Min 25 sieve, find the sum of prime numbers within 1 billion

const int N = 1000010;

typedef long long LL;

namespace Min25
{
    
    

int prime[N], id1[N], id2[N], flag[N], ncnt, m;

LL g[N], sum[N], a[N], T, n;

inline int ID(LL x)
{
    
    
    return x <= T ? id1[x] : id2[n / x];
}

inline LL calc(LL x)
{
    
    
    return x * (x + 1) / 2 - 1;
}

inline LL f(LL x)
{
    
    
    return x;
}

inline void init()
{
    
    
    T = sqrt(n + 0.5);
    ncnt=0;
    m=0;
    for (int i = 2; i <= T; i++)
    {
    
    
        if (!flag[i])
            prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
        for (int j = 1; j <= ncnt && i * prime[j] <= T; j++)
        {
    
    
            flag[i * prime[j]] = 1;
            if (i % prime[j] == 0)
                break;
        }
    }
    for (LL l = 1; l <= n; l = n / (n / l) + 1)
    {
    
    
        a[++m] = n / l;
        if (a[m] <= T)
            id1[a[m]] = m;
        else
            id2[n / a[m]] = m;
        g[m] = calc(a[m]);
    }
    for (int i = 1; i <= ncnt; i++)
        for (int j = 1; j <= m && (LL)prime[i] * prime[i] <= a[j]; j++)
            g[j] = g[j] - (LL)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
}

inline LL solve(LL x)
{
    
    
    if (x <= 1)
        return x;
    return n = x, init(), g[ID(n)];
}
}