【SDOI 2008】 仪仗队

【题目链接】

https://www.lydsy.com/JudgeOnline/problem.php?id=2190

【算法】

同POJ3090

值得注意的是此题数据规模较大，建议使用用线性筛筛出欧拉函数

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 40010

int i,n,TC,T;
int sum[MAXN];

inline void init()
{
        int i,j,tmp,cnt = ;
        static int f[MAXN],prime[MAXN],phi[MAXN];
        for (i = ; i < MAXN; i++)
        {
                if (!f[i])
                {
                        f[i] = prime[++cnt] = i;
                        phi[i] = i - ;
                }
                for (j = ; j <= cnt; j++)
                {
                        tmp = i * prime[j];
                        if (tmp >= MAXN) break;
                        f[tmp] = prime[j];
                        if (prime[j] == f[i])
                        {
                                phi[tmp] = phi[i] * prime[j];
                                break;
                        }    else phi[tmp] = phi[i] * (prime[j] - );
                }
        }
        for (i = ; i < MAXN; i++) sum[i] = sum[i-] + phi[i];
}

int main()
{

        init();
        scanf("%d",&n);
        printf("%d\n",*sum[n-]+);

        return ;

}