HDU5382 GCD?LCM!

description

Let's define

[F(n)=sum_{i=1}^nsum_{j=1}^n[ ext{lcm}(i,j)+gcd(i,j)ge n]\ S(n)=sum_{i=1}^nF(i) ]

Find S(n)(mod 258280327)

data range

(Tle 10^5,nle 10^6)

solution

It's obvious that if we can calculate (F(n)) quickly, the problem is solved.

How to calculate (F(n))?Let's define (G(n)) as follows:

[G(n)=sum_{i=1}^nsum_{j=1}^n[ ext{lcm}(i,j)+gcd(i,j)= n] ]

then

[F(n)=F(n-1)+2n-1-G(n-1) ]

Now the key point is to calculate (G(n)) quickly.

[G(n)=sum_{i=1}^nsum_{j=1}^n[ ext{lcm}(i,j)+gcd(i,j)= n]\ =sum_{d=1}^nsum_{i=1}^nsum_{j=1}^n[frac{ij}d+d=n][gcd(i,j)=d]\ =sum_{d=1}^nsum_{i=1}^{frac nd}sum_{j=1}^{frac nd}[ijd+d=n][gcd(i,j)=1]\ =sum_{d=1}^nsum_{i=1}^{frac nd}sum_{j=1}^{frac nd}[ij=frac nd-1][gcd(i,j)=1] ]

Let's define

[t(n)=sum_{i=1}^nsum_{j=1}^n[gcd(i,j)=1][ij=n] ]

then

[G(n)=sum_{d=1}^nt(frac nd-1)=sum_{dmid n}^nt(frac nd-1) ]

To my surprise, (t(n)) is easy to calculate . In fact, (t(n)=2^k),in which (k) means the number of different types of prime factors in (n) .That's because to make sure the condition (gcd(i,j)=1) ,prime factors of the same kind should be assigned to either (i) or (j),which are only (2) ways , so for (k) different types of prime factors there are (2^k) ways .We can use linear sieve to find (t(n)) .Then, we can find (G(n)) through enumerating the multiples . After that ,the problem is solved successfully .

time complexity

(mathcal O(nln n))

code

#include<cstdio>
using namespace std;
const int N=1e6+5,mod=258280327;
bool flag[N];int pr[N],pcnt,t[N],g[N],f[N];
inline int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline int dec(int x,int y){return x-y<0?x-y+mod:x-y;}
inline void pre()
{
	t[1]=1;int n=N-5;
	for(int i=2;i<=n;++i)
	{
		if(!flag[i])pr[++pcnt]=i,t[i]=2;
		for(int j=1;j<=pcnt;++j)
		{
			int num=i*pr[j];if(num>n)break;
			flag[num]=1;
			if(i%pr[j])t[num]=add(t[i],t[i]);
			else{t[num]=t[i];break;}
		}
	}
	for(int i=1;i<=n;++i)
		for(int j=i;j<=n;j+=i)
			g[j]=add(g[j],t[j/i-1]);
	for(int i=1;i<=n;++i)
		f[i]=add(f[i-1],dec(2*i-1,g[i-1]));
	for(int i=1;i<=n;++i)f[i]=add(f[i],f[i-1]);
}
int main()
{
	pre();int T,n;scanf("%d",&T);
	while(T-->0)scanf("%d",&n),printf("%d
",f[n]);
	return 0;
}

inspiration

It's of vital importance to consider combinatorial meaning while deduction .