LightOJ 1007 Mathematically Hard

题意，给定 \(l\) , \(r\)

求
\[ \sum_{i=l}^{r} \varphi(i) \]

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欧拉函数筛

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

根据定义可得

\(\varphi(p)=p-1\) (\(p\)为质数)

\(\varphi(i \times p_j)=\varphi(i) \times (p_j-1)\) (\(p_j\)为质数且 \(i \ge 1\) 且 \(i \bmod p_j \ne 0\))

\(\varphi(i \times p_j)=\varphi(i) \times p_j\) \((p_j\)为质数且 \(i \ge 1\) 且 \(i \bmod p_j =0\))

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 \(O(n\sqrt n)\) 的筛法就可以了

// This code writed by chtholly_micromaker(MicroMaker)
#include 
#define reg register
#define int long long
#define U unsigned
using namespace std;
const int MaxN=5000050;
template  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)
                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>T;
    reg int a,b;
    for(int i=1;i<=T;++i)
    {
        scanf("%lld %lld",&a,&b);
        printf("Case %lld: %llu\n",i,phi[b]-phi[a-1]);
    }
    return 0;
}