题目链接:

1239:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1239

1244:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1244

 

杜教筛裸题,不过现在我也只会筛这俩前缀和...

$$s(n)=\sum _{i=1}^{n}f(i)$$

那么就有:

$$\sum_{i=1}^{n}f(i)\lfloor \frac{n}{i} \rfloor=\sum_{i=1}^{n}s(\lfloor \frac{n}{i} \rfloor)=s(n)+\sum_{i=2}^{n}s(\lfloor \frac{n}{i} \rfloor)$$

移项得到:

$$s(n)=\sum_{i=1}^{n}f(i)\lfloor \frac{n}{i} \rfloor-\sum_{i=2}^{n}s(\lfloor \frac{n}{i} \rfloor)$$

 

对于欧拉函数,$f(n)=\phi(n)$

$$\sum_{i=1}^{n}\phi(i)\lfloor \frac{n}{i} \rfloor=\sum_{i=1}^{n}\sum_{d|n}\phi(d)=\sum_{i=1}^{n}i=\frac{n*(n+1)}{2}$$

 

对于莫比乌斯函数,$f(n)=\mu(n)$

$$\sum_{i=1}^{n}\mu(i)\lfloor \frac{n}{i} \rfloor=\sum_{i=1}^{n}\sum_{d|n}\mu(d)=\sum_{i=1}^{n}[i=1]=1$$

 

然后这两个公式就可以在线筛预处理$n^{\frac{2}{3}}$只后记忆化达到$O(n^{\frac{2}{3}})$的效率.

 

值得注意的就是,记忆化要写hash,以及不要忘了取模,筛欧拉函数前缀和时牵扯取模和除2,可以先讨论奇偶除掉2再计算。

 

1239:

#include
#include
#include
#include
#include
using namespace std;
#define LL long long
#define N 5000000
#define P 233333
#define MAXN 250000
#define MO 1000000007
int cnt,prime[N+10],flag[N+10];
LL X,phi[N+10];
inline void Pre(LL n)
{
	flag[1]=1; phi[1]=1;
	for (LL i=2; i<=n; i++)
		{
			if (!flag[i]) prime[++cnt]=i,phi[i]=i-1;
			for (int j=1; j<=cnt && i*prime[j]<=n; j++)
				{
					flag[i*prime[j]]=1;
					if (!(i%prime[j])) {phi[i*prime[j]]=phi[i]*prime[j]; break;}
					phi[i*prime[j]]=phi[i]*(prime[j]-1);
				}
		}
	for (LL i=1; i<=n; i++) phi[i]=(phi[i]+phi[i-1])%MO;
}
struct Hash{
	int next; LL i,x;
}mp[MAXN];
int head[MAXN],tot;
inline void Add(LL i,LL x) {int pos=i%P; tot++; mp[tot].next=head[pos]; head[pos]=tot; mp[tot].i=i; mp[tot].x=x;}
inline LL Sum(LL x)
{
	if (x<=N) return phi[x];
	else 
		{
			int pos=x%P;
			for (int i=head[pos]; i; i=mp[i].next)
				if (mp[i].i==x) {return mp[i].x;}	
		} 
	LL sum=0,s=0;
	for (LL i=2,j; i<=x; i=j+1) 
		j=x/(x/i),(sum+=Sum(x/i)%MO*(j-i+1)%MO)%=MO;
	if (x&1) s=(((x+1)/2)%MO)*(x%MO)%MO; else s=((x/2)%MO)*((x+1)%MO)%MO;
	sum=(s-sum+MO)%MO;
	Add(x,sum);
	return sum; 
}
int main()
{
	scanf("%lld",&X);
	Pre(N);
	printf("%lld\n",Sum(X));
	return 0;
}

1244

#include
#include
#include
#include
#include
using namespace std;
#define LL long long
#define P 233333
#define N 5000000
#define MAXN 250000
int cnt,prime[N+10],flag[N+10];
LL L,R,mu[N+10];
inline void Pre(LL n)
{
	flag[1]=1; mu[1]=1;
	for (LL i=2; i<=n; i++)
		{
			if (!flag[i]) prime[++cnt]=i,mu[i]=-1;
			for (int j=1; j<=cnt && i*prime[j]<=n; j++)
				{
					flag[i*prime[j]]=1;
					if (!(i%prime[j])) {mu[i*prime[j]]=0; break;}
					mu[i*prime[j]]=-mu[i];
				}
		}
	for (LL i=1; i<=n; i++) mu[i]+=mu[i-1];
}
struct Hash{
	int next; LL i,x;
}mp[MAXN];
int head[MAXN],tot;
inline void Add(LL i,LL x) {int pos=i%P; tot++; mp[tot].next=head[pos]; head[pos]=tot; mp[tot].i=i; mp[tot].x=x;}
inline LL Sum(LL x)
{
	if (x<=N) return mu[x];
	else 
		{
			int pos=x%P;
			for (int i=head[pos]; i; i=mp[i].next)
				if (mp[i].i==x) {return mp[i].x;}	
		} 
	LL sum=0;
	for (LL i=2,j; i<=x; i=j+1)
		j=x/(x/i),sum+=Sum(x/i)*(j-i+1);
	Add(x,1LL-sum);
	return 1LL-sum;
}
int main()
{
	scanf("%lld%lld",&L,&R);
	Pre(N);
	printf("%lld\n",Sum(R)-Sum(L-1));
	return 0;
}