LightOJ 1007 Mathematically Hard

Posted 2021-03-10 chinesepikaync

题意，给定 (l,r)

求
[ sum_{i=l}^{r} varphi(i) ]

---

欧拉函数筛

欧拉函数是积性函数，即 (varphi(a)*varphi(b)=varphi(ab))

根据定义可得

(varphi(p)=p-1) ((p)为质数)

(varphi(i*p_j)=varphi(i)*(p_j-1)) ((p_j)为质数且 (i ge 1) 且 (i \%p_j >0))

(varphi(i*p_j)=varphi(i)*p_j) ((p_j)为质数且 (i ge 1) 且 (i \%p_j =0))

于是预处理线性筛出范围内的每个数的欧拉函数，做一遍前缀和然后 (O(1)) 查询即可

时间复杂度 (O(n))

然后你会发现你一会RE一会MLE

其实这题空间不够，然后还要开 unsigned long long

用 (O(nsqrt n)) 的筛法就可以了

// This code writed by chtholly_micromaker(MicroMaker)
#include <bits/stdc++.h>
#define reg register
#define int long long
#define U unsigned
using namespace std;
const int MaxN=5000050;
template <class t> inline void read(t &s)
{
    s=0;
    reg int f=1;
    reg char c=getchar();
    while(!isdigit(c))
    {
        if(c=='-')
            f=-1;
        c=getchar();
    }
    while(isdigit(c))
        s=(s<<3)+(s<<1)+(c^48),c=getchar();
    s*=f;
    return;
}
U int phi[MaxN];
// ,p[MaxN];
// int pn;
// bool vis[MaxN];
/*
inline void Init_phi()
{
    phi[1]=1;
    for(int i=2;i<MaxN;++i)
    {
        if(!vis[i])
            p[++pn]=i,phi[i]=i-1;
        for(int j=1;j<=pn;++j)
        {
            if(i*p[j]>=MaxN)
                break;
            vis[i*p[j]]=true;
            if(!(i%p[j]))
            {
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
            else
            {
                phi[i*p[j]]=phi[i]*(p[j]-1);
            }
        }
    }
    return;
}
*/
inline void Init_phi()
{
    phi[1]=1;
    for(int i=2;i<MaxN;++i)
        phi[i]=i;
    for(int i=2;i<MaxN;++i)
        if(phi[i]==(U)i)
            for(int j=i;j<MaxN;j+=i)
                phi[j]=phi[j]/i*(i-1);
    return;
}
signed main(void)
{
    Init_phi();
    for(int i=1;i<MaxN;++i)
        phi[i]=phi[i-1]+phi[i]*phi[i];
    int T;cin>>T;
    reg int a,b;
    for(int i=1;i<=T;++i)
    {
        scanf("%lld %lld",&a,&b);
        printf("Case %lld: %llu
",i,phi[b]-phi[a-1]);
    }
    return 0;
}