloor frac{n+m}{k} ight floor-left lfloor frac{n}{k} ight floor-left lfloor frac{m}{k} ight floor=1$

$sum_ {left lfloor frac{n+m}{k} ight floor-left lfloor frac{n}{k} ight floor-left lfloor frac{m}{k} ight floor=1} phi(k)$

$=sum_ {k=1}^{n+m} phi(k)*leftlfloor frac{n+m}{k} ight floor-sum_ {k=1}^{n} phi(k)*leftlfloor frac{n}{k} ight floor-sum_ {k=1}^{m} phi(k)*leftlfloor frac{m}{k} ight floor$

然后我们看$sum_ {k=1}^{n} phi(k)*leftlfloor frac{n}{k} ight floor$是什么...

因为一个数的因子的欧拉函数之和为它本身...所以可以有以下的式子...

$sum_ {k=1}^{n} phi(k)*leftlfloor frac{n}{k} ight floor=sum_ {i=1}^{n} sum_ {kmid i} phi(k)=sum_ {i=1}^{n} i$

那么答案就是$phi(n)*phi(m)*n*m$...

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
//by NeighThorn
#define int long long
using namespace std;

int n,m,mod=998244353;

inline int phi(int x){
	int ans=x,y=sqrt(x);
	for(int i=2;i<=y;i++)
		if(x%i==0){
			ans=ans/i*(i-1);
			while(x%i==0)
				x/=i;
		}
	if(x!=1)
		ans=ans/x*(x-1);
	return ans%mod;
}

signed main(void){
	scanf("%lld%lld",&n,&m);
	int ans=phi(n)*phi(m)%mod*(n%mod)%mod*(m%mod)%mod;
	printf("%lld
",ans);
	return 0;
}

　　

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By NeighThorn