牛客周赛84 题解 Java ABCDE 仅供参考

多多*2025-03-27 9:59

A 小苯跑外卖

除一下看有没有余数

有余数得多一天

没余数正好

复制代码 
// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;

/**
 * 题目地址
 * 牛客周赛 Round 86 A
 */

// xixi♡西
public class Main {

    static IoScanner sc = new IoScanner();
    static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);

    static int n;
    static int arr[];
    static boolean visited[];
    static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();

    /**
     * @throws IOException
     */
    private static void solve() throws IOException {
        // todo
        long n=sc.nextLong();
        long m=sc.nextInt();
        if(m%n==0){
            dduoln(m/n);
        }else{
            dduoln(m/n+1);
        }
        
    }

    public static void main(String[] args) throws Exception {
        int t = 1;
//        t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static <T> void dduo(T t) {
        System.out.print(t);
    }

    static <T> void dduoln() {
        System.out.println("");
    }

    static <T> void dduoln(T t) {
        System.out.println(t);
    }
}

/**
 * IoScanner类
 *
 * @author Dduo
 * @version 1.0
 * @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。
 */
class IoScanner {
    BufferedReader bf;
    StringTokenizer st;
    BufferedWriter bw;

    public IoScanner() {
        bf = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

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

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

    public char nextChar() throws IOException {
        return next().charAt(0);
    }

    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 float nextFloat() throws IOException {
        return Float.parseFloat(next());
    }

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

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

B 小苯的区间删除

一次删一个元素(选择一个长度为 1 的区间)

把负数全删掉

求和

复制代码 
// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;

/**
 * 题目地址
 * 牛客周赛 Round 86 B
 */

// xixi♡西
public class Main {

    static IoScanner sc = new IoScanner();
    static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);

    static int n;
    static int arr[];
    static boolean visited[];
    static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();

    /**
     * @throws IOException
     */
    private static void solve() throws IOException {
        // todo
        long n=sc.nextLong();
        long k=sc.nextInt();

        long arr[]=new long[(int) n];
        for (long i = 0; i < n; i++) {
            arr[(int) i]=sc.nextLong();
        }
        long sum=0;
        for (long l : arr) {
            if(l>0)sum+=l;
        }
        dduoln(sum);
    }

    public static void main(String[] args) throws Exception {
        int t = 1;
        t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static <T> void dduo(T t) {
        System.out.print(t);
    }

    static <T> void dduoln() {
        System.out.println("");
    }

    static <T> void dduoln(T t) {
        System.out.println(t);
    }
}

/**
 * IoScanner类
 *
 * @author Dduo
 * @version 1.0
 * @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。
 */
class IoScanner {
    BufferedReader bf;
    StringTokenizer st;
    BufferedWriter bw;

    public IoScanner() {
        bf = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

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

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

    public char nextChar() throws IOException {
        return next().charAt(0);
    }

    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 float nextFloat() throws IOException {
        return Float.parseFloat(next());
    }

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

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

C 小苯的数字消除

这题如果要考虑对字符 串 处理 我感觉是很难的

我们不难想到可以用双端队列

如果队列尾部的元素跟要添加的元素一样

那么不用添加了 队尾的元素也出来吧

如果队列尾部的元素跟要添加的元素不一样

那么把这个元素到队列尾部

最后得到的队列肯定是 01010101 101010... 类似的

满足消除相同的

之后再对队列长度/2 即可得出答案

复制代码 
// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;

/**
 * 题目地址
 * 牛客周赛 Round 86 C
 */

// xixi♡西
public class Main {

    static IoScanner sc = new IoScanner();
    static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);

    static int n;
    static int arr[];
    static boolean visited[];
    static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();

    static ArrayList<Integer>list=new ArrayList<>();
    /**
     * @throws IOException
     */
    private static void solve() throws IOException {
        // todo
        int n=sc.nextInt();
        String str=sc.next();

        Deque<Character>queue= new LinkedList<>();
        queue.add(str.charAt(0));
        for(int i=1;i<str.length();i++){
            if(queue.isEmpty()){
                queue.addLast(str.charAt(i));
                continue;
            }
            if(queue.getLast().equals(str.charAt(i))){
//                dduoln(i);
                queue.removeLast();
            }else{
                queue.addLast(str.charAt(i));
            }
        }

//        dduoln(queue.size());
        dduoln(queue.size()/2);

    }

    // 计算两个数的最大公约数
    public static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    // 计算两个数的最小公倍数
    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static void main(String[] args) throws Exception {
        int t = 1;
        t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static <T> void dduo(T t) {
        System.out.print(t);
    }

    static <T> void dduoln() {
        System.out.println("");
    }

    static <T> void dduoln(T t) {
        System.out.println(t);
    }
}

/**
 * IoScanner类
 *
 * @author Dduo
 * @version 1.0
 * @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。
 */
class IoScanner {
    BufferedReader bf;
    StringTokenizer st;
    BufferedWriter bw;

    public IoScanner() {
        bf = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

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

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

    public char nextChar() throws IOException {
        return next().charAt(0);
    }

    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 float nextFloat() throws IOException {
        return Float.parseFloat(next());
    }

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

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

D 小苯的数字集合

首先我们知道的是 两个数如果相同 异或运算得到的数为 0

那么如果现在集合里面有两个一样的数 那么只要再操作一次

之后就不难想到

最多操作 3 次就得出 0

因为前两次我们完全可以进行同样的两次运算

得到两个一样的数 然后第三次运算异或

所以最多经过 3 轮就能得出 0

之后我们完全可以枚举

如果要经过 1 轮得出答案 有 4 种组合 看一下这些组合有没有得出 0

如果要经过 2 轮得出答案 有 4*4=16 种组合 看一下这些组合有没有得出 0

复制代码 
// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;

/**
 * 题目地址
 * 牛客周赛 Round 86 D
 */

// xixi♡西
public class Main {

    static IoScanner sc = new IoScanner();
    static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);

    static int n;
    static int arr[];
    static boolean visited[];
    static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();

    static ArrayList<Integer>list=new ArrayList<>();
    /**
     * @throws IOException
     */
    private static void solve() throws IOException {
        // todo
        long x = sc.nextLong();
        long y = sc.nextLong();

        if (x == 0 || y == 0) {
            dduoln(0);
            return;
        } else if ( (x & y) == 0 || (x | y) == 0 || (x ^ y) == 0 || gcd(x,y)==0) {
            dduoln(1);
            return;
        }
        
        long z=0;

        // &
        z=x&y;
        if(judge(x,z)||judge(x,y)||judge(z,y)){
            dduoln("2");
            return;
        }

        // |
        z=x|y;
        if(judge(x,z)||judge(x,y)||judge(z,y)){
            dduoln("2");
            return;
        }

        // ^
        z=x^y;
        if(judge(x,z)||judge(x,y)||judge(z,y)){
            dduoln("2");
            return;
        }

        // gcd
        z=gcd(x,y);
        if(judge(x,z)||judge(x,y)||judge(z,y)){
            dduoln("2");
            return;
        }

        dduoln(3);

    }

    public static boolean judge(long x,long y){
        if((x&y)==0||(x|y)==0||(x^y)==0||gcd(x,y)==0){
            return true;
        }else return false;
    }

    // 计算两个数的最大公约数
    public static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    // 计算两个数的最小公倍数
    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static void main(String[] args) throws Exception {
        int t = 1;
        t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static <T> void dduo(T t) {
        System.out.print(t);
    }

    static <T> void dduoln() {
        System.out.println("");
    }

    static <T> void dduoln(T t) {
        System.out.println(t);
    }
}

/**
 * IoScanner类
 *
 * @author Dduo
 * @version 1.0
 * @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。
 */
class IoScanner {
    BufferedReader bf;
    StringTokenizer st;
    BufferedWriter bw;

    public IoScanner() {
        bf = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

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

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

    public char nextChar() throws IOException {
        return next().charAt(0);
    }

    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 float nextFloat() throws IOException {
        return Float.parseFloat(next());
    }

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

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

E 小苯的Polygon

封闭凸多边形内角都小于 180 度

首先要知道 要构成封闭凸多边形

那么得最长那条边比其余的边加起来要小就行

即 max<sum

所以我们先对数组排序

因为数据量很小 我考虑的暴力了一点 直接枚举了每条边都是最长边的情况

然后 就是很经典 dp 模版 选一些边要求和大于 sum 而且这些和要最小

复制代码 
// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structure

import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;

/**
 * 题目地址
 * 牛客周赛 Round 86 E
 */

// xixi♡西
public class Main {

    static IoScanner sc = new IoScanner();
    static final int mod = (int) (1e9 + 7);
//    static final int mod = (int) (1e9 + 7);

    static int n;
    static int arr[];
    static boolean visited[];
    static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();

    /**
     * @throws IOException
     */
    private static void solve() throws IOException {
        // todo
        int n = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        Arrays.sort(a);
        int min = Integer.MAX_VALUE;

        for (int i = 0; i < n; i++) {
            int max = a[i];

            if (max >= min) {
                break;
            }
            List<Integer> list = new ArrayList<>();
            for (int j = 0; j < i; j++) {
                list.add(a[j]);
            }
            if (list.isEmpty()) {
                continue;
            }
            int sum = 0;
            for (int x : list) {
                sum += x;
            }
            if (sum <= max) {
                continue;
            }
            boolean[] dp = new boolean[sum + 5];
            dp[0] = true;

            // 当前最大值
            int c = 0;

            for (int x : list) {
                for (int j = c; j >= 0; j--) {
                    if (dp[j] == true) {
                        int newSum = j + x;
                        if (newSum > sum) {
                            continue;
                        }
                        if (dp[newSum] == false) {
                            dp[newSum] = true;
                            if (newSum > c) {
                                c = newSum;
                            }
                        }
                    }
                }
            }

            for (int sumOther = max + 1; sumOther <= c; sumOther++) {
                if (dp[sumOther]) {
                    int perimeter = sumOther + max;
                    if (perimeter < min) {
                        min = perimeter;
                    }
                    break;
                }
            }
        }
        dduoln(min == 2147483647 ? "-1" : min);
    }


    public static void main(String[] args) throws Exception {
        int t = 1;
        t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static <T> void dduo(T t) {
        System.out.print(t);
    }

    static <T> void dduoln() {
        System.out.println("");
    }

    static <T> void dduoln(T t) {
        System.out.println(t);
    }

}


/**
 * IoScanner类
 *
 * @author Dduo
 * @version 1.0
 * @description 通过IO流操作缓冲区减少了与底层输入输出设备的交互次数，旨在简化 Java 中的标准输入读取操作。
 */
class IoScanner {
    BufferedReader bf;
    StringTokenizer st;
    BufferedWriter bw;

    public IoScanner() {
        bf = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

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

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

    public char nextChar() throws IOException {
        return next().charAt(0);
    }

    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 float nextFloat() throws IOException {
        return Float.parseFloat(next());
    }

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

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

F 小苯的线性dp

不会

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