【11.27】做题日记

杜教筛模板

又写了一遍，发现之前写的有错误，竟然水过去了，我线筛今天写错了两遍，我以后再写错我是傻逼。

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

const int N = 3e6 + 10;
typedef long long ll;

ll prime[N], cnt, mu[N], phi[N], sm[N], sp[N];
bool vis[N];

void init() {
	mu[1] = phi[1] = 1;
	for (int i = 2; i < N; i++) {
		if (!vis[i])prime[++cnt] = i, mu[i] = -1, phi[i] = i - 1;
		for (int j = 1; j <= cnt and prime[j] * i < N; j++) {
			vis[prime[j] * i] = 1;
			mu[prime[j] * i] = -mu[i];
			phi[prime[j] * i] = (prime[j] - 1) * phi[i];
			if (i % prime[j] == 0) {
				mu[prime[j] * i] = 0;
				phi[prime[j] * i] = phi[i] * prime[j];
				break;
			}
		}
	}
	for (int i = 1; i < N; i++) sm[i] = sm[i - 1] + mu[i], sp[i] = sp[i - 1] + phi[i];
}

unordered_map<ll, ll>M, P;
ll m(ll n) {
	if (n < N)return sm[n];
	if (M.count(n)) return M[n];
	ll ans = 1;
	for (ll l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		ans -= (r - l + 1) * m(n / l);
	}
	return M[n] = ans;
}

ll p(ll n) {
	if (n < N)return sp[n];
	if (P.count(n)) return P[n];
	ll ans = n * (n + 1) / 2;
	for (ll l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		ans -= (r - l + 1) * p(n / l);
	}
	return P[n] = ans;
}

ll T, n;
int main() {
	init();
	scanf("%lld", &T);
	while (T--) {
		scanf("%lld", &n);
		printf("%lld %lld
", p(n), m(n));
	}
}

P1390 公约数的和 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

[sum_{i=1}^nsum_{j = i + 1}^ngcd(i,j) ]

求解：

我们求 (t = sum_{i=1}^nsum_{j=1}^ngcd(i,j)=2sum_{i=1}^nsum_{j = i + 1}^ngcd(i,j) + sum_{i=1}^n)

[sum_{i=1}^nsum_{j=1}^ngcd(i,j)=sum_{i=1}^nsum_{j=1}^nsum_{d|gcd(i,j)}varphi(d)\=sum_{d=1}^nvarphi(d) lfloor dfrac ni floorlfloor dfrac ni floor ]

直接 (O(n)) 求就可以了，不需要分块，因为预处理就需要 (O(n)) 的时间差不了多少

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

typedef long long ll;
const int N = 2e6 + 10;
ll phi[N], prime[N], cnt, vis[N];
ll sum[N];
void init(){
	phi[1] = 1;
	for(int i = 2; i < N;i++){
		if (!vis[i])prime[++cnt] = i, phi[i] = i - 1;
		for(int j = 1; j <= cnt and prime[j] * i < N;j++){
			
			phi[prime[j] * i] = phi[i] * (prime[j] - 1);
			vis[prime[j] * i] = 1;
			if(i % prime[j] == 0){
				phi[prime[j] * i] = phi[i] * prime[j];
				break;
			}
		}
	}
	for (int i = 1;i < N;i++)sum[i] = sum[i - 1] + phi[i];
}
int main(){
	init();
	int n;scanf("%d", &n);
	long long ans = 0;
	for(int i = 1;i <= n;i++){
		ans -= i;
		ans += phi[i] * (n / i) * (n / i);
	}
	ans >>= 1;
	printf("%lld
", ans);
}

还偷来了一个别人的筛素数的东西，但是跑的在 51nod 上跑的好像根 min25 差不了多少

#6235. 区间素数个数 - 题目 - LibreOJ (loj.ac)

P3912 素数个数 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

min25

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

const int maxn = 1e6 + 10;
int prime[maxn], cnt, vis[maxn];
int g[maxn], tot, w[maxn], idx1[maxn], idx2[maxn];
int sqr, n;


signed main() {
    cin >> n;
    sqr = sqrt(n + 0.5);

    for (int i = 2; i <= sqr; i++) {
        if (!vis[i]) {
            prime[++cnt] = i;
        }

        for (int j = 1; j <= cnt && prime[j] * i <= sqr; j++) {
            vis[prime[j]*i] = 1;

            if (!i % prime[j])
                break;
        }
    }

    for (int l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        w[++tot] = n / l;
        g[tot] = w[tot] - 1;

        if (w[tot] <= sqr)
            idx1[w[tot]] = tot;
        else
            idx2[n / w[tot]] = tot;
    }

    for (int i = 1; i <= cnt; i++) {
        for (int j = 1; j <= tot && prime[i] * prime[i] <= w[j]; j++) {
            int k = (w[j] / prime[i] <= sqr) ? idx1[w[j] / prime[i]] : idx2[n / (w[j] / prime[i])];
            g[j] = g[j] - (g[k] - i + 1);
        }
    }

    cout << g[1] << endl;
}

别人的

暂时没懂

/*
 * @Author: zhl
 * @LastEditTime: 2020-11-27 10:24:11
 */

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

using ll = long long;
int count(ll N) {
  int v = sqrt(N);
  vector<int> smalls(v + 1), larges(v + 1);
  for (int i = 1; i <= v; ++i) smalls[i] = i - 1;
  for (int i = 1; i <= v; ++i) larges[i] = N / i - 1;
  for (int p = 2; p <= v; ++p) if (smalls[p] > smalls[p - 1]) {
    int q = p * p, c = smalls[p - 1], e = min(v, N / q);
    for (int i = 1; i <= e; ++i) {
      larges[i] -= ((i * p <= v) ? larges[i * p] : smalls[N / (i * p)]) - c;
    }
    for (int i = v; i >= q; --i) smalls[i] -= smalls[i / p] - c;
  }
  return larges[1];
}

signed main(){
    ll n;
    cin >> n;
    cout << count(n) << endl;
}

下面的是min25

这个代码是真的牛皮

偷过来的跑了第一

#include <cstdio>
#include <cassert>
#include <cmath>
#include <vector>
//Min_25
using namespace std;

using i64 = long long;

int isqrt(i64 n) {
  return sqrtl(n);
}

i64 prime_pi(const i64 N) {
  if (N <= 1) return 0;
  if (N == 2) return 1;
  const int v = isqrt(N);
  int s = (v + 1) / 2;
  vector<int> smalls(s); for (int i = 1; i < s; ++i) smalls[i] = i;
  vector<int> roughs(s); for (int i = 0; i < s; ++i) roughs[i] = 2 * i + 1;
  vector<i64> larges(s); for (int i = 0; i < s; ++i) larges[i] = (N / (2 * i + 1) - 1) / 2;
  vector<bool> skip(v + 1);
  const auto divide = [] (i64 n, i64 d) -> int { return double(n) / d; };
  const auto half = [] (int n) -> int { return (n - 1) >> 1; };
  int pc = 0;
  for (int p = 3; p <= v; p += 2) if (!skip[p]) {
    int q = p * p;
    if (i64(q) * q > N) break;
    skip[p] = true;
    for (int i = q; i <= v; i += 2 * p) skip[i] = true;
    int ns = 0;
    for (int k = 0; k < s; ++k) {
      int i = roughs[k];
      if (skip[i]) continue;
      i64 d = i64(i) * p;
      larges[ns] = larges[k] - (d <= v ? larges[smalls[d >> 1] - pc] : smalls[half(divide(N, d))]) + pc;
      roughs[ns++] = i;
    }
    s = ns;
    for (int i = half(v), j = ((v / p) - 1) | 1; j >= p; j -= 2) {
      int c = smalls[j >> 1] - pc;
      for (int e = (j * p) >> 1; i >= e; --i) smalls[i] -= c;
    }
    ++pc;
  }
  larges[0] += i64(s + 2 * (pc - 1)) * (s - 1) / 2;
  for (int k = 1; k < s; ++k) larges[0] -= larges[k];
  for (int l = 1; l < s; ++l) {
    int q = roughs[l];
    i64 M = N / q;
    int e = smalls[half(M / q)] - pc;
    if (e < l + 1) break;
    i64 t = 0;
    for (int k = l + 1; k <= e; ++k) t += smalls[half(divide(M, roughs[k]))];
    larges[0] += t - i64(e - l) * (pc + l - 1);
  }
  return larges[0] + 1;
}

int main() {
  i64 N;
  while (~scanf("%lld", &N)) {
    printf("%lld
", prime_pi(N));
  }
}