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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

// @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 {// todolong 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

// @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 {// todolong 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

// @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 {// todoint 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

// @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 {// todolong 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;}// gcdz=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

// @github https://github.com/Dddddduo
// @github https://github.com/Dddddduo/acm-java-algorithm
// @github https://github.com/Dddddduo/Dduo-mini-data_structureimport 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 {// todoint 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