【数学】质数的k次方和

来源的claris的模板，应该按我的水平不能超过claris。

质数的0次方和，也就是质数个数。

已通过wolfram alpha验证。

int mod;

inline ll add_mod(ll x, ll y) {
    return (x + y >= mod) ? (x + y - mod) : (x + y);
}

inline ll sub_mod(ll x, ll y) {
    return (x < y) ? (x - y + mod) : (x - y);
}

inline ll sum(ll n) {
    n %= mod;
    return n;
}

const int MAXN = 1e6 + 5;

ll ssum[MAXN];
ll lsum[MAXN];
bool mark[MAXN];

ll prime_cnt(ll n) {
    const int v = sqrt(n);
    ssum[0] = 0;
    lsum[0] = 0;
    memset(mark, 0, sizeof(mark[0]) * (v + 1));
    for(ll i = 1; i <= v; ++i) {
        ssum[i] = sum(i) - 1;
        lsum[i] = sum(n / i) - 1;
    }
    for(ll p = 2; p <= v; ++p) {
        if(ssum[p] == ssum[p - 1])
            continue;
        ll psum = ssum[p - 1];
        ll q = p * p;
        ll ed = min((ll)v, n / q);
        ll delta1 = (p & 1) + 1;
        for(ll i = 1; i <= ed; i += delta1) {
            if(!mark[i]) {
                ll d = i * p;
                ll tmp = (d <= v) ? lsum[d] : ssum[n / d];
                tmp = sub_mod(tmp, psum);
                lsum[i] = sub_mod(lsum[i], tmp);
            }
        }
        ll delta2 = p * delta1;
        for(ll i = q; i <= ed; i += delta2)
            mark[i] = 1;
        for(ll i = v; i >= q; --i) {
            ll tmp = ssum[i / p];
            tmp = sub_mod(tmp, psum);
            ssum[i] = sub_mod(ssum[i], tmp);
        }
    }
    return lsum[1];
}

质数的1次方和，也就是质数求和。

已通过wolfram alpha验证。
已通过HDU的2020CCPC的1002验证。

int mod;

inline ll add_mod(ll x, ll y) {
    return (x + y >= mod) ? (x + y - mod) : (x + y);
}

inline ll sub_mod(ll x, ll y) {
    return (x < y) ? (x - y + mod) : (x - y);
}

inline ll mul_mod(ll x, ll y) {
    return x * y % mod;
}

inline ll sum(ll n) {
    n %= mod;
    return (n * (n + 1)) / 2 % mod;
}

const int MAXN = 1e6 + 5;

ll ssum[MAXN];
ll lsum[MAXN];
bool mark[MAXN];

ll prime_sum(ll n) {
    const int v = sqrt(n);
    ssum[0] = 0;
    lsum[0] = 0;
    memset(mark, 0, sizeof(mark[0]) * (v + 1));
    for(ll i = 1; i <= v; ++i) {
        ssum[i] = sum(i) - 1;
        lsum[i] = sum(n / i) - 1;
    }
    for(ll p = 2; p <= v; ++p) {
        if(ssum[p] == ssum[p - 1])
            continue;
        ll psum = ssum[p - 1];
        ll q = p * p;
        ll ed = min((ll)v, n / q);
        ll delta1 = (p & 1) + 1;
        for(ll i = 1; i <= ed; i += delta1) {
            if(!mark[i]) {
                ll d = i * p;
                ll tmp = (d <= v) ? lsum[d] : ssum[n / d];
                tmp = sub_mod(tmp, psum);
                tmp = mul_mod(tmp, p);
                lsum[i] = sub_mod(lsum[i], tmp);
            }
        }
        ll delta2 = p * delta1;
        for(ll i = q; i <= ed; i += delta2)
            mark[i] = 1;
        for(ll i = v; i >= q; --i) {
            ll tmp = ssum[i / p];
            tmp = sub_mod(tmp, psum);
            tmp = mul_mod(tmp, p);
            ssum[i] = sub_mod(ssum[i], tmp);
        }
    }
    return lsum[1];
}

低次的求和，用拉格朗日插值求出来，然后再改动一下tmp = mul_mod(tmp, p);这里。