2024年实验班选拔考试（正式赛）

2,4(1H,3H)-PD are mine

1430人浏览 · 2024-08-11 10:03:03

2,4(1H,3H)-PD are mine · 2024-08-11 10:03:03 发布

PS:

经过 0_binary_beast_1 同学的允许，把她的 java 解题代码一起放这里了，供 java 选手参考。

0_binary_beast_1帖子传送门

c++ 代码框架

#include<bits/stdc++.h>
#define rep(i,a,b) for (int i=a;i<b;++i)
#define per(i,a,b) for (int i=a;i>b;--i)
#define se second 
#define fi first
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define pii pair<int,int>
#define pli pair<LL,int>
#define pll pair<LL,LL>
#define MEM(a,x) memset(a,x,sizeof(a))
#define OJBK {cout<<"ok"<<endl;return;}
inline int Ls(int p){return p<<1;}
inline int Rs(int p){return p<<1|1;}
typedef long long LL;
typedef unsigned long long ULL;
const int MOD=1e9+7;
const int N=1e7+10,M=9999973;
using namespace std;

inline void Solve()
{
    
}
int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //INIT();
    int _=1;
    cin>>_;
    while(_--){
        Solve(); 
    }
    return 0;
}

A. 不会写就伪造

直接打印就行，Ged_id 函数的变量都要用 long long。

c++

LL Get_id(LL n)
{
    LL seed=0x5DEECE66D,mask=(1ll<<48)-1;
    LL state=(n^seed)&mask;
    rep(_,0,5){
        state=(state*seed+0xB)&mask;
        state^=(state>>17);
        state=(state<<31)|(state>>17);
        state&=mask;
    }
    return (state>>16)&((1ll<<32)-1);
}
inline void Solve()
{
    int n;cin>>n;
    LL x=Get_id(n);
    cout<<"Start Running Processes"<<endl<<"------------------------------"<<endl;
    rep(i,0,3) cout<<"process "<<i<<" is running, id: "<<x+i<<endl;
    cout<<"process 2 finished."<<endl;
    cout<<"process 3 is running, id: "<<x+2<<endl;
    cout<<"process 1 finished."<<endl;
    cout<<"process 4 is running, id: "<<x+1<<endl;
    cout<<"process 0 finished."<<endl;
    cout<<"process 3 finished."<<endl;
    cout<<"process 4 finished."<<endl;
    cout<<"------------------------------"<<endl;
    cout<<"Processes Finished"<<endl;
}

python


def Get_id(n):
        seed=0x5DEECE66D
        mask=(1<<48)-1
        state=(n^seed)&mask
        for _ in range(5):
                state=(state*seed+0xB)&mask
                state^=(state>>17)
                state=(state<<31)|(state >> 17)
                state&=mask
        result=(state>>16)&((1<<32)-1)
        return result
def Solve():
        n=int(input())
        x=Get_id(n)
        print('Start Running Processes')
        print('------------------------------')
        for i in range(3): print('process {} is running, id: {}'.format(i,x+i))
        print('process 2 finished.')
        print('process 3 is running, id: {}'.format(x+2))
        print('process 1 finished.')
        print('process 4 is running, id: {}'.format(x+1))
        print('process 0 finished.')
        print('process 3 finished.')
        print('process 4 finished.')
        print('------------------------------')
        print('Processes Finished')
if __name__ == '__main__':
        t=1
        t=int(input())
        while t:
                Solve()
                t-=1

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.StringTokenizer;
public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws Exception {
        int t = 1;
        t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }
    public static long getId(int n) {
        long seed = 0x5DEECE66DL;
        long mask = (1L << 48) - 1;
        long state = (n ^ seed) & mask;
        for (int i = 0; i < 5; i++) {
            state = (state * seed + 0xBL) & mask;
            state ^= (state >> 17);
            state = (state << 31) | (state >> 17);
            state &= mask;
        }
        return ((state >> 16) & ((1L << 32) - 1));
    }

    public static void solve() throws IOException {
        int n = in.nextInt();
        long x = getId(n);
        out.println("Start Running Processes");
        out.println("------------------------------");
        for (int i = 0; i < 3; i++) {
            out.println(String.format("process %d is running, id: %d", i, x + i));
        }
        out.println("process 2 finished.");
        out.println(String.format("process 3 is running, id: %d", x + 2));
        out.println("process 1 finished.");
        out.println(String.format("process 4 is running, id: %d", x + 1));
        out.println("process 0 finished.");
        out.println("process 3 finished.");
        out.println("process 4 finished.");
        out.println("------------------------------");
        out.println("Processes Finished");
    }
    static class InputReader{
        private StringTokenizer st;
        private BufferedReader bf;
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }
        public String next() throws IOException{
            while(st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException{
            return bf.readLine();
        }

        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException{
             return Long.parseLong(next());
        }

        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException{
            return new BigDecimal(next());
        }
    } 
}

B. 美味生日茶

把加密函数倒过来写就行了，注意原来函数中类型是 unsigned int，运算中产生溢出相当于对 $2^{32}$ 取模，如果用其他类型变量需要考虑取模。

c++

void De(unsigned int* v)
{
    unsigned int v0 = v[0], v1 = v[1];
    unsigned int delta = 1312;
    unsigned int sum = delta*32;
    unsigned int k0 = 1, k1 = 3, k2 = 1, k3 = 4;
    for(int i=0;i<32;i++)
    {
        v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
        v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
        sum -= delta;
    }
    v[0] = v0, v[1] = v1;
}
inline void Solve()
{
    unsigned int v[4];
    cin>>v[0]>>v[1];
    De(v);
    cout<<v[0]<<" "<<v[1]<<endl;
}

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.StringTokenizer;
public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws Exception {
        long[] v = new long[2];
        v[0] = in.nextLong();
        v[1] = in.nextLong();
        decrypt(v);
        out.printf("%d %d\n", v[0], v[1]);
        out.close();
    }

    public static void decrypt(long[] v) {
        long MOD = (1L<<32);
        long v0 = v[0], v1 = v[1];
        int delta = 1312;
        int sum = delta * 32;
        int k0 = 1, k1 = 3, k2 = 1, k3 = 4;

        for (int i = 0; i < 32; i++) {
            v1 -= ((v0 << 4)%MOD + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3) %MOD;
            v1 = (v1%MOD + MOD)%MOD;
            v0 -= ((v1 << 4)%MOD + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1) %MOD;
            v0 = (v0%MOD + MOD)%MOD;
            sum -= delta;
        }
        v[0] = v0; v[1] = v1;
    }
    static class InputReader{
        private StringTokenizer st;
        private BufferedReader bf;
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }
        public String next() throws IOException{
            while(st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException{
            return bf.readLine();
        }

        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException{
             return Long.parseLong(next());
        }

        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException{
            return new BigDecimal(next());
        }
    } 
}

C. 114514,1919810

数 $i$ 在序列中的最后一个位置是 $\frac{i(i+1)}{2}$ ，很容易想到二分答案，找到 $\frac{i(i+1)}{2} \ge n$ 的最小 $i$ 就是答案。（或者找到 $\frac{i(i+1)}{2} \le n$ 的最大 $i$ ，最后判断 $\frac{i(i+1)}{2} \ne n$ 就加1。）

c++

inline void Solve()
{
    int n;
    cin>>n;
    int l=1,r=n;
    while(l<r){
        int m=l+r>>1;
        if(1ll*m*(m+1)/2>=n) r=m;
        else l=m+1;
    }
    cout<<l<<endl;
}

python

def Solve():
    n=int(input())
    l,r=1,n
    while l<r:
        m=(l+r+1)//2
        if m*(m+1)//2<=n: l=m
        else:r=m-1
    if n==l*(l+1)//2: print(l)
    else: print(l+1)
if __name__ == '__main__':
        t=1
        t=int(input())
        while t:
                Solve()
                t-=1

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    
    public static void main(String[] args) throws Exception {
        int t = 1;
        t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    public static void solve() throws IOException {
        int n = in.nextInt();
        int l = 1, r = n;
        while (l < r) {
            int mid = (l + r) / 2;
            long sum = (1L * mid * (mid + 1) )/ 2;
            if (sum < n) l = mid + 1;  else r = mid;
        }
        out.println(l);

    }

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;
        
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return bf.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException {
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException {
            return new BigDecimal(next());
        }
    }
}

D. IP黑名单

简单模拟，日志中含有 "Invalid" 直接封 IP，含有 "Failded" 的提取 IP 计算一下出现的次数，超过 m 才禁封。c++ 中可以使用正则匹配提取 IP，哈希表计算 IP 出现的次数，禁封过的 IP 加入集合中，每次要禁封前判断一下即可。

c++

void Solve()
{
    int n,m;cin>>n>>m;
    unordered_set<string>st;
    unordered_map<string,int>mp;
    regex ip_re(R"(\b\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\b)");
    smatch ip;
    rep(_,0,n){
        string s;
        while(getline(cin,s)&&s=="");
        if(s.find("Invalid")!=string::npos){
            regex_search(s,ip,ip_re);
            if(st.find(ip[0])!=st.end()) continue;
            st.insert(ip[0]);
            cout<<"sshd:"<<ip[0]<<endl;
            continue;
        }
        if(s.find("Failed")!=string::npos){
            regex_search(s,ip,ip_re);
            if(st.find(ip[0])!=st.end()) continue;
            mp[ip[0]]+=1;
            if(mp[ip[0]]>m){
                st.insert(ip[0]);
                cout<<"sshd:"<<ip[0]<<endl;
            }
        }
    }
}

python

import re
def Solve():
        n,m=map(int,input().split())
        ban={}
        ctip={}
        for _ in range(n):
                line=input()
                gp=re.search(r'Invalid user \w+ from (\d+\.\d+\.\d+\.\d+)',line)
                if gp and not ban.get(gp[1]):
                        ban[gp[1]]=1
                        print('sshd:{}'.format(gp[1]))
                gp=re.search(r'Failed password for \w+ from (\d+\.\d+\.\d+\.\d+)',line)
                if gp and not ban.get(gp[1]):
                        ip=gp[1]
                        ctip[ip]=ctip.get(ip,0)+1
                        if ctip[ip]>m:
                                del ctip[ip]
                                ban[ip]=1
                                print('sshd:{}'.format(ip))
if __name__ == '__main__':
        t=1
        #t=int(input())
        while t:
                Solve()
                t-=1

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.util.regex.*;

public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    
    public static void main(String[] args) throws Exception {
        int t = 1;
        // t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    public static void solve() throws IOException {
        int n = in.nextInt();
        int m = in.nextInt();
        Set<String> st = new HashSet<>();
        Map<String, Integer> hs = new HashMap<>();
        Pattern ipRe = Pattern.compile("\\b\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\b");

        for (int i = 0; i < n; i++) {
            String s;
            while ((s = in.nextLine()) != null && s.equals(""));
            if (s.contains("Accepted")) continue;
            Matcher ma = ipRe.matcher(s);
            if (s.contains("Invalid")) {
                if (ma.find()) {
                    String ip = ma.group();
                    if (st.contains(ip)) continue;
                    st.add(ip);
                    out.println("sshd:" + ip);
                }
            }
            else if (s.contains("Failed")) {
                if (ma.find()) {
                    String ip = ma.group();
                    if (st.contains(ip)) continue;
                    hs.put(ip, hs.getOrDefault(ip, 0) + 1);
                    if (hs.get(ip) > m) {
                        st.add(ip);
                        out.println("sshd:" + ip);
                    }
                }
            }
        }
    }

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;
        
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return bf.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException {
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException {
            return new BigDecimal(next());
        }
    }
}

E. 桨声灯影

当 n 为素数时，因数只有 1 和 n， $S_n=n$ ，显然满足条件。

当 n 为和数时， $k> 2$ ，设 p 为 n 的最小素因子，有 $S_n > d_{k-1}d_k=\frac{n}{p} \cdot n=\frac{n^2}{p}$ 。假设 $S_n \mid n^2$ ，则有 $\frac{n^2}{S_n} < p$ ，而且注意到 $S_n=\sum_{i=1}^{k-1}d_id_{i+1}=n^2 \sum_{i=1}^{k-1}\frac{1}{d_id_{i+1}} < n^2$ ，故 $\frac{n^2}{S_n}$ 为 $n^2$ 的一个大于 1 且比 p 小的因子，与前提 p 为 $n^2$ 的最小素因子矛盾，合数情况不满足。

写一个线性素数筛即可通过所有数据。

c++

const int N=1e7+10,M=9999973;
using namespace std;
int v[N],pr[N],cnt;
void INIT(int n)
{
    rep(i,2,n+1){
        if(!v[i]) pr[++cnt]=i;
        rep(j,1,cnt+1){
            if(i*pr[j]>n) break;
            v[i*pr[j]]=1;
            if(i%pr[j]==0) break;
        }
    }
}
void Solve()
{
    int n;cin>>n;
    rep(i,2,n+1) if(v[i]) cout<<"NO"<<endl;else cout<<"YES"<<endl;
}
int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    INIT(10000000);
    int _=1;
    //cin>>_;
    while(_--){
        Solve(); 
    }
    return 0;
}

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.StringTokenizer;
public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static final int N = (int) 1e7;
    static int[] vis = new int[N+5];
    static int[] primes = new int[N+5];
    public static void main(String[] args) throws Exception {
        int t = 1;
        //t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }
    public static void solve() throws IOException {
        int n = in.nextInt(), cnt = 0;
        for (int i = 2; i <= N; i++) {
            if (vis[i] == 0) primes[++cnt] = i;
            for (int j = 1; j <= cnt && i * primes[j] <= N; j++) {
                vis[i * primes[j]] = 1;
                if (i % primes[j] == 0) break;
            }
        }
        for (int i = 2; i <= n; i++) {
            if (vis[i] != 0) out.println("NO"); else out.println("YES");
        }
    }
    static class InputReader{
        private StringTokenizer st;
        private BufferedReader bf;
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }
        public String next() throws IOException{
            while(st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException{
            return bf.readLine();
        }

        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException{
             return Long.parseLong(next());
        }

        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException{
            return new BigDecimal(next());
        }
    } 
}

F. 文件系统树

构造的链的数据可以卡掉 dfs 的做法，考虑拓扑排序，建立父节点到子节点的有向边，记录出度，把出度为 0 的子节点加入队列中，往上更新权值的同时删边，中途把产生的新子节点加入队列中，直到只剩下根节点为止。

c++

const int N=1e6+10,M=9999973;
using namespace std;
int fa[N],d[N];
LL w[N];
void Solve()
{
    int n;
    cin>>n;
    int rt=0;
    rep(i,0,n){
        int a,b,c;
        cin>>a>>b>>c;
        w[a]=c;
        if(b==-1) rt=a;
        else{
            fa[a]=b;++d[b];
        }
    }
    queue<int>q;
    rep(i,0,n) if(d[i]==0) q.push(i);
    while(q.size()){
        int t=q.front();q.pop();
        if(t==rt) continue;
        int u=fa[t];
        w[u]+=w[t];
        if(--d[u]==0) q.push(u);
    }
    rep(i,0,n) cout<<w[i]<<" ";
}

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static final int N = (int) 1e6 + 10;
    static int[] fa = new int[N];
    static int[] deg = new int[N];
    static long[] val = new long[N];
    public static void main(String[] args) throws Exception {
        int t = 1;
        //t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }
    public static void solve() throws IOException {
        int n = in.nextInt();
        for (int i = 0; i < n; i++) {
            int a = in.nextInt(), b = in.nextInt(), c = in.nextInt();
            val[a] = c; fa[a] = b;
            if (b != -1) ++deg[b];
        }
        Queue<Integer> que = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (deg[i] == 0) que.offer(i);
        }
        while (!que.isEmpty()) {
            int v = que.poll(), u = fa[v];
            if (u == -1) break;
            val[u] += val[v];
            if (--deg[u] == 0) que.offer(u);
        }
        for (int i = 0; i < n; i++) out.print(val[i] + " ");
    }
    static class InputReader{
        private StringTokenizer st;
        private BufferedReader bf;
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }
        public String next() throws IOException{
            while(st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException{
            return bf.readLine();
        }

        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException{
             return Long.parseLong(next());
        }

        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException{
            return new BigDecimal(next());
        }
    } 
}

G. 10cm距离

不妨设 $a_i \le a_j$ ，则有 $a_j-a_i \ge b_i+b_j$ ，即 $a_j-b_j \ge b_i+a_i$ 。对于每一个节点 i 都维护出 $l_i=a_i-b_i,r_i=a_i+b_i$ 两个值。把节点信息看作线段 $[l_i,r_i]$ ，两者不干扰条件就可以表示成 $l_j \ge r_i$ ，可以发现只有线段不相交（包含端点相交的情况）才能满足 $l_j \ge r_i$ 。有人问如果两个线段不相交且 $l_i > r_j$ 也不满足条件，但是注意根据前提 $a_i \le a_j$ 有 $l_i \le r_j$ ，前提条件下并不会出现这种情况，只需要保证尽可能多的线段不相交（包含端点相交）即可。

问题转换为经典贪心问题做法，按照右端点的值从小到大排序，优先选择右端点的值小的。

c++

const int N=1e6+10;
using namespace std;
pii a[N];
bool cmp(pii a,pii b)
{
    return a.se<b.se;
}
void Solve()
{
    int n;cin>>n;
    rep(i,1,n+1){
        int x,y;cin>>x>>y;
        a[i].fi=x-y,a[i].se=x+y;
    }
    sort(a+1,a+1+n,cmp);
    int ans=0,r=-2e9;
    rep(i,1,n+1){
        if(a[i].fi>=r) ++ans,r=a[i].se;
    }
    cout<<ans<<endl;
}

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static final int N = (int) 1e6 + 10;
    static class Pair implements Comparable<Pair> {
        int x, y;
        Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
         public int compareTo(Pair to) {
             return Integer.compare(y, to.y);
         }
    }
    static Pair[] p = new Pair[N];
    public static void main(String[] args) throws Exception {
        int t = 1;
        //t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }
    public static void solve() throws IOException {
        int n = in.nextInt();
        for (int i = 0; i < n; i++) {
            int x = in.nextInt(), y = in.nextInt();
            p[i] = new Pair(x - y, x + y);
        }
        Arrays.sort(p, 0, n);
        int ans = 0;
        int now = (int) -2e9;
        for (int i = 0; i < n; i++) {
            if (p[i].x >= now) {
                ans++;
                now = p[i].y;
            }
        }
        out.println(ans);
    }
    static class InputReader{
        private StringTokenizer st;
        private BufferedReader bf;
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }
        public String next() throws IOException{
            while(st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException{
            return bf.readLine();
        }

        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException{
             return Long.parseLong(next());
        }

        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException{
            return new BigDecimal(next());
        }
    } 
}

H. 退避时间管理带师

直接计算不现实，考虑计算 n 个数当中时间为 i 的个数，若 x 的时间为 i 需要满足 $j(1 \le j \le i-1) \mid x,i \nmid x$ ，相当于求满足 $\left [ 1,2,...,i-1\right ] \mid x,\left [ 1,2,...,i\right ] \nmid x$ 的个数（ $\left [ \ \right ]$ 表示最小公倍数），即为 $\left \lfloor \frac{n}{\left [ 1,2,...,i-1 \right ]} \right \rfloor - \left \lfloor \frac{n}{\left [ 1,2,...,i \right ]} \right \rfloor$ ，先预处理出 1~i 前缀最小公倍数，由于最小公倍数增长很快，处理到40到50就可以了。

c++

const int MOD=998244353;
const int N=1e6+10;
using namespace std;
LL s[45];
LL Gcd(LL a,LL b)
{
    while(b^=a^=b^=a%=b);
    return a;
}
void INIT()
{
    s[1]=1;
    rep(i,2,41) s[i]=i/Gcd(s[i-1],i)*s[i-1];
}
void Solve()
{
    LL n,res=0;cin>>n;
    for(int i=2;s[i-1]<=n;++i){
        res=(res+i*(n/s[i-1]-n/s[i])%MOD)%MOD;
    }
    cout<<res%MOD<<endl;
}
int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int _=1;
    cin>>_;
    INIT();
    while(_--){
        Solve(); 
    }
    return 0;
}

java

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
public class Main {
    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static final int N = (int) 1e6 + 10;
    static final int MOD = 998244353;
    static long[] lcm = new long[50];
    public static void main(String[] args) throws Exception {
        int t = 1;
        t = in.nextInt();
        initial();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }
    static long gcd(long a, long b) {
        while (b != 0) {
            long t = b;
            b = a % b;
            a = t;
        }
        return a;
    }
    static void initial() {
        lcm[1] = 1;
        for (int i = 2; i < 50; ++i) {
            lcm[i] = i / gcd(lcm[i - 1], i) * lcm[i - 1];
        }
    }
    public static void solve() throws IOException {
        long n = in.nextLong();
        long ans=0;
        for (int i = 2; lcm[i-1] <= n; ++i) {
            ans = (ans + i * ((n / lcm[i-1]) - (n / lcm[i])) % MOD) % MOD;
        }
        out.println(ans % MOD);
    }
    static class InputReader{
        private StringTokenizer st;
        private BufferedReader bf;
        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }
        public String next() throws IOException{
            while(st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException{
            return bf.readLine();
        }

        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException{
             return Long.parseLong(next());
        }

        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() throws IOException{
            return new BigDecimal(next());
        }
    } 
}

I. Packet Subset Selection

相当于从集合 $A = \{1,2,3,...,2p\}$ 中选择 p 个元素，记选择的 p 个元素为 $1 \le x_1 < x_2 ... < x_p \le 2p$ ，求满足 $\sum_{x=1}^{p}x_i \equiv 0 \pmod{p}$ 的解的个数。

考虑将一个合法解按照 p 为分界做划分，即：

$1 \le x_1 < x_2 < ... < x_t \le p < x_{t+1} < ... < x_p \le 2p$

下面可以分为三种情况：

（1）如果 $t=0$ ，即所有数都比 p 大，显然存在唯一解： $x_i=i+p$

（2）如果 $t=p$ ，即所有数都小于等于 p ，显然存在唯一解： $x_i=i$

（3）剩下 $1\le t \le p-1$ ，含 p 个元素的子集总个数是 $\binom{2p}{p}$ ，除去前面两种合法的情况就是

$\binom{2p}{p} -2$ ，考虑把子集再次做划分，按照下面的规则划分为不相交的等价类：

定义两个子集 $\{x_1,x_2,...,x_p\}$ 和 $\{x^{'}_1,x^{'}_2,...,x^{'}_p\}$ 属于同一个等价类，当且仅当它们的 $t$ 划分相等，当 $t+1 \le i \le p$ 时满足 $x^{'}_i = x_i$ ；当 $1 \le i \le t$ 时，满足

$\exists m,x^{'}_i \equiv x_i + m \pmod{p}$

这样，每一个等价类按照 m 的取值都可以表示 p 个子集，等价类的总个数就是 $\frac{\binom{2p}{p} -2}{p}$ 。

考虑某一个等价类的元素和，选择其中的一个子集 $\{x_1,x_2,...,x_p\}$ ，记 $s=\sum_{i=1}^{p} x_i$ ，这个等价类模 p 的和就可以表示成 $s+m \cdot t \pmod{p}$ ，则需要满足条件 $s+m \cdot t \equiv 0 \pmod{p}$ ，对于固定了 $s$ 和 $t$ 的等价类，p 为奇素数，该方程有唯一解为 $m \equiv (p-s)\cdot t^{p-2} \pmod{p}$ 。所以满足条件解的个数就是等价类的总个数。

综上所述，总方案个数就是 $\frac{\binom{2p}{p} -2}{p} +2$ 。

预处理阶乘逆元计算组合数，加上卢卡斯定理即可通过。

const int N=1e7+10,M=9999973;
using namespace std;
int fac[N],inv[N];
int qp(int a,int b)
{
    int res=1;
    while(b){
        if(b&1) res=1ll*res*a%M;
        a=1ll*a*a%M;
        b>>=1;
    }
    return res%M;
}
void INIT()
{
    int n=10000000;
    fac[0]=inv[0]=1;
    rep(i,1,n+1) fac[i]=1ll*i*fac[i-1]%M,inv[i]=1ll*inv[i-1]*qp(i,M-2)%M;
}
int C(LL a,LL b)
{
    if(b>a) return 0;
    if(a==b||!b) return 1;
    if(a<M&&b<M){
        return 1ll*fac[a]*inv[b]%M*inv[a-b]%M;
    }
    return 1ll*C(a/M,b/M)*C(a%M,b%M)%M;
}
int sfmod(int x)
{
    return (x%M+M)%M;
}
void Solve()
{
    LL p;cin>>p;
    int ans=C(2*p,p)-2;ans=sfmod(ans);
    ans=1ll*ans*qp(p%M,M-2)%M;ans+=2;ans=sfmod(ans);
    cout<<ans<<endl;
}
int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int _=1;
    cin>>_;
    INIT();
    while(_--){
        Solve(); 
    }
    return 0;
}