Project Euler 010 Summation of primes

题意：求 2×106 以内的质数的和。
分析：欧拉筛法即可。

#include 

#define ll long long

#define pii std::pair
#define mp std::make_pair
#define fi first
#define se second

#define SZ(x) (int)(x).size()
#define pb push_back

template<class T>inline void chkmax(T &x, const T &y) {if(x < y) x = y;}
template<class T>inline void chkmin(T &x, const T &y) {if(x > y) x = y;}

template<class T>
inline void read(T &x) {
    char c;int f = 1;x = 0;
    while(((c=getchar()) < '0' || c > '9') && c != '-');
    if(c == '-') f = -1;else x = c-'0';
    while((c=getchar()) >= '0' && c <= '9') x = x*10+c-'0';
    x *= f;
}
static int outn;
static char out[(int)2e7];
template<class T>
inline void write(T x) {
    if(x < 0) out[outn++] = '-', x = -x;
    if(x) {
        static int tmpn;
        static char tmp[20];
        tmpn = 0;
        while(x) tmp[tmpn++] = x%10+'0', x /= 10;
        while(tmpn) out[outn++] = tmp[--tmpn];
    }
    else out[outn++] = '0';
}

const int N = 2e6;

int p[N + 9], pn;
bool is[N + 9];

inline void sieve(int n) {
    for(int i = 2; i <= n; ++i) {
        if(!is[i]) p[++pn] = i;
        for(int j = 1; j <= pn && p[j] * i <= n; ++j) {
            is[p[j] * i] = true;
            if(i % p[j] == 0) break;
        }
    }
}

int main() {
    sieve(N);
    ll sum = 0;
    for(int i = 1; i <= pn; ++i)
        sum += p[i];
    std::cout << sum << std::endl;
    return 0;
}