根据质数的性质：$k \equiv 1\ (\bmod\ 4), k \in \mathbb P$（$2$ 为特例，需特判）可得以下代码：

AC code：

#include <bits/stdc++.h>
using namespace std;
#define ll long long

vector<ll> sieve(ll n) {
    vector<bool> is_p(n + 1, true);
    is_p[0] = is_p[1] = false;
    for (ll i = 2; i * i <= n; ++i) {
        if (is_p[i]) {
            for (ll j = i * i; j <= n; j += i) {
                is_p[j] = false;
            }
        }
    }
    vector<ll> res;
    for (ll i = 2; i <= n; ++i) {
        if (is_p[i]) {
            res.push_back(i);
        }
    }
    return res;
}

vector<bool> seg_sieve(ll L, ll R, const vector<ll>& primes) {
    vector<bool> is_p(R - L + 1, true);
    if (L == 1) {
        is_p[0] = false;
    }
    for (ll p : primes) {
        ll start = max(p * p, ((L + p - 1) / p) * p);
        if (start > R) {
            continue;
        }
        for (ll j = start; j <= R; j += p) {
            is_p[j - L] = false;
        }
    }
    return is_p;
}

int main() {
    ll l, r;
    cin >> l >> r;
    if (r < 2) {
        cout << 0 << endl;
        return 0;
    }
    ll sqrt_r = sqrt(r);
    vector<ll> primes = sieve(sqrt_r);
    vector<bool> is_p = seg_sieve(l, r, primes);
    int ans = 0;
    for (ll i = 0; i <= r - l; ++i) {
        if (is_p[i]) {
            ll num = l + i;
            if (num == 2) {
                ans++;
            } else if (num % 4 == 1) {
                ans++;
            }
        }
    }
    cout << ans << endl;
    return 0;
}