A. Codeforces Checking
题意
每个案例给一个字符，如果在 ”codeforces“ 中出现过，输出 YES，否则输出 NO
code
/**
* @author :Changersh
* @date : 2023/2/3 22:37
*/
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
private static boolean[] a = new boolean[26];
public static void main(String[] args) {
String s = "codeforces";
for (int i = 0; i < s.length(); i++)
a[s.charAt(i) - 'a'] = true;
int n = sc.nextInt();
for (int i = 0; i < n; i++) {
char t = sc.next().charAt(0);
if (a[t - 'a']) out.println("YES");
else out.println("NO");
}
out.close();
}
static class FastScanner{
// 看看有没有溢出，是否要用 long
// sc.xxx;
// out.print();
// out.flush();
// out.close();
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream in) {
br=new BufferedReader( new InputStreamReader(System.in));
eat("");
}
public void eat(String s) {
st=new StringTokenizer(s);
}
public String nextLine() {
try {
return br.readLine();
}catch(IOException e) {
return null;
}
}
public boolean hasNext() {
while(!st.hasMoreTokens()) {
String s=nextLine();
if(s==null)return false;
eat(s);
}
return true;
}
public String next() {
hasNext();
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static FastScanner sc=new FastScanner(System.in);
static PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
B. Following Directions
题意
每个案例给一串字符串，包含 U、D、L、R，是上下左右四个方向，从 (0， 0) 开始，问是否经过 （1， 1）
code
直接模拟即可
/**
* @author :Changersh
* @date : 2023/2/3 22:43
*/
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
private static int N = 55, n, T;
public static void main(String[] args) {
T = sc.nextInt();
while (T-- > 0) {
out.println(solve() ? "YES" : "NO");
}
out.close();
}
private static boolean solve() {
n = sc.nextInt();
char[] c = sc.next().toCharArray();
int x = 0, y = 0;
for (int i = 0; i < n; i++) {
if (c[i] == 'U') x++;
else if (c[i] == 'D') x--;
else if (c[i] == 'L') y--;
else y++;
if (x == 1 && y == 1) return true;
}
return false;
}
static class FastScanner {
// 看看有没有溢出，是否要用 long
// sc.xxx;
// out.print();
// out.flush();
// out.close();
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream in) {
br = new BufferedReader(new InputStreamReader(System.in));
eat("");
}
public void eat(String s) {
st = new StringTokenizer(s);
}
public String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
return null;
}
}
public boolean hasNext() {
while (!st.hasMoreTokens()) {
String s = nextLine();
if (s == null) return false;
eat(s);
}
return true;
}
public String next() {
hasNext();
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static FastScanner sc = new FastScanner(System.in);
static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
C. Prepend and Append
题意
给你一串由 0 1 组成的字符串，如果第一个和最后一个是 0 1或者 1 0 ，可以消掉，这种操作可以进行任意次，问，任意次操作后，字符串剩下的最短长度是多少？
code
双指针判断第一个和最后一个字符，模拟即可
/**
* @author :Changersh
* @date : 2023/2/3 22:51
*/
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
private static int T;
public static void main(String[] args) {
T = sc.nextInt();
while (T-- > 0)
solve();
out.close();
}
private static void solve() {
int n = sc.nextInt();
char[] c = sc.next().toCharArray();
int l = 0, r = n - 1;
while (l < r) {
if ((c[l] == '0' && c[r] == '1') || (c[l] == '1' && c[r] == '0')) {
l++;
r--;
}
else break;
}
out.println(r - l + 1);
}
static class FastScanner{
// 看看有没有溢出，是否要用 long
// sc.xxx;
// out.print();
// out.flush();
// out.close();
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream in) {
br=new BufferedReader( new InputStreamReader(System.in));
eat("");
}
public void eat(String s) {
st=new StringTokenizer(s);
}
public String nextLine() {
try {
return br.readLine();
}catch(IOException e) {
return null;
}
}
public boolean hasNext() {
while(!st.hasMoreTokens()) {
String s=nextLine();
if(s==null)return false;
eat(s);
}
return true;
}
public String next() {
hasNext();
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static FastScanner sc=new FastScanner(System.in);
static PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
D. Distinct Split
题意
每个案例给定一个由小写字母组成的字符串，求将字符串从中间劈开，分成两个字串中，出现的不同的字符数的和的最大值？
code
暴力写会tle，所以先预处理一下前缀和后缀，前缀是到 第 i 个字符截止，前面 i 个字符共出现多少个不同的字符
后缀：从
/**
* @author :Changersh
* @date : 2023/2/3 23:04
*/
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
private static int T, n;
private static char[] c;
public static void main(String[] args) {
T = sc.nextInt();
while (T-- > 0)
solve();
out.close();
}
private static void solve() {
n = sc.nextInt();
c = sc.next().toCharArray();
int ans = 0;
int[] l = new int[n + 2];
int[] r = new int[n + 2];
get(l, r);
for (int i = 0; i < n; i++) {
ans = Math.max(ans, l[i + 1] + r[i + 2]);
}
out.println(ans);
}
private static void get(int[] l, int[] r) {
int ans = 0;
HashSet<Character> vis = new HashSet<>();
for (int i = 0; i < n; i++) {
if (!vis.contains(c[i])) {
vis.add(c[i]);
ans++;
}
l[i + 1] = ans;
}
vis.clear();
ans = 0;
for (int i = n - 1; i >= 0; i--) {
if (!vis.contains(c[i])) {
vis.add(c[i]);
ans++;
}
r[i + 1] = ans;
}
}
static class FastScanner {
// 看看有没有溢出，是否要用 long
// sc.xxx;
// out.print();
// out.flush();
// out.close();
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream in) {
br = new BufferedReader(new InputStreamReader(System.in));
eat("");
}
public void eat(String s) {
st = new StringTokenizer(s);
}
public String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
return null;
}
}
public boolean hasNext() {
while (!st.hasMoreTokens()) {
String s = nextLine();
if (s == null) return false;
eat(s);
}
return true;
}
public String next() {
hasNext();
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static FastScanner sc = new FastScanner(System.in);
static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
E. Negatives and Positives
题意
给定一串数字，你可以进行任意次以下的操作：
选择两个不同的数字，变成相反数
求，进行任意次之后，数组数字和的最大值是多少？
code
数字有正数、负数、0
翻转的时候，最佳的方法是把所有的负数都变成正数，是最佳情况
- 负数个数是偶数，完美，全部都可以翻转成正数
- 负数个数是奇数：
1. 有 0，依然完美
2. 找绝对值最小的，取负即可
将第二种情况的两种小情况和一
遍历数组的时候，统计负数个数，并且把负数都翻转，求和
如果是偶数，返回答案
如果是奇数，说明不能完美翻转，数组排序，得到最小的数字，无论是正数还是负数。抑或是 0，减两次即可。
因为最小的如果是 负数翻转的，减两次相当于没有翻转
如果是正数，相当于把 落单的负数和绝对值小于它的正数翻转，和变大
如果是 0，也是完美的情况
/**
* @author :Changersh
* @date : 2023/2/4 9:20
*/
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
private static int T, n, N = 200010;
private static int[] a;
public static void main(String[] args) {
T = sc.nextInt();
while (T-- > 0)
solve();
out.close();
}
private static void solve() {
long sum = 0;
int cnt = 0;
n = sc.nextInt();
a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
if (a[i] < 0) {
cnt++;
a[i] = -a[i];
}
sum += a[i];
}
if ((cnt & 1) == 1) {
Arrays.sort(a);
sum -= 2 * a[0];
}
out.println(sum);
}
static class FastScanner{
// 看看有没有溢出，是否要用 long
// sc.xxx;
// out.print();
// out.flush();
// out.close();
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream in) {
br=new BufferedReader( new InputStreamReader(System.in));
eat("");
}
public void eat(String s) {
st=new StringTokenizer(s);
}
public String nextLine() {
try {
return br.readLine();
}catch(IOException e) {
return null;
}
}
public boolean hasNext() {
while(!st.hasMoreTokens()) {
String s=nextLine();
if(s==null)return false;
eat(s);
}
return true;
}
public String next() {
hasNext();
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static FastScanner sc=new FastScanner(System.in);
static PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
G1. Teleporters (Easy Version)
题意
有 0~n，n + 1个点，我们可以从 1 ~ n 号点用 传送器瞬间移动到 0号点，花费 a[i] 块钱
从一个点移动到隔壁，花费 1 块钱
给你一串 传送的花费 和现在总共有的钱 ，问从 0 出发最多能用几次传送
由题意得，传送的使用条件是：
- 先花费 i 块钱，从 0 走到 i
- 花费 a[i] 传送
code