[BZOJ2818] Gcd （数论，欧拉函数，线性筛）

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2818

必须用线性筛。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 10001001;
LL phi[maxn], sum[maxn], n;
bool isprime[maxn];
LL prime[maxn];
int tot;
bool mark[maxn+10];
void getphi()
{
    int i,j;
    phi[1]=1;
    for(i=2;i<=maxn;i++)
    {
        if(!mark[i])
        {
            prime[++tot]=i;//筛素数的时候首先会判断i是否是素数。    
            phi[i]=i-1;//当 i 是素数时 phi[i]=i-1    
        }
        for(j=1;j<=tot;j++)    
        {
            if(i*prime[j]>maxn)  break;
            mark[i*prime[j]]=1;//确定i*prime[j]不是素数    
            if(i%prime[j]==0)//接着我们会看prime[j]是否是i的约数    
                {phi[i*prime[j]]=phi[i]*prime[j];break;}
            else phi[i*prime[j]]=phi[i]*(prime[j]-1);//其实这里prime[j]-1就是phi[prime[j]]，利用了欧拉函数的积性
        }
    }
}

namespace fastIO {
    #define BUF_SIZE 100000
    //fread -> read
    bool IOerror = 0;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if (p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if (pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch == ' ' || ch == '
' || ch == '
' || ch == '	';
    }
    inline int read(LL &x) {
        char ch;
        while (blank(ch = nc()));
        if (IOerror)
            return 0;
        for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
        return 1;
    }
    #undef BUF_SIZE
};

inline void out(LL x) { 
    if (x > 9) out(x / 10); 
    putchar(x % 10 + '0');
}

signed main() {
    // freopen("in", "r", stdin);
    getphi();
    sum[1] = 0;
    for(int i = 2; i < maxn; i++) sum[i] = sum[i-1] + phi[i];
    while(fastIO::read(n)) {
        LL ret = 0;
        for(int i = 1; i < tot && prime[i] <= n; i++) {
            LL p = prime[i];
            ret += 2LL * sum[n/p] + 1;
        }
        out(ret); putchar('
');
    }
    return 0;
}