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牛客周赛 Round 114 Java题解

news 2025/11/9 6:53:49

目录

一、A题 小彩找数

二、B题 小彩的好字符串

三、C题 小彩的字符串交换

四、D题 小彩的数组选数

五、E题 小彩的数组构造

六、F题 小彩的好数构

一、A题 小彩找数

枚举 ，可能写的有点复杂

import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;public class Main {static IoScanner sc = new IoScanner();
//    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 {// todoint arr[]=new int[4];int a=sc.nextInt();if(a==1){arr[1]=1;}else if(a==2){arr[2]=1;}else if(a==3){arr[3]=1;}int b=sc.nextInt();if(b==1){arr[1]=2;}else if(b==2){arr[2]=2;}else if(b==3){arr[3]=2;}int c=sc.nextInt();if(c==1){arr[1]=3;}else if(c==2){arr[2]=3;}else if(c==3){arr[3]=3;}for (int i = 1; i < arr.length; i++) {System.out.print(arr[i]+" ");}}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}}/*** 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());}
}

二、B题 小彩的好字符串

暴力，判断 3 的倍数的长度的 子串 即可

import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;public class Main {static IoScanner sc = new IoScanner();
//    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 {int n=sc.nextInt();String str=sc.next();long cnt=0;for(int i=0;i<n;i++){for(int j=i+2;j<n;j++){String substring = str.substring(i, j + 1);if(substring.length()%3!=0){continue;}int cnt1=0;int cnt2=0;int cnt3=0;for(int k=0;k<substring.length();k++){if(substring.charAt(k)=='1'){cnt1++;}else if(substring.charAt(k)=='2'){cnt2++;}else if(substring.charAt(k)=='3'){cnt3++;}}if(cnt1==cnt2 && cnt2==cnt3){cnt++;}}}System.out.println(cnt);}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}}/*** 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());}
}

三、C题 小彩的字符串交换

因为最多只需要操作 1 次

所以其实也就三种答案 -1 0 1

先检查 如果没有 1 2 3 这三个数 结果是 -1

再检查直接存在 123 靠一起的情况 结果是 0

剩下的结果都是 1

import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;public class Main {static IoScanner sc = new IoScanner();
//    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 {// todoint n=sc.nextInt();String str=sc.next();if(count(str)==false){System.out.println("-1");return;}for(int i=0;i<n-3;i++){String substring = str.substring(i, i + 3);if(count(substring)==true){System.out.println("0");return;}}System.out.println("1");}static boolean count(String str){int cnt1=0;int cnt2=0;int cnt3=0;for (int i = 0; i < str.length(); i++) {char c=str.charAt(i);if(c=='1'){cnt1++;}else if(c=='2'){cnt2++;}else if(c=='3'){cnt3++;}}
//        System.out.println(cnt1+" "+cnt2+" "+cnt3);if(cnt1!=0&&cnt2!=0&&cnt3!=0){return true;}return false;}public static void main(String[] args) throws Exception {int t = 1;t = sc.nextInt();while (t-- > 0) {solve();}}}/*** 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());}
}

四、D题 小彩的数组选数

动态规划

dp[] 数组的值为当前最大贪心值

import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;public class Main {static IoScanner sc = new IoScanner();
//    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 {// todoint n=sc.nextInt();long arr[]=new long[n];for (int i = 0; i < n; i++) {arr[i]=sc.nextLong();}// dp[i]表示到第i个元素的最大贪心值long dp[]=new long[n+1];//        long max=0;dp[1]=arr[0];for (int i = 2; i <= n; i++) {// 不操作long num1 = dp[i - 1];// 操作long num2 = dp[i - 2] + arr[i - 1];dp[i] = Math.max(num1, num2);}System.out.println(dp[n]);}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}}/*** 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());}
}

五、E题 小彩的数组构造

因为每个子数组肯定都是 1 的倍数

所以由 a 可知 n

即构造数组的长度为 a+2

我们可以一位一位的构造 这样就能每次构造出一个子数组

我们的目的是一步步消耗 b 和 c 的值

构造和为 2 的倍数的子数组但不是 3 的倍数的数的话 b--

构造和为 3 的倍数的子数组但不是 2 的倍数的数的话 c--

构造和为 6 的倍数的子数组的话 b-- c--

构造和为 5 的倍数的子数组 不变

最后要使 b 和 c 得为 0

import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;public class Main {static IoScanner sc = new IoScanner();
//    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 {// todolong a=sc.nextInt();long b=sc.nextInt();long c=sc.nextInt();if(b>a||c>a){System.out.println("-1");return;}long n=a+2;long arr[]=new long[(int)n];arr[0]=2;arr[1]=3;for(int i=2;i<n;i++){if(a==b+c){// 构造和为2的倍数的子数组但不是3的倍数的数if (b != 0) {long sumPrev = arr[i - 1] + arr[i - 2];long total = sumPrev;// 找到第一个是2的倍数且不是3的倍数的totaldo {total++;} while (total % 2 != 0 || total % 3 == 0);arr[i] = total - sumPrev;a--;b--;}// 构造和为3的倍数的子数组但不是2的倍数的数else if (c != 0) {long sumPrev = arr[i - 1] + arr[i - 2];long base = (sumPrev / 3 + 1) * 3;long ans = (base % 2 == 0) ? base + 3 : base;arr[i] = ans - sumPrev;a--;c--;}}else if(a<b+c){// 构造和为6的子数组long ans = (((arr[i-1]+arr[i-2])/6)+1)*6;arr[i]=ans-arr[i-1]-arr[i-2];a--;b--;c--;}else if (a > b + c) {// 构造和为5的倍数，但不是2或3的倍数的子数组long sumPrev = arr[i - 1] + arr[i - 2];long base = ((sumPrev / 5) + 1) * 5;while (base % 2 == 0 || base % 3 == 0) {base += 5;}arr[i] = (int) (base - sumPrev);a--;}}System.out.println(n);StringBuilder sb=new StringBuilder();for (long i : arr) {sb.append(i).append(" ");}System.out.println(sb);}public static void main(String[] args) throws Exception {int t = 1;
//        t = sc.nextInt();while (t-- > 0) {solve();}}}/*** 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());}
}

六、F题 小彩的好数构

构造题

我是打表打出来的

这是结果

如果 n 是偶数

构造如下 1000000001 1212121212

如果 n 是奇数

构造如下 10000000001 12121212312

import java.math.BigInteger;
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.time.*;public class Main {static IoScanner sc = new IoScanner();
//    static final int mod = (int) (1e9 + 7);static int n;static int arr[];static boolean visited[];static ArrayList<ArrayList<Integer>> adj = new ArrayList<>();public static void main(String[] args) throws IOException {//        long testStart = 100000000;
//        long testEnd = 999999999;
//        System.out.println("开始搜索...（范围：" + testStart + " ~ " + testEnd + "）");
//        long count = 0;
//        for (long a = testStart; a <= testEnd; a++) {
//            for (long b = a; b <= testEnd; b++) {
//                count++;
//                BigInteger product = BigInteger.valueOf(a).multiply(BigInteger.valueOf(b));
//                if (checkProduct(product)==true) {
//                    System.out.println(a + " × " + b + " = " + product);
//                    return;
//                }
//            }
//        }int n = sc.nextInt();if (n == 1) {System.out.println("-1");return;}StringBuilder sb = new StringBuilder();sb.append(1);for (int i = 0; i < n - 2; i++) {sb.append(0);}sb.append(1);String num1=sb.toString();if(n%2==0){sb = new StringBuilder();for (int i = 0; i < n; i+=2) {sb.append("12");}String num2 = sb.toString();System.out.println(num1+" "+num2);}else {if(n==3){System.out.println(num1+" 131");}else{sb = new StringBuilder();for (int i = 0; i < n-3; i+=2) {sb.append("12");}sb.append("312");String num2 = sb.toString();System.out.println(num1+" "+num2);}}}// 检查乘积是否符合条件（仅含1、2、3，无连续相同字符，且同时存在1、2、3）private static boolean checkProduct(BigInteger product) {String numStr = product.toString();boolean has1 = false;boolean has2 = false;boolean has3 = false;for (char c : numStr.toCharArray()) {if (c < '1' || c > '3') {return false;}if (c == '1') {has1 = true;} else if (c == '2') {has2 = true;} else if (c == '3') {has3 = true;}}if (!has1 || !has2 || !has3) {return false;}for (int i = 0; i < numStr.length() - 1; i++) {if (numStr.charAt(i) == numStr.charAt(i + 1)) {return false;}}return true;}
}/*** 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());}
}
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