exec_outcome stringclasses 1 value | code_uid stringlengths 32 32 | file_name stringclasses 111 values | prob_desc_created_at stringlengths 10 10 | prob_desc_description stringlengths 63 3.8k | prob_desc_memory_limit stringclasses 18 values | source_code stringlengths 117 65.5k | lang_cluster stringclasses 1 value | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_time_limit stringclasses 27 values | prob_desc_sample_outputs stringlengths 2 796 | prob_desc_notes stringlengths 4 3k ⌀ | lang stringclasses 5 values | prob_desc_input_from stringclasses 3 values | tags listlengths 0 11 | src_uid stringlengths 32 32 | prob_desc_input_spec stringlengths 28 2.37k ⌀ | difficulty int64 -1 3.5k ⌀ | prob_desc_output_spec stringlengths 17 1.47k ⌀ | prob_desc_output_to stringclasses 3 values | hidden_unit_tests stringclasses 1 value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | b5776a223d0c77c1971b27779904359b | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | //package codeforces;
import java.util.Scanner;
public class Codechef {
public static Scanner scan = new Scanner(System.in);
public static void main(String[] args) {
int testCases = scan.nextInt();
for (int vedant = 0; vedant < testCases; vedant++) {
solve();
}
}
private static void solve() {
// Solution goes over here.
String str = scan.next();
boolean flag = str.charAt(0) == 'A' && str.charAt(str.length() - 1) == 'B';
int aCount = 0;
int bCount = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'A') {
aCount++;
} else {
if (aCount > bCount) {
bCount++;
} else {
flag = false;
break;
}
}
}
if (flag && aCount >= bCount) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 872a0c231956359ae9be8c017a9a9e19 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | //Some of the methods are copied from GeeksforGeeks Website
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
//static Scanner sc=new Scanner(System.in);
//static Reader sc=new Reader();
static FastReader sc=new FastReader(System.in);
static long mod = (long)(1e9)+ 7;
static int max_num=(int)1e5+5;
static int count(String a, String b)
{
int m = a.length();
int n = b.length();
// Create a table to store
// results of sub-problems
int lookup[][] = new int[m + 1][n + 1];
// If first string is empty
for (int i = 0; i <= n; ++i)
lookup[0][i] = 0;
// If second string is empty
for (int i = 0; i <= m; ++i)
lookup[i][0] = 1;
// Fill lookup[][] in
// bottom up manner
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
// If last characters are
// same, we have two options -
// 1. consider last characters
// of both strings in solution
// 2. ignore last character
// of first string
if (a.charAt(i - 1) == b.charAt(j - 1))
lookup[i][j] = lookup[i - 1][j - 1] +
lookup[i - 1][j];
else
// If last character are
// different, ignore last
// character of first string
lookup[i][j] = lookup[i - 1][j];
}
}
return lookup[m][n];
}
public static void main (String[] args) throws java.lang.Exception
{
try{
/*
Collections.sort(al,(a,b)->a.x-b.x);
Collections.sort(al,Collections.reverseOrder());
long n=sc.nextLong();
String s=sc.next();
char a[]=s.toCharArray();
StringBuilder sb=new StringBuilder();
map.put(a[i],map.getOrDefault(a[i],0)+1);
out.println("Case #"+tt+": "+ans );
*/
int t = sc.nextInt();
for(int tt=1;tt<=t;tt++)
{
String s=sc.next();
char a[]=s.toCharArray();
int n=a.length;
int b=0;
boolean f=true;
for(char c : a)
{
if(c=='B')
b++;
}
if(b==0) f=false;
if(a[0]!='A') f=false;
if(a[n-1]!='B') f=false;
if(b>n-b) f=false;
int x=0,y=0;
for(int i=0;i<n;i++)
{
if(a[i]=='A') x++;
else y++;
if(y>x) f=false;
}
flag(f);
}
out.flush();
out.close();
}
catch(Exception e)
{}
}
/*
...SOLUTION ENDS HERE...........SOLUTION ENDS HERE...
*/
static void flag(boolean flag)
{
out.println(flag ? "YES" : "NO");
out.flush();
}
/*
Map<Long,Long> map=new HashMap<>();
for(int i=0;i<n;i++)
{
if(!map.containsKey(a[i]))
map.put(a[i],1);
else
map.replace(a[i],map.get(a[i])+1);
}
Set<Map.Entry<Long,Long>> hmap=map.entrySet();
for(Map.Entry<Long,Long> data : hmap)
{
}
Iterator<Integer> itr = set.iterator();
while(itr.hasNext())
{
int val=itr.next();
}
*/
// static class Pair
// {
// int x,y;
// Pair(int x,int y)
// {
// this.x=x;
// this.y=y;
// }
// }
// Arrays.sort(p, new Comparator<Pair>()
// {
// @Override
// public int compare(Pair o1,Pair o2)
// {
// if(o1.x>o2.x) return 1;
// else if(o1.x==o2.x)
// {
// if(o1.y>o2.y) return 1;
// else return -1;
// }
// else return -1;
// }});
static void print(int a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print(long a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print_int(ArrayList<Integer> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void print_long(ArrayList<Long> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static int LowerBound(int a[], int x)
{ // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
static int UpperBound(int a[], int x)
{// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
static void DFS(ArrayList<Integer> graph[],boolean[] visited, int u)
{
visited[u]=true;
int v=0;
for(int i=0;i<graph[u].size();i++)
{
v=graph[u].get(i);
if(!visited[v])
DFS(graph,visited,v);
}
}
static boolean[] prime(int num)
{
boolean[] bool = new boolean[num];
for (int i = 0; i< bool.length; i++) {
bool[i] = true;
}
for (int i = 2; i< Math.sqrt(num); i++) {
if(bool[i] == true) {
for(int j = (i*i); j<num; j = j+i) {
bool[j] = false;
}
}
}
if(num >= 0) {
bool[0] = false;
bool[1] = false;
}
return bool;
}
static long nCr(long a,long b,long mod)
{
return (((fact[(int)a] * modInverse(fact[(int)b],mod))%mod * modInverse(fact[(int)(a - b)],mod))%mod + mod)%mod;
}
static long fact[]=new long[max_num];
static void fact_fill()
{
fact[0]=1l;
for(int i=1;i<max_num;i++)
{
fact[i]=(fact[i-1]*(long)i);
if(fact[i]>=mod)
fact[i]%=mod;
}
}
static long modInverse(long a, long m)
{
return power(a, m - 2, m);
}
static long power(long x, long y, long m)
{
if (y == 0)
return 1;
long p = power(x, y / 2, m) % m;
p = (long)((p * (long)p) % m);
if (y % 2 == 0)
return p;
else
return (long)((x * (long)p) % m);
}
static long sum_array(int a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static long sum_array(long a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static void sort(int[] a)
{
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void sort(long[] a)
{
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void reverse_array(int a[])
{
int n=a.length;
int i,t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static void reverse_array(long a[])
{
int n=a.length;
int i; long t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static PrintWriter out=new PrintWriter(System.out);
static int int_max=Integer.MAX_VALUE;
static int int_min=Integer.MIN_VALUE;
static long long_max=Long.MAX_VALUE;
static long long_min=Long.MIN_VALUE;
}
// Thank You ! | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | f992af8ad9c1a238dc22c06c69508ae3 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | /* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t>0)
{
String s=sc.next();
int l=s.length();
int flag=0;
if(l>1 && s.charAt(0)=='A' && s.charAt(l-1)=='B')
{
int counta=1, countb=0;
for(int i=1; i<l-1 && flag==0 ;i++)
{
if(s.charAt(i)=='A') counta++;
if(s.charAt(i)=='B') countb++;
if(countb>counta )
flag=1;
}
if(countb+1>counta) flag=1;
}
else
flag=1;
if(flag==1) System.out.println("no");
else System.out.println("yes");
t--;
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 1abce4ba7f0220856d69319e9d22c4dc | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class IloveAAAB {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
String s = br.readLine();
if(s.length() >= 2) {
if(s.charAt(0) == 'A' && s.charAt(s.length()-1) == 'B') {
int a = 0;
int b = 0;
Boolean correct = true;
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i) == 'A')
a++;
else
b++;
if(b>a) {
out.println("NO");
correct = false;
break;
}
}
if(correct)
out.println("YES");
}
else {
out.println("NO");
}
}
else {
out.println("NO");
}
}
out.flush();
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 6672838f6e7d2fac46bf3fa0ae99503e | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class IloveAAAB {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
String s = br.readLine();
if(s.length() >= 2) {
if(s.charAt(0) == 'A' && s.charAt(s.length()-1) == 'B') {
int a = 0;
Boolean correct = true;
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i) == 'A')
a++;
else
a--;
if(a == -1) {
out.println("NO");
correct = false;
break;
}
}
if(correct)
out.println("YES");
}
else {
out.println("NO");
}
}
else {
out.println("NO");
}
}
out.flush();
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 205b660f59008fa893f02e1b9c4b1c2b | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
public class B
{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
String str=sc.next();
boolean flag=true;
if(str.length()==1 || str.charAt(0)=='B' || str.charAt(str.length()-1)=='A')
flag=false;
int count=1;
for(int i=1;i<str.length();i++)
{
if(str.charAt(i)=='A')
count++;
else
count--;
if(count<0)
{
flag=false;
break;
}
}
System.out.println(flag?"YES":"NO");
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | a22daba449262ee9618757f9b1fcaad1 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
// when can't think of anything -->>
// 1. In sorting questions try to think about all possibilities like starting from start, end, middle.
// 2. Two pointers, brute force.
// 3. In graph query questions try to solve it reversely or try to process all the queries in a single parse.
// 4. If order does not matter then just sort the data if constraints allow. It'll never harm.
// 5. In greedy problems if you are just overwhelmed by the possibilities and stuck, try to code whatever you come up with.
// 6. Try to solve it from back or reversely.
// 7. When can't prove something take help of contradiction/pigeon hole principle.
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
PrintWriter writer = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0) {
String s = sc.next();
int count = 0;
if(s.length() < 2 || s.charAt(s.length() - 1) != 'B')writer.println("NO");
else {
for(char c : s.toCharArray()) {
if(c=='A')count++;
else count--;
if(count < 0)break;
}
if(count < 0)writer.println("NO");
else writer.println("YES");
}
}
writer.flush();
writer.close();
}
private static boolean palindrome(String ss) {
int i = 0;
int j = ss.length() - 1;
while(i<j) {
if(ss.charAt(i) == ss.charAt(j)) {
i++; j--;
}else return false;
}
return true;
}
private static long power (long a, long n, long p) {
long res = 1;
while(n!=0) {
if(n%2==1) {
res=(res*a)%p;
n--;
}else {
a= (a*a)%p;
n/=2;
}
}
return res;
}
private static boolean isPrime(int c) {
for (int i = 2; i*i <= c; i++) {
if(c%i==0)return false;
}
return true;
}
private static int find(int a , int arr[]) {
if(arr[a] != a) return arr[a] = find(arr[a], arr);
return arr[a];
}
private static void union(int a, int b, int arr[]) {
int aa = find(a,arr);
int bb = find(b, arr);
arr[aa] = bb;
}
private static int gcd(int a, int b) {
if(a==0)return b;
return gcd(b%a, a);
}
public static int[] readIntArray(int size, FastReader s) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = s.nextInt();
}
return array;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
class Pair implements Comparable<Pair>{
int a;
int b;
Pair(int a, int b){
this.a = a;
this.b = b;
}
@Override
public boolean equals(Object obj) {
if(this == obj) return true;
if(obj == null || obj.getClass()!= this.getClass()) return false;
Pair pair = (Pair) obj;
return (pair.a == this.a && pair.b == this.b);
}
@Override
public int hashCode()
{
return Objects.hash(a,b);
}
@Override
public int compareTo(Pair o) {
if(this.a == o.a) {
return Integer.compare(this.b, o.b);
}else {
return Integer.compare(this.a, o.a);
}
}
@Override
public String toString() {
return this.a + " " + this.b;
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 92b0a737b10e66ce907d1416e0443519 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class AAAB {
public static void main(String[] args) throws IOException {
BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
int B = Integer.parseInt(buff.readLine());
for (int N = 0; N < B; N++) {
String str = buff.readLine();
boolean works = true;
boolean b = false;
int ACount = 0;
for (int i = 0; i < str.length(); i++) {
if(str.substring(i, i + 1).equals("B")) {
b = true;
}
if (str.substring(i, i + 1).equals("A")) {
ACount++;
}
if(str.substring(i, i+1).equals("B")) {
ACount--;
}
if(ACount<0) {
works = false;
}
}
if(works&&b&&str.charAt(str.length()-1)=='B') {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 473e4131a4ccffa302335626ae465ec6 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.Scanner;
// https://codeforces.com/problemset/problem/1672/B
public class CF1672B {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
String str = sc.next();
if(str.length()==1 || str.charAt(0)=='B' || str.charAt(str.length()-1)=='A') System.out.println("NO");
else {
int a = 0, b = 0;
byte humt = 1;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'A') a++;
else {
b++;
if (b > a) {
humt = 0;
break;
}
}
}
System.out.println(humt == 1 ? "YES" : "NO");
}
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 89182927791cc24af80abe671995816d | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main{
static void solve() {
String line=in.next();
if(line.charAt(line.length()-1)=='A'){out.println("NO");return;}
int a=0,b=0;
for(int i=0;i<line.length();i++){
if(line.charAt(i)=='A'){
a++;
}else{
if(a==0){out.println("NO"); return;}
b++;
a--;
}
}
if(b==0){out.println("NO"); return;}
out.println("YES");
}
public static void main(String[] args) throws Exception {
int cnt=in.nextInt();
while (cnt-->0){
solve();
}
out.flush();
}
static class FastReader{
StringTokenizer st;
BufferedReader br;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st==null||!st.hasMoreElements()){
try {
st=new StringTokenizer(br.readLine());
}catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 6d46f1409f96aeba375fe4f3d285bfae | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | /*-----------------------------------------
Created By : Piyush
On 29-04-2022 at 14:52
--------------------------------------------*/
import java.util.*;
import java.io.*;
import java.math.*;
public class AAAB {
static final int MOD = (int) 1e9 + 7;
static FastScanner fs = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
int t = fs.nextInt();
while (t-- > 0)
solve();
out.close();
}
private static void solve() {
String s = fs.next();
int a = 0 , b = 0;
for (int i = 0; i <s.length(); i++) {
if(s.charAt(i) == 'A') a++;
else b++;
if(b > a){
out.println("NO");
return;
}
}
if(a == 0 && b != 0 || b == 0 && a != 0 || s.endsWith("A"))
out.println("NO");
else
out.println("YES");
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens()) try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | d2c67e11b184f8259d4a81ef75c95568 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import javax.swing.plaf.synth.SynthDesktopIconUI;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
sc.nextLine();
for(int i=0;i<n;i++){
String s = sc.nextLine();
int cnt=0;
int c = 0;
char ch[] = new char[s.length()];
for(int j=0;j<s.length();j++){
ch[j] = s.charAt(j);
if(ch[j]=='A'){
cnt++;
}
else{
cnt--;
}
if(cnt<0){
c=1;
}
}
if(c==0 && ch[s.length()-1]=='B'){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
public static String Ssort(String s) {
char tempArray[] = s.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
public static int smallest(int x, int y,int z ) {
return Math.min(Math.min(x, y), z);
}
public static int largest(int x,int y, int z) {
return Math.max(Math.max(x, y), z);
}
public static int average(int x,int y,int z) {
return(x+y+z)/3;
}
public static int sumdigit( long x) {
int result = 0;
while(x>0){
result+=x%10;
x/=10;
}
return result;
}
public static String isPowerOfTwo(long n)
{
if (n == 0)
return ("Yes");
while (n != 1)
{
if (n % 2 != 0)
return ("Yes");
n = n / 2;
}
return ("No");
}
public static char[] swap(String s)
{
char ch[] = s.toCharArray();
char temp = ch[0];
ch[0] = ch[s.length()-1];
ch[s.length()-1] = temp;
return ch;
}
public static boolean perfectsquare(int x)
{
if (x >= 0) {
int sr = (int)Math.sqrt(x);
return ((sr * sr) == x);
}
return false;
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 97048cce98e6b2f76508ba768f0035e1 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;
/**
* iloveaaab
*/
public class iloveaaab {
static String tolove(String str) {
int sum = 0;
if (str.charAt(str.length() - 1) != 'B')
return "No";
for (char ch : str.toCharArray()) {
if (ch == 'A')
sum++;
if (ch == 'B')
sum--;
if (sum < 0)
return "No";
}
if (sum < 0)
return "No";
else
return "yes";
}
// static String tolove(String str) {
// if (str.length() == 1)
// return "NO";
// if (str.charAt(str.length() - 1) != 'B' || str.charAt(0) == 'B')
// return "NO";
// for (int i = 1; i < str.length(); i++) {
// if (str.charAt(i - 1) == 'B' && str.charAt(i) == 'B') {
// return "NO";
// }
// }
// return "YES";
// }
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int testcase = sc.nextInt();
sc.nextLine();
List<String> res = new ArrayList<>();
while (testcase-- > 0) {
String str = sc.nextLine();
res.add(tolove(str));
}
Iterator<String> it = res.iterator();
while (it.hasNext()) {
System.out.println(it.next());
}
sc.close();
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | bbade857273e98d4fee42a198d61528c | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static Scanner sc = new Scanner(System.in);
public static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) throws Exception {
int t = sc.nextInt();
while( t > 0 ) {
solve();
t--;
}
pw.flush();
}
static void solve(){
String S = sc.next();
char[] s = S.toCharArray();
int N = S.length();
int[] cnt = {0, 0};
for( char c : s ) {
cnt[c-'A']++;
if( cnt[0] < cnt[1] ) {
pw.println("NO");
return;
}
}
if( s[0] == 'A' && s[N-1] == 'B' && cnt[0] >= cnt[1] ) pw.println("YES");
else pw.println("NO");
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | a6d2a79b3011ab991b62af2c9bc830eb | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = Integer.parseInt(sc.nextLine());
// System.out.println("------------");
for(int i = 0; i < n; i++) {
String input = sc.nextLine();
String[] inputArr = input.trim().split("");
if(inputArr.length == 1) {
System.out.println("NO");
}else if(!inputArr[inputArr.length - 1].equals("B")) {
System.out.println("NO");
}else {
System.out.println(stackChecker(inputArr));
}
// if(input.contains("BB")) {
// System.out.println("NO");
// }else if(input.length() == 1) {
// System.out.println("NO");
// }else if(!inputArr[inputArr.length - 1].equals("B")) {
// System.out.println("NO");
// }
// else if(inputArr[0].equals("B")) {
// System.out.println("NO");
// }
// else {
// System.out.println("YES");
// }
}
}
public static String stackChecker(String[] arr) {
String[] stack = new String[arr.length];
int sp = -1;
for(int i = 0; i < arr.length; i++) {
if(arr[i].equals("A")) {
sp += 1;
stack[sp] = arr[i];
}else {
if(sp == -1) {
return "NO";
}else {
sp--;
}
}
}
return "YES";
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 95c1b69ee4bdbd842d409fa8a7b7a1e7 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | // 1672B
import java.util.*;
public class ILoveAAAB {
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
in.nextLine();
for (int i = 0; i < t; i++) {
String s = in.nextLine();
if (s.length() < 2 || s.charAt(0) == 'B' || s.charAt(s.length() - 1) != 'B') {
System.out.println("NO");
continue;
}
int c = 1;
for (int j = 1; j < s.length(); j++) {
if (s.charAt(j) == 'B') {
c--;
} else if (s.charAt(j) == 'A') {
c++;
}
if (c < 0) {
break;
}
}
if (c >= 0) System.out.println("YES");
else System.out.println("NO");
}
in.close();
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 26bc26ce4207e3f7b269e701ec8f617a | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B_I_love_AAAB {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long t = Long.parseLong(br.readLine());
while (t-- > 0) {
boolean can = true;
String s = br.readLine();
long ca = 0, cb = 0;
if (s.charAt(0) == 'B' || s.charAt(s.length()-1)=='A') {
can = false;
}
if(can)
for (int i = 0; i < s.length()-1; i++) {
if(s.charAt(i)=='A'){
ca++;
}
if(s.charAt(i+1)!='A'){
cb++;
}else{
if(cb > ca){
can = false;
break;
}
}
}
if(can && cb <= ca){
System.out.println("YES");
}else System.out.println("NO");
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 22e446f15f1dcbc91041e1bc2fc073e8 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B_I_love_AAAB {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long t = Long.parseLong(br.readLine());
while (t-- > 0) {
boolean can = true;
String s = br.readLine();
long ca = 0, cb = 0;
if (s.charAt(0) == 'B' || s.charAt(s.length()-1)=='A') {
can = false;
}
if(can)
for (int i = 0; i < s.length()-1; i++) {
if(s.charAt(i)=='A'){
ca++;
}
if(s.charAt(i+1)!='A'){
cb++;
}else{
if(cb > ca){
can = false;
break;
}else{
continue;
}
}
}
if(can && cb <= ca && cb!=0){
System.out.println("YES");
}else System.out.println("NO");
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 254caf970b189cfe4d3b7d58a5b6f2b0 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.math.BigInteger;
public final class Main{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() { this(System.in, System.out); }
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(new FileWriter(problemName + ".out"));
r = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) { }
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
static Kattio sc = new Kattio();
static long mod = 998244353l;
static PrintWriter out =new PrintWriter(System.out);
//Heapify function to maintain heap property.
public static void swap(int i,int j,int arr[]) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void swap(int i,int j,long arr[]) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void swap(int i,int j,char arr[]) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static String endl = "\n" , gap = " ";
public static void main(String[] args)throws IOException {
int t =ri();
// int MAK = 2500 * 2500;
// long seive[] = new long[MAK + 1];
// for (int i = 1; i <= MAK; i++) {
// for (int j = i; j <= MAK; j+=i) {
// seive[j]++;
// }
// }
// for (int i = 1; i <= MAK; i++) {
// seive[i] += seive[i - 1];
// }
int test_case = 1;
while(t-->0) {
// out.write("Case "+(test_case++)+":" + endl);
// System.out.println();
solve();
}
out.close();
}
public static void solve()throws IOException {
char s[] = rac();
boolean ans = true;
int n = s.length;
if(s[n-1] == 'A') {
System.out.println("NO");
return;
}
int a = 0 , b =0;
for(int i =0;i<n;i++) {
if(s[i] == 'A') a++;
else b++;
if(b > a) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
public static boolean valid(String s) {
System.out.println(s);
if(s.length() == 1) return false;
if(s.length() == 0) return true;
if(s.charAt(s.length()-1) != 'B') return false;
for(int i =0;i<s.length()-1;i++) {
if(s.charAt(i) != 'A') return false;
}
return true;
}
public static boolean collinearr(long a[] , long b[] , long c[]) {
return (b[1]-a[1])*(b[0]-c[0]) == (b[0]-a[0])*(b[1]-c[1]);
}
public static boolean callfun(int index , int n,int neg , int pos , String s) {
if(neg < 0 || pos < 0) return false;
if(index >= n) return true;
if(s.charAt(0) == 'P') {
if(neg <= 0) return false;
return callfun(index + 1,n , neg-1 , pos , s.charAt(1) + "N");
}
else {
if(pos <= 0) return false;
return callfun(index + 1 , n , neg , pos-1 , s.charAt(1) + "P");
}
}
public static void getPerm(int n , char arr[] , String s ,List<String>list) {
if(n == 0) {
list.add(s);
return;
}
for(char ch : arr) {
getPerm(n-1 , arr , s+ ch,list);
}
}
public static int getLen(int i ,int j , char s[]) {
while(i >= 0 && j < s.length && s[i] == s[j]) {
i--;
j++;
}
i++;
j--;
if(i>j) return 0;
return j-i + 1;
}
public static int getMaxCount(String x) {
char s[] = x.toCharArray();
int max = 0;
int n = s.length;
for(int i =0;i<n;i++) {
max = Math.max(max,Math.max(getLen(i , i,s) , getLen(i ,i+1,s)));
}
return max;
}
public static double getDis(int arr[][] , int x, int y) {
double ans = 0.0;
for(int a[] : arr) {
int x1 = a[0] , y1 = a[1];
ans += Math.sqrt((x-x1)*(x-x1) + (y-y1)*(y-y1));
}
return ans;
}
public static long callfun(int day , int k, int limit,int n) {
if(k > limit) return 0;
if(day > n) return 1;
long ans = callfun(day + 1 , k , limit, n)*k + callfun(day + 1,k+1,limit,n)*(k+1);
return ans;
}
public static void primeDivisor(HashMap<Long , Long >cnt , long num) {
for(long i = 2;i*i<=num;i++) {
while(num%i == 0) {
cnt.put(i ,(cnt.getOrDefault(i,0l) + 1));
num /= i;
}
}
if(num > 2) {
cnt.put(num ,(cnt.getOrDefault(num,0l) + 1));
}
}
public static int callfun(int arr[],int move , int index , Integer memo[][]) {
// System.out.println("AT " + index);
if(index >= arr.length-1) return 0;
if(memo[index][move] != null) return memo[index][move];
if(move == 0) {
int next = getNext(arr[index] , arr , index + 1);
int dif = getNextDIF(arr[index] ,arr , index + 1);
int ans = Integer.MAX_VALUE;
if(next != -1) {
int curans = callfun(arr,move ,next , memo);
if(curans != Integer.MAX_VALUE) ans = Math.min(ans , curans + 1);
}
if(dif != -1){
int curans = callfun(arr , move + 1 , dif ,memo);
if(curans != Integer.MAX_VALUE) ans = Math.min(ans , curans + 1);
}
return memo[index][move] = ans;
}
else {
int next = getNext(arr[index] , arr , index + 1);
int ans = Integer.MAX_VALUE;
if(next != -1) {
ans = callfun(arr,move , next , memo);
if(ans != Integer.MAX_VALUE) ans++;
}
return memo[index][move] = ans;
}
}
public static int getNext(int same , int arr[],int index) {
for(int i = index;i<arr.length;i++) {
if(arr[i]%2 == same%2) return i;
}
return -1;
}
public static int getNextDIF(int same , int arr[],int index) {
for(int i = index;i<arr.length;i++) {
if(arr[i]%2 != same%2) return i;
}
return -1;
}
public static boolean hasAll(String a , String b) {
for(int i =0;i<b.length();i++) {
boolean cur = false;
for(int j =0;j<a.length();j++) {
if(b.charAt(i) == a.charAt(j)) cur = true;
}
if(cur == false) return false;
}
return true;
}
public static String getAbsent(String s) {
char arr[] = {'a' , 'e' , 'i' , 'o' ,'u'};
StringBuilder res = new StringBuilder();
for(int i =0;i<5;i++) {
boolean has = false;
for(int j =0;j<s.length();j++) {
if(s.charAt(j) == arr[i]) has = true;
}
if(has == false)res.append(arr[i]);
}
return res.toString();
}
public static String getCompressed(String x) {
char s[] = x.toCharArray();
Arrays.sort(s);
StringBuilder res = new StringBuilder();
HashSet<Character> set = new HashSet<>();
for(char ch : s) {
if(set.contains(ch)) continue;
set.add(ch);
res.append(ch);
}
return res.toString();
}
public static boolean callfun(char s[] , char t[] , int f1[] , int f2[]) {
for(int i = 0;i<26;i++) {
if(f1[i] > 0 && f2[i] == 0) {
f1[i]--;
return !callfun(s , t,f2 , f1);
}
}
return false;
}
public static void primeFactors(long x ,HashMap<Long, Integer> map ) {
int cnt= 0;
while(x%2 == 0) {
cnt++;
x /=2;
}
if(cnt > 0) map.put(2l , cnt );
cnt = 0;
for(long i =3;i*i<=x;i+=2) {
cnt = 0;
while(x%i == 0) {
cnt++;
x /= i;
}
if(cnt > 0) {
map.put(i ,cnt);
}
cnt = 0;
}
if(x > 2) map.put(x , 1);
}
public static long callfun(long x,long c,HashMap<Long,Long> set , HashSet<Long> vis) {
if(x == 1) return 1;
long ans = x;
long i = 1;
while (pow(i ,c) <= pow(10l , 3l) * x) {
long gc = gcd(pow(i,c), x);
ans = Math.min(ans, pow(i,c)*x/(gc*gc));
i += 1;
}
return ans;
// for(long i =1;i<=x;i++) {
// long val = pow(i , c);
// if(val > 10000000000l) {
// return ans;
// }
// long _gcd = gcd(val , x);
// long _lcm = val/_gcd*x;
// long newx = _lcm/_gcd;
// ans = Math.min(ans , newx);
// }
// return ans;
}
public static long callfun(int cursum , int n , int k , int m ,int index ,Long memo[][]) {
if(n == 0 ) {
if(cursum > k) return 0;
return 1;
}
if(memo[index][cursum] != null) return memo[index][cursum];
long mod = (long)998244353;
long ans = 0;
for(int i =1;i<=m;i++) {
long curans = callfun(cursum + i , n-1 ,k , m,index + 1,memo )%mod;
ans = (ans + curans)%mod;
}
return memo[index][cursum] = ans;
}
public static boolean isSubsequene(char a[], char b[] ) {
int i =0 , j = 0;
while(i < a.length && j <b.length) {
if(a[i] == b[j]) {
j++;
}
i++;
}
return j >= b.length;
}
public static long fib(int n ,long M) {
if (n == 0) {
return 0;
} else if (n == 1) {
return 1;
} else {
long[][] mat = {{1, 1}, {1, 0}};
mat = pow(mat, n-1 , M);
return mat[0][0];
}
}
public static long[][] pow(long[][] mat, int n ,long M) {
if (n == 1) return mat;
else if (n % 2 == 0) return pow(mul(mat, mat , M), n/2 , M);
else return mul(pow(mul(mat, mat,M), n/2,M), mat , M);
}
static long[][] mul(long[][] p, long[][] q,long M) {
long a = (p[0][0]*q[0][0] + p[0][1]*q[1][0])%M;
long b = (p[0][0]*q[0][1] + p[0][1]*q[1][1])%M;
long c = (p[1][0]*q[0][0] + p[1][1]*q[1][0])%M;
long d = (p[1][0]*q[0][1] + p[1][1]*q[1][1])%M;
return new long[][] {{a, b}, {c, d}};
}
public static long[] kdane(long arr[]) {
int n = arr.length;
long dp[] = new long[n];
dp[0] = arr[0];
long ans = dp[0];
for(int i = 1;i<n;i++) {
dp[i] = Math.max(dp[i-1] + arr[i] , arr[i]);
ans = Math.max(ans , dp[i]);
}
return dp;
}
public static void update(int low , int high , int l , int r, int val , int treeIndex ,int tree[]) {
if(low > r || high < l || high < low) return;
if(l <= low && high <= r) {
System.out.println("At " +low + " and " + high + " ans ttreeIndex " + treeIndex);
tree[treeIndex] += val;
return;
}
int mid = low + (high - low)/2;
update(low , mid , l , r , val , treeIndex*2 + 1, tree);
update(mid + 1 , high , l , r , val , treeIndex*2 + 2 , tree);
}
static int colx[] = {1 ,-1, 0,0 , 1,1,-1,-1};
static int coly[] = {0 ,0, 1,-1,1,-1,1,-1};
public static void reverse(char arr[]) {
int i =0 , j = arr.length-1;
while(i < j) {
swap(i , j , arr);
i++;
j--;
}
}
public static long[] reverse(long arr[]) {
long newans[] = arr.clone();
int i =0 , j = arr.length-1;
while(i < j) {
swap(i , j , newans);
i++;
j--;
}
return newans;
}
public static long inverse(long x , long mod) {
return pow(x , mod -2 , mod);
}
public static int maxArray(int arr[]) {
int ans = arr[0] , n = arr.length;
for(int i =1;i<n;i++) {
ans = Math.max(ans , arr[i]);
}
return ans;
}
public static long maxArray(long arr[]) {
long ans = arr[0];
int n = arr.length;
for(int i =1;i<n;i++) {
ans = Math.max(ans , arr[i]);
}
return ans;
}
public static int minArray(int arr[]) {
int ans = arr[0] , n = arr.length;
for(int i =0;i<n;i++ ) {
ans = Math.min(ans ,arr[i]);
}
return ans;
}
public static long minArray(long arr[]) {
long ans = arr[0];
int n = arr.length;
for(int i =0;i<n;i++ ) {
ans = Math.min(ans ,arr[i]);
}
return ans;
}
public static int sumArray(int arr[]) {
int ans = 0;
for(int x : arr) {
ans += x;
}
return ans;
}
public static long sumArray(long arr[]) {
long ans = 0;
for(long x : arr) {
ans += x;
}
return ans;
}
public static long rl() {
return sc.nextLong();
}
public static char[] rac() {
return sc.next().toCharArray();
}
public static String rs() {
return sc.next();
}
public static char rc() {
return sc.next().charAt(0);
}
public static int [] rai(int n) {
int ans[] = new int[n];
for(int i =0;i<n;i++) {
ans[i] = sc.nextInt();
}
return ans;
}
public static long [] ral(int n) {
long ans[] = new long[n];
for(int i =0;i<n;i++) {
ans[i] = sc.nextLong();
}
return ans;
}
public static int ri() {
return sc.nextInt();
}
public static int getValue(int num ) {
int ans = 0;
while(num > 0) {
ans++;
num = num&(num-1);
}
return ans;
}
public static boolean isValid(int x ,int y , int n,char arr[][],boolean visited[][][][]) {
return x>=0 && x<n && y>=0 && y <n && !(arr[x][y] == '#');
}
// public static Pair join(Pair a , Pair b) {
// Pair res = new Pair(Math.min(a.min , b.min) , Math.max(a.max , b.max) , a.count + b.count);
// return res;
// }
// segment tree query over range
// public static int query(int node,int l , int r,int a,int b ,Pair tree[] ) {
// if(tree[node].max < a || tree[node].min > b) return 0;
// if(l > r) return 0;
// if(tree[node].min >= a && tree[node].max <= b) {
// return tree[node].count;
// }
// int mid = l + (r-l)/2;
// int ans = query(node*2 ,l , mid ,a , b , tree) + query(node*2 +1,mid + 1, r , a , b, tree);
// return ans;
// }
// // segment tree update over range
// public static void update(int node, int i , int j ,int l , int r,long value, long arr[] ) {
// if(l >= i && j >= r) {
// arr[node] += value;
// return;
// }
// if(j < l|| r < i) return;
// int mid = l + (r-l)/2;
// update(node*2 ,i ,j ,l,mid,value, arr);
// update(node*2 +1,i ,j ,mid + 1,r, value , arr);
// }
public static long pow(long a , long b , long mod) {
if(b == 1) return a;
if(b == 0) return 1;
long ans = pow(a , b/2 , mod)%mod;
if(b%2 == 0) {
return (ans*ans)%mod;
}
else {
return ((ans*ans)%mod*a)%mod;
}
}
public static long pow(long a , long b ) {
if(b == 1) return a;
if(b == 0) return 1;
long ans = pow(a , b/2);
if(b%2 == 0) {
return (ans*ans);
}
else {
return ((ans*ans)*a);
}
}
public static boolean isVowel(char ch) {
if(ch == 'a' || ch == 'e'||ch == 'i' || ch == 'o' || ch == 'u') return true;
return false;
}
public static int getFactor(int num) {
if(num==1) return 1;
int ans = 2;
int k = num/2;
for(int i = 2;i<=k;i++) {
if(num%i==0) ans++;
}
return Math.abs(ans);
}
public static int[] readarr()throws IOException {
int n = sc.nextInt();
int arr[] = new int[n];
for(int i =0;i<n;i++) {
arr[i] = sc.nextInt();
}
return arr;
}
public static boolean isPowerOfTwo (long x) {
return x!=0 && ((x&(x-1)) == 0);
}
public static boolean isPrime(long num) {
if(num==1) return false;
if(num<=3) return true;
if(num%2==0||num%3==0) return false;
for(long i =5;i*i<=num;i+=6) {
if(num%i==0 || num%(i+2) == 0) return false;
}
return true;
}
public static boolean isPrime(int num) {
// System.out.println("At pr " + num);
if(num==1) return false;
if(num<=3) return true;
if(num%2==0||num%3==0) return false;
for(int i =5;i*i<=num;i+=6) {
if(num%i==0 || num%(i+2) == 0) return false;
}
return true;
}
// public static boolean isPrime(long num) {
// if(num==1) return false;
// if(num<=3) return true;
// if(num%2==0||num%3==0) return false;
// for(int i =5;i*i<=num;i+=6) {
// if(num%i==0) return false;
// }
// return true;
// }
public static long gcd(long a , long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static int gcd(int a , int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static int get_gcd(int a , int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static long get_gcd(long a , long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
// public static long fac(long num) {
// long ans = 1;
// int mod = (int)1e9+7;
// for(long i = 2;i<=num;i++) {
// ans = (ans*i)%mod;
// }
// return ans;
// }
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 67b08a9133cb3be30670c692d010bf37 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | /*LoudSilence*/
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Solution {
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static FastScanner s = new FastScanner();
static FastWriter out = new FastWriter();
final static int mod = (int)1e9 + 7;
final static int INT_MAX = Integer.MAX_VALUE;
final static int INT_MIN = Integer.MIN_VALUE;
final static long LONG_MAX = Long.MAX_VALUE;
final static long LONG_MIN = Long.MIN_VALUE;
final static double DOUBLE_MAX = Double.MAX_VALUE;
final static double DOUBLE_MIN = Double.MIN_VALUE;
final static float FLOAT_MAX = Float.MAX_VALUE;
final static float FLOAT_MIN = Float.MIN_VALUE;
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static class FastScanner{BufferedReader br;StringTokenizer st;
public FastScanner() {if(System.getProperty("ONLINE_JUDGE") == null){try {br = new BufferedReader(new FileReader("E:\\Competitive Coding\\input.txt"));}
catch (FileNotFoundException e) {br = new BufferedReader(new InputStreamReader(System.in));}}else{br = new BufferedReader(new InputStreamReader(System.in));}}
String next(){while (st == null || !st.hasMoreElements()){try{st = new StringTokenizer(br.readLine());}catch (IOException e){e.printStackTrace();}}return st.nextToken();}
int nextInt(){return Integer.parseInt(next());}
long nextLong(){return Long.parseLong(next());}
double nextDouble(){return Double.parseDouble(next());}
List<Integer> readIntList(int n){List<Integer> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextInt()); return arr;}
List<Long> readLongList(int n){List<Long> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextLong()); return arr;}
int[] readIntArr(int n){int[] arr = new int[n]; for (int i = 0; i < n; i++) arr[i] = s.nextInt(); return arr;}
long[] readLongArr(int n){long[] arr = new long[n]; for (int i = 0; i < n; i++) arr[i] = s.nextLong(); return arr;}
String nextLine(){String str = "";try{str = br.readLine();}catch (IOException e){e.printStackTrace();}return str;}}
static class FastWriter{private BufferedWriter bw;public FastWriter(){if(System.getProperty("ONLINE_JUDGE") == null){try {this.bw = new BufferedWriter(new FileWriter("E:\\Competitive Coding\\output.txt"));}
catch (IOException e) {this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}else{this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}
public void print(Object object) throws IOException{bw.append(""+ object);}
public void println(Object object) throws IOException{print(object);bw.append("\n");}
public void debug(int object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void debug(long object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void close() throws IOException{bw.close();}}
public static void println(Object str) throws IOException{out.println(""+str);}
public static void println(Object str, int nextLine) throws IOException{out.print(""+str);}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static ArrayList<Integer> seive(int n){ArrayList<Integer> list = new ArrayList<>();int arr[] = new int[n+1];for(long i = 2; i <= n; i++) {if(arr[(int)i] == 1) {continue;}else {list.add((int)i);for(long j = i*i; j <= n; j = j + i) {arr[(int)j] = 1;}}}return list;}
public static long gcd(long a, long b){if(a > b) {a = (a+b)-(b=a);}if(a == 0L){return b;}return gcd(b%a, a);}
public static void swap(int[] arr, int i, int j) {arr[i] = arr[i] ^ arr[j]; arr[j] = arr[j] ^ arr[i]; arr[i] = arr[i] ^ arr[j];}
public static boolean isPrime(long n){if(n < 2){return false;}if(n == 2 || n == 3){return true;}if(n%2 == 0 || n%3 == 0){return false;}long sqrtN = (long)Math.sqrt(n)+1;for(long i = 6L; i <= sqrtN; i += 6) {if(n%(i-1) == 0 || n%(i+1) == 0) return false;}return true;}
public static long mod_add(long a, long b){ return (a%mod + b%mod)%mod;}
public static long mod_sub(long a, long b){ return (a%mod - b%mod + mod)%mod;}
public static long mod_mul(long a, long b){ return (a%mod * b%mod)%mod;}
public static long modInv(long a, long b){ return expo(a, b-2)%b;}
public static long mod_div(long a, long b){return mod_mul(a, modInv(b, mod));}
public static long expo (long a, long n){if(n == 0){return 1;}long recAns = expo(mod_mul(a,a), n/2);if(n % 2 == 0){return recAns;}else{return mod_mul(a, recAns);}}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Pair class
public static class Pair<X extends Comparable<X>,Y extends Comparable<Y>> implements Comparable<Pair<X, Y>>{
X first;
Y second;
public Pair(X first, Y second){
this.first = first;
this.second = second;
}
public String toString(){
return "( " + first+" , "+second+" )";
}
@Override
public int compareTo(Pair<X, Y> o) {
int t = first.compareTo(o.first);
if(t == 0) return second.compareTo(o.second);
return t;
}
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Code begins
public static void solve() throws IOException {
String str = s.nextLine();
int n = str.length();
if(n == 1){
println("NO");
return;
}
if(str.charAt(0) == 'B' || str.charAt(n-1) == 'A'){
println("NO");
return;
}
int a = 0;
for(int i = 0; i < n; i++){
if(str.charAt(i) == 'A') a++;
else{
if(a < 1){
println("NO");
return;
}
a--;
}
}
println("YES");
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static void main(String[] args) throws IOException {
int test = s.nextInt();
for(int t = 1; t <= test; t++) {
solve();
}
out.close();
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 06b227e3f366b2169eb00e4f7bf7333e | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
import java.lang.Math;
public class aaab {
public static void main(String[] args) throws Exception{
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-- > 0){
String s = in.next();
boolean b = false;
int n = 0;
if(s.length() < 2) System.out.println("NO");
else if(s.charAt(0) == 'B' || s.charAt(s.length() - 1) == 'A') System.out.println("NO");
else{
for(int i = 0; i < s.length(); i++){
if(s.charAt(i) == 'A') n++;
else {
n--;
if(n < 0){
System.out.println("NO");
b = true;
break;
}
};
}
if(!b && n < 0) System.out.println("NO");
else if(!b && n >= 0) System.out.println("YES");
}
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | f81a05ba71d589d22fd27255b06127b5 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
char arr[] = br.readLine().toCharArray();
int n = arr.length;
int af = 0;
int bf = 0;
for(int i = 0; i < n; i++)
{
if(arr[i] == 'A')
af = 1;
else
bf = 1;
}
if((af+bf)!=2 || (arr[n-1]!='B'))
{
output.write("NO\n");
continue;
}
int ac = 0;
int bc = 0;
// af = 0;
// bf = 0;
boolean flag = true;
if(arr[0] == 'A')
ac++;
else
ac--;
for(int i = 1; i < n; i++)
{
if(ac < 0)
flag = false;
if(arr[i] == 'A')
ac++;
else
ac--;
}
if(ac < 0)
flag = false;
if(flag)
output.write("YES\n");
else
output.write("NO\n");
}
output.flush();
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | e6cec5bf07d45409b9bc6d148a4f3022 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.Scanner;
public class I_love_AAAB {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
while(t-- > 0)
{
String str = sc.nextLine();
boolean is_possible = true;
int len = str.length();
if(str.charAt(len-1) != 'B')
{
is_possible = false;
}
else
{
int count_of_A = 0, count_of_B = 0;
for(int i = 0 ; i < len ; i++)
{
if(str.charAt(i) == 'B')
{
count_of_B++;
}
else
{
count_of_A++;
}
if(count_of_A < count_of_B)
{
is_possible = false;
break;
}
}
}
if(is_possible)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
sc.close();
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 4e4c108086d3bc7b0339986d2e563aa4 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.Scanner;
public class Codeforces {
public static void main(String[] args){
Scanner sc = new Scanner (System.in);
int t = sc.nextInt();
while(t-- > 0){
String s = sc.next();
boolean b = true;
if(s.length()==1){
b = false;
}
else{
if(s.charAt(s.length()-1)=='A'){
b = false;
}
if(s.charAt(0)=='B'){
b = false;
}
else{int k =0;
int m =0;
for(int i =0;i<s.length();i++){
if(s.charAt(i)=='A'){
k++;
}
else{m++;
if(m>k){
b = false;
}
}
}
}
}
if(b){
System.out.println("Yes");
}
else{
System.out.println("No");
}
}
}} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 571f6f00d4630192dd428614e1d6399c | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution{
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
static StringTokenizer st;
public static void main(String[] YDSV) throws IOException{
//int t=1;
int t=Integer.parseInt(br.readLine());
while(t-->0) Solve();
bw.flush();
}
public static void Solve() throws IOException{
// st=new StringTokenizer(br.readLine());
// int n=Integer.parseInt(st.nextToken());
// st=new StringTokenizer(br.readLine());
// int[] ar=new int[n];
// for(int i=0;i<n;i++) ar[i]=Integer.parseInt(st.nextToken());
char[] ch=br.readLine().toCharArray();
int n=ch.length;
if(ch[n-1]=='A' || ch[0]=='B'){
bw.write("NO\n");
return ;
}
int cb=0,ca=0;
for(int i=0;i<n;i++){
if(ch[i]=='B') cb++;
if(ch[i]=='A') ca++;
if(ca<cb){
bw.write("NO\n"); return;
}
}
bw.write("YES\n");
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 82640436216b862e6fd11ce603a62233 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
public class AAAB {
public static void main(String[] args) {
Scanner sn=new Scanner(System.in);
int t=Integer.parseInt(sn.nextLine());
while(t-->0){
String s=sn.nextLine();
int len=s.length();
if(check(s, len))
System.out.println("YES");
else
System.out.println("NO");
}
}
public static boolean check(String s,int len) {
int na=0;
int nb=0;
if(len==1)return false;
if(s.charAt(len-1)=='A')return false;
if(s.charAt(0)=='B')return false;
for(int i=0;i<len;i++)
{
if(s.charAt(i)=='A'){
na++;
if(na<nb)return false;
}
else{
nb++;
if(nb>na)return false;
}
}
if(nb>na)return false;
else
return true;
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | d6c376b93033ce3c78100de30e6f3221 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
static PrintWriter out=new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
O : while(t-->0)
{
char a[]=sc.next().toCharArray();
int acount=0,bcount=0;
boolean flag=true;
for(int i=0;i<a.length;i++){
if(a[i]=='A')acount++;
else bcount++;
if(bcount>acount)flag=false;
}
if(a[a.length-1]!='B')printN();
else if(!flag)printN();
else printY();
}
out.flush();
}
static void printN()
{
System.out.println("NO");
}
static void printY()
{
System.out.println("YES");
}
static int findfrequencies(int a[],int n)
{
int count=0;
for(int i=0;i<a.length;i++)
{
if(a[i]==n)
{
count++;
}
}
return count;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
static int[] EvenOddArragement(int a[])
{
ArrayList<Integer> list=new ArrayList<>();
for(int i=0;i<a.length;i++)
{
if(a[i]%2==0)
{
list.add(a[i]);
}
}
for(int i=0;i<a.length;i++)
{
if(a[i]%2!=0)
{
list.add(a[i]);
}
}
for(int i=0;i<a.length;i++)
{
a[i]=list.get(i);
}
return a;
}
static int gcd(int a, int b) {
while (b != 0) {
int t = a;
a = b;
b = t % b;
}
return a;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static int DigitSum(int n)
{
int r=0,sum=0;
while(n>=0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
return sum;
}
static boolean checkPerfectSquare(int number)
{
double sqrt=Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static boolean isPowerOfTwo(int n)
{
if(n==0)
return false;
return (int)(Math.ceil((Math.log(n) / Math.log(2)))) == (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static boolean isPrime2(int n)
{
if (n <= 1)
{
return false;
}
if (n == 2)
{
return true;
}
if (n % 2 == 0)
{
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
static String minLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for(int i=0;i<n;i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[0];
}
static String maxLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for (int i = 0; i < n; i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[arr.length-1];
}
static class P implements Comparable<P> {
int i, j;
public P(int i, int j) {
this.i=i;
this.j=j;
}
public int compareTo(P o) {
return Integer.compare(i, o.i);
}
}
static class pair{
int i,j;
pair(int x,int y){
i=x;
j=y;
}
}
static int binary_search(int a[],int value)
{
int start=0;
int end=a.length-1;
int mid=start+(end-start)/2;
while(start<=end)
{
if(a[mid]==value)
{
return mid;
}
if(a[mid]>value)
{
end=mid-1;
}
else
{
start=mid+1;
}
mid=start+(end-start)/2;
}
return -1;
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 267f3e9a9acdb0e95c6aa5be4acf95c5 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static boolean check(int arr[] , int x)
{
for(int i = 0 ; i < arr.length ; i++)
{
if(Math.abs(x+i-arr[i]) > 1)
return false;
}
return true;
}
public static void main(String []args) throws IOException
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
//int t = sc.nextInt();
while(t-- > 0)
{
String s = sc.nextLine();
int n = s.length();
if(s.length() < 2 || s.charAt(0) == 'B' || s.charAt(n-1) == 'A')
System.out.println("NO");
else
{
int a = 0;
boolean bl = true;
for(int i = 0 ; i < n ; i++)
{
if(s.charAt(i) == 'A')
a++;
else
a--;
if(a < 0)
{
bl = false;
break;
}
}
if(bl)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | c229fc23756aa1fd850dfc2786e5848d | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class Practice {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new BufferedWriter(new PrintWriter(System.out)));
StringBuilder sb = new StringBuilder();
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
while (t --> 0) {
st = new StringTokenizer(br.readLine());
String s = st.nextToken();
int len = s.length();
if (len == 1) sb.append("No");
else if (len == 2) sb.append(s.equals("AB") ? "Yes" : "No");
else if (len == 3) sb.append(s.equals("AAB") ? "Yes" : "No");
else {
if (s.charAt(0) == 'B' || s.charAt(len - 1) == 'A') sb.append("No");
else {
boolean flag = false;
int a = 0;
for (int i = 0; i < len; i++) {
if (s.charAt(i) == 'A') a++;
else {
if (i == len - 1 && a == 0) flag = true;
else {
if (a == 0) flag = true;
else {
a--;
}
}
}
}
sb.append(flag ? "No" : "Yes");
}
}
sb.append("\n");
}
pw.println(sb.toString().trim());
pw.close();
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 650950ccae89184f2777471bd9a926f3 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String[] args) throws IOException {
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
String s = br.readLine();
int n = s.length();
if (n == 1) {
bw.write("NO\n");
continue;
}
boolean can = true;
int b = 0, a = 0;
for (int i = 0; i < n; i++) {
if (s.charAt(i) == 'B') {
if (a == 0) {
can = false;
break;
}
b++;
a--;
if (a < 0) {
can = false;
break;
}
} else {
b = 0;
a++;
}
}
if (b == 0) {
can = false;
}
if (can) {
bw.write("YES\n");
} else {
bw.write("NO\n");
}
}
bw.flush();
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | e62452cc80ef74c72f714b4fae664403 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class JaiShreeRam{
static Scanner in=new Scanner();
static long systemTime;
static long mod = 1000000007;
static ArrayList<ArrayList<Integer>> adj;
static int seive[]=new int[1000001];
static long C[][];
static ArrayList<Integer> ans=new ArrayList<>();
public static void main(String[] args) throws Exception{
int z=in.readInt();
for(int test=1;test<=z;test++) {
//setTime();
solve();
//printTime();
//printMemory();
}
}
static void solve() {
char c[]=in.readString().toCharArray();
int n=c.length;
if(c[n-1]=='A'||c[0]=='B') {
print("NO");
return;
}
int a=0,b=0;
for(int i=0;i<n;i++) {
if(c[i]=='B') {
b++;
if(a<b) {
print("NO");
return;
}
}
else {
a++;
}
}
if(a<b) {
print("NO");
return;
}
print("YES");/*
int n=in.readInt();
int a[]=nia(n);
int cnt=0;
for(int i=1;i<n;i++) {
if(a[i]==a[i-1]) {
cnt++;
a[i]=-1;
}
}
print(cnt);*/
}
static long pow(long n, long m) {
if(m==0)
return 1;
else if(m==1)
return n;
else {
long r=pow(n,m/2);
if(m%2==0)
return (r*r)%mod;
else
return (r*r*n)%mod;
}
}
static long maxsumsub(ArrayList<Long> al) {
long max=0;
long sum=0;
for(int i=0;i<al.size();i++) {
sum+=al.get(i);
if(sum<0) {
sum=0;
}
max=Math.max(max,sum);
}
return max;
}
static long abs(long a) {
return Math.abs(a);
}
static void ncr(int n, int k){
C= new long[n + 1][k + 1];
int i, j;
for (i = 0; i <= n; i++) {
for (j = 0; j <= Math.min(i, k); j++) {
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
}
static boolean isPalin(String s) {
int i=0,j=s.length()-1;
while(i<=j) {
if(s.charAt(i)!=s.charAt(j)) {
return false;
}
i++;
j--;
}
return true;
}
static int knapsack(int W, int wt[],int val[], int n){
int []dp = new int[W + 1];
for (int i = 1; i < n + 1; i++) {
for (int w = W; w >= 0; w--) {
if (wt[i - 1] <= w) {
dp[w] = Math.max(dp[w],dp[w - wt[i - 1]] + val[i - 1]);
}
}
}
return dp[W];
}
static void seive() {
Arrays.fill(seive, 1);
seive[0]=0;
seive[1]=0;
for(int i=2;i*i<1000001;i++) {
if(seive[i]==1) {
for(int j=i*i;j<1000001;j+=i) {
if(seive[j]==1) {
seive[j]=0;
}
}
}
}
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a)
l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++)
a[i]=l.get(i);
}
static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a)
l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++)
a[i]=l.get(i);
}
static int[] nia(int n){
int[] arr= new int[n];
int i=0;
while(i<n){
arr[i++]=in.readInt();
}
return arr;
}
static long[] nla(int n){
long[] arr= new long[n];
int i=0;
while(i<n){
arr[i++]=in.readLong();
}
return arr;
}
static int[] nia1(int n){
int[] arr= new int[n+1];
int i=1;
while(i<=n){
arr[i++]=in.readInt();
}
return arr;
}
static Integer[] nIa(int n){
Integer[] arr= new Integer[n];
int i=0;
while(i<n){
arr[i++]=in.readInt();
}
return arr;
}
static Long[] nLa(int n){
Long[] arr= new Long[n];
int i=0;
while(i<n){
arr[i++]=in.readLong();
}
return arr;
}
static long gcd(long a, long b) {
if (b==0) return a;
return gcd(b, a%b);
}
static void print(long i) {
System.out.println(i);
}
static void print(Object o) {
System.out.println(o);
}
static void print(int a[]) {
for(int i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long a[]) {
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(ArrayList<Long> a) {
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(Object a[]) {
for(Object i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void setTime() {
systemTime = System.currentTimeMillis();
}
static void printTime() {
System.err.println("Time consumed: " + (System.currentTimeMillis() - systemTime));
}
static void printMemory() {
System.err.println("Memory consumed: " + (Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()) / 1000 + "kb");
}
static class Scanner{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String readString() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
double readDouble() {
return Double.parseDouble(readString());
}
int readInt() {
return Integer.parseInt(readString());
}
long readLong() {
return Long.parseLong(readString());
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | c0e7dc6718dcf2d718df86d1a73bdf82 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static long nod(long a, long b) {
while (Math.min(a, b) != 0) {
if (a > b) {
a = a % b;
} else b = b % a;
}
return a + b;
}
public static void main(String[] args) throws IOException {
// br = new BufferedReader(new FileReader("pencils.in"));
// out = new PrintWriter("pencils.out");
int t = nextInt();
for (int i = 0; i < t; i++) {
char[] a = nextToken().toCharArray();
String ans = "YES";
if (a[0] == 'B' || a[a.length - 1] == 'A') ans = "NO";
int l=0;
for (int j = 0; j < a.length && ans.equals("YES"); j++) {
if(a[j]=='A'){
l++;
}
else {
l--;
if(l<0)ans="NO";
}
}
if(l<0)ans="NO";
out.println(ans);
}
out.close();
}
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(System.out);
static StringTokenizer in = new StringTokenizer("");
public static boolean hasNext() throws IOException {
if (in.hasMoreTokens()) return true;
String s;
while ((s = br.readLine()) != null) {
in = new StringTokenizer(s);
if (in.hasMoreTokens()) return true;
}
return false;
}
public static String nextToken() throws IOException {
while (!in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public static int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public static long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
public static double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
static class Maincraft implements Comparable<Maincraft> {
int x;
int y;
int z;
int type;
public Maincraft(int x, int y, int z, int type) {
this.x = x;
this.y = y;
this.z = z;
this.type = type;
}
@Override
public int compareTo(Maincraft o) {
if (x != o.x) return Integer.compare(x, o.x);
else if (y != o.y) return -Integer.compare(y, o.y);
else return -Integer.compare(z, o.z);
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | b683791605f64054ab2b5c5030834a17 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | // Generated by Code Flattener.
// https://plugins.jetbrains.com/plugin/9979-idea-code-flattener
import java.io.*;
import java.util.StringTokenizer;
public class Main {
public void solve(FastReader in, FastWriter out) {
String str = in.nextLine();
int n = str.length();
if (str.charAt(n - 1) == 'A') {
out.println("NO");
return;
}
int count = 0;
for (int i = 0; i < n; i++) {
char ch = str.charAt(i);
if (ch == 'A') {
count++;
} else {
count--;
if (count < 0) {
out.println("NO");
return;
}
}
}
out.println("YES");
}
public static void main(String[] args) {
Main main = new Main();
boolean onlineJudge = System.getProperty("ONLINE_JUDGE") != null;
Problem.Builder builder = Problem
.builder()
.solution(main::solve)
.multiTest(true);
if (onlineJudge) {
builder.io(StreamPair.standard());
} else {
builder.io(StreamPair.debug(main.getClass()));
}
builder.build().execute();
}
private static class Problem {
private final Executable solution;
private final Executable init;
private final boolean multiTest;
private final InputStream inputStream;
private final OutputStream outputStream;
public Problem(Executable solution, Executable init, boolean multiTest,
InputStream inputStream, OutputStream outputStream) {
this.solution = solution;
this.init = init;
this.multiTest = multiTest;
this.inputStream = inputStream;
this.outputStream = outputStream;
}
public void execute() {
if (solution == null) {
throw new RuntimeException("please, set solution");
}
FastReader in = new FastReader(inputStream);
FastWriter out = new FastWriter(outputStream);
if (init != null) {
init.run(in, out);
}
if (!multiTest) {
solution.run(in, out);
} else {
int times = in.nextInt();
for (int i = 0; i < times; i++) {
solution.run(in, out);
}
}
out.close();
}
public static Builder builder() {
return new Builder();
}
public static class Builder {
private Executable solution;
private Executable init;
private boolean multiTest;
private InputStream inputStream;
private OutputStream outputStream;
private Builder() {
}
public Builder solution(Executable solution) {
this.solution = solution;
return Builder.this;
}
public Builder init(Executable init) {
this.init = init;
return Builder.this;
}
public Builder multiTest(boolean multiTest) {
this.multiTest = multiTest;
return Builder.this;
}
public Builder io(StreamPair pair) {
this.inputStream = pair.getInputStream();
this.outputStream = pair.getOutputStream();
return Builder.this;
}
public Problem build() {
return new Problem(solution, init, multiTest, inputStream, outputStream);
}
}
}
private static class FastReader {
private final BufferedReader br;
private StringTokenizer st;
public FastReader(InputStream inputStream) {
br = new BufferedReader(new InputStreamReader(inputStream));
}
public String next() {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(nextLine());
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public String nextLine() {
try {
String line = br.readLine();
if (line == null) {
throw new RuntimeException("empty line");
}
st = null;
return line;
} catch (IOException e) {
throw new RuntimeException(e);
}
}
}
private static class FastWriter {
final private PrintWriter printWriter;
public FastWriter(OutputStream outputStream) {
printWriter = new PrintWriter(new OutputStreamWriter(outputStream));
}
public void println(String s) {
printWriter.println(s);
}
public void close() {
printWriter.close();
}
}
private static class StreamPair {
private final InputStream inputStream;
private final OutputStream outputStream;
public StreamPair(InputStream inputStream, OutputStream outputStream) {
this.inputStream = inputStream;
this.outputStream = outputStream;
}
public static StreamPair standard() {
return new StreamPair(System.in, System.out);
}
public static StreamPair debug(Class resourceClass) {
try {
File file = new File(resourceClass.getClassLoader().getResource("input.txt").getFile());
return new StreamPair(new FileInputStream(file), System.out);
} catch (FileNotFoundException e) {
throw new RuntimeException(e);
}
}
public InputStream getInputStream() {
return inputStream;
}
public OutputStream getOutputStream() {
return outputStream;
}
}
private interface Executable {
void run(FastReader in, FastWriter out);
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 9ddebc2b6e0c8bd534ae024dd4de6043 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.io.*;
public class B {
static class Scan {
private byte[] buf=new byte[1024];
private int index;
private InputStream in;
private int total;
public Scan()
{
in=System.in;
}
public int scan()throws IOException
{
if(total<0)
throw new InputMismatchException();
if(index>=total)
{
index=0;
total=in.read(buf);
if(total<=0)
return -1;
}
return buf[index++];
}
public int scanInt()throws IOException
{
int integer=0;
int n=scan();
while(isWhiteSpace(n))
n=scan();
int neg=1;
if(n=='-')
{
neg=-1;
n=scan();
}
while(!isWhiteSpace(n))
{
if(n>='0'&&n<='9')
{
integer*=10;
integer+=n-'0';
n=scan();
}
else throw new InputMismatchException();
}
return neg*integer;
}
public double scanDouble()throws IOException
{
double doub=0;
int n=scan();
while(isWhiteSpace(n))
n=scan();
int neg=1;
if(n=='-')
{
neg=-1;
n=scan();
}
while(!isWhiteSpace(n)&&n!='.')
{
if(n>='0'&&n<='9')
{
doub*=10;
doub+=n-'0';
n=scan();
}
else throw new InputMismatchException();
}
if(n=='.')
{
n=scan();
double temp=1;
while(!isWhiteSpace(n))
{
if(n>='0'&&n<='9')
{
temp/=10;
doub+=(n-'0')*temp;
n=scan();
}
else throw new InputMismatchException();
}
}
return doub*neg;
}
public String scanString()throws IOException
{
StringBuilder sb=new StringBuilder();
int n=scan();
while(isWhiteSpace(n))
n=scan();
while(!isWhiteSpace(n))
{
sb.append((char)n);
n=scan();
}
return sb.toString();
}
private boolean isWhiteSpace(int n)
{
if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
return true;
return false;
}
}
public static void sort(int arr[],int l,int r) { //sort(arr,0,n-1);
if(l==r) {
return;
}
int mid=(l+r)/2;
sort(arr,l,mid);
sort(arr,mid+1,r);
merge(arr,l,mid,mid+1,r);
}
public static void merge(int arr[],int l1,int r1,int l2,int r2) {
int tmp[]=new int[r2-l1+1];
int indx1=l1,indx2=l2;
//sorting the two halves using a tmp array
for(int i=0;i<tmp.length;i++) {
if(indx1>r1) {
tmp[i]=arr[indx2];
indx2++;
continue;
}
if(indx2>r2) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
if(arr[indx1]<arr[indx2]) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
tmp[i]=arr[indx2];
indx2++;
}
//Copying the elements of tmp into the main array
for(int i=0,j=l1;i<tmp.length;i++,j++) {
arr[j]=tmp[i];
}
}
public static void sort(long arr[],int l,int r) { //sort(arr,0,n-1);
if(l==r) {
return;
}
int mid=(l+r)/2;
sort(arr,l,mid);
sort(arr,mid+1,r);
merge(arr,l,mid,mid+1,r);
}
public static void merge(long arr[],int l1,int r1,int l2,int r2) {
long tmp[]=new long[r2-l1+1];
int indx1=l1,indx2=l2;
//sorting the two halves using a tmp array
for(int i=0;i<tmp.length;i++) {
if(indx1>r1) {
tmp[i]=arr[indx2];
indx2++;
continue;
}
if(indx2>r2) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
if(arr[indx1]<arr[indx2]) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
tmp[i]=arr[indx2];
indx2++;
}
//Copying the elements of tmp into the main array
for(int i=0,j=l1;i<tmp.length;i++,j++) {
arr[j]=tmp[i];
}
}
public static void main(String args[]) throws IOException {
Scan input=new Scan();
StringBuilder ans=new StringBuilder("");
int test=input.scanInt();
for(int tt=1;tt<=test;tt++) {
String str=input.scanString();
if(str.length()==1) {
ans.append("NO\n");
continue;
}
if(str.charAt(0)=='B') {
ans.append("NO\n");
continue;
}
if(!str.contains("B")) {
ans.append("NO\n");
continue;
}
if(str.charAt(str.length()-1)!='B') {
ans.append("NO\n");
continue;
}
boolean is_pos=true;
int a=0,b=0;
for(int i=0;i<str.length();i++) {
if(str.charAt(i)=='A') {
a++;
}
else {
b++;
}
if(b>a) {
is_pos=false;
}
}
if(b>a) {
is_pos=false;
}
if(is_pos) {
ans.append("YES\n");
}
else {
ans.append("NO\n");
}
}
System.out.println(ans);
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 93555dfd6e20c249e4ba1d2ecc6173cf | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
import java.lang.*;
import java.math.BigInteger;
import java.io.*;
public class Main implements Runnable {
public static void main(String[] args) {
new Thread(null, new Main(), "whatever", 1 << 26).start();
}
private FastScanner sc;
private PrintWriter pw;
public void run() {
try {
boolean isSumitting = true;
// isSumitting = false;
if (isSumitting) {
pw = new PrintWriter(System.out);
sc = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
} else {
pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
sc = new FastScanner(new BufferedReader(new FileReader("input.txt")));
}
} catch (Exception e) {
throw new RuntimeException();
}
int t = sc.nextInt();
// int t = 1;
while (t-- > 0) {
// sc.nextLine();
// System.out.println("for t=" + t);
solve();
}
pw.close();
}
public long mod = 1_000_000_007;
private class Pair {
int first;
long second;
Pair(int first, long second) {
this.first = first;
this.second = second;
}
}
private void solve() {
String s = sc.next();
int n = s.length();
if (s.charAt(n - 1) != 'B' || s.charAt(0) == 'B') {
pw.println("NO");
return;
}
int countA = 0;
int countB = 0;
for (int i = 0; i < n; i++) {
if (s.charAt(i) == 'A') countA++;
if (s.charAt(i) == 'B') {
countB++;
if (countA < countB) {
pw.println("NO");
return;
}
}
}
pw.println("YES");
}
class FastScanner {
private BufferedReader reader = null;
private StringTokenizer tokenizer = null;
public FastScanner(BufferedReader bf) {
reader = bf;
tokenizer = null;
}
public String next() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public String nextLine() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
return reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken("\n");
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public float nextFloat() {
return Float.parseFloat(next());
}
public int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String[] nextStringArray(int n) {
String[] a = new String[n];
for (int i = 0; i < n; i++) {
a[i] = next();
}
return a;
}
public long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = nextLong();
}
return a;
}
}
private static class Sorter {
public static <T extends Comparable<? super T>> void sort(T[] arr) {
Arrays.sort(arr);
}
public static <T> void sort(T[] arr, Comparator<T> c) {
Arrays.sort(arr, c);
}
public static <T> void sort(T[][] arr, Comparator<T[]> c) {
Arrays.sort(arr, c);
}
public static <T extends Comparable<? super T>> void sort(ArrayList<T> arr) {
Collections.sort(arr);
}
public static <T> void sort(ArrayList<T> arr, Comparator<T> c) {
Collections.sort(arr, c);
}
public static void normalSort(int[] arr) {
Arrays.sort(arr);
}
public static void normalSort(long[] arr) {
Arrays.sort(arr);
}
public static void sort(int[] arr) {
timSort(arr);
}
public static void sort(int[] arr, Comparator<Integer> c) {
timSort(arr, c);
}
public static void sort(int[][] arr, Comparator<Integer[]> c) {
timSort(arr, c);
}
public static void sort(long[] arr) {
timSort(arr);
}
public static void sort(long[] arr, Comparator<Long> c) {
timSort(arr, c);
}
public static void sort(long[][] arr, Comparator<Long[]> c) {
timSort(arr, c);
}
private static void timSort(int[] arr) {
Integer[] temp = new Integer[arr.length];
for (int i = 0; i < arr.length; i++) temp[i] = arr[i];
Arrays.sort(temp);
for (int i = 0; i < arr.length; i++) arr[i] = temp[i];
}
private static void timSort(int[] arr, Comparator<Integer> c) {
Integer[] temp = new Integer[arr.length];
for (int i = 0; i < arr.length; i++) temp[i] = arr[i];
Arrays.sort(temp, c);
for (int i = 0; i < arr.length; i++) arr[i] = temp[i];
}
private static void timSort(int[][] arr, Comparator<Integer[]> c) {
Integer[][] temp = new Integer[arr.length][arr[0].length];
for (int i = 0; i < arr.length; i++)
for (int j = 0; j < arr[0].length; j++)
temp[i][j] = arr[i][j];
Arrays.sort(temp, c);
for (int i = 0; i < arr.length; i++)
for (int j = 0; j < arr[0].length; j++)
temp[i][j] = arr[i][j];
}
private static void timSort(long[] arr) {
Long[] temp = new Long[arr.length];
for (int i = 0; i < arr.length; i++) temp[i] = arr[i];
Arrays.sort(temp);
for (int i = 0; i < arr.length; i++) arr[i] = temp[i];
}
private static void timSort(long[] arr, Comparator<Long> c) {
Long[] temp = new Long[arr.length];
for (int i = 0; i < arr.length; i++) temp[i] = arr[i];
Arrays.sort(temp, c);
for (int i = 0; i < arr.length; i++) arr[i] = temp[i];
}
private static void timSort(long[][] arr, Comparator<Long[]> c) {
Long[][] temp = new Long[arr.length][arr[0].length];
for (int i = 0; i < arr.length; i++)
for (int j = 0; j < arr[0].length; j++)
temp[i][j] = arr[i][j];
Arrays.sort(temp, c);
for (int i = 0; i < arr.length; i++)
for (int j = 0; j < arr[0].length; j++)
temp[i][j] = arr[i][j];
}
}
public long fastPow(long x, long y, long mod) {
if (y == 0) return 1;
if (y == 1) return x % mod;
long temp = fastPow(x, y / 2, mod);
long ans = (temp * temp) % mod;
return (y % 2 == 1) ? (ans * (x % mod)) % mod : ans;
}
public long fastPow(long x, long y) {
if (y == 0) return 1;
if (y == 1) return x;
long temp = fastPow(x, y / 2);
long ans = (temp * temp);
return (y % 2 == 1) ? (ans * x) : ans;
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 54091ca300b9ec7ab01ae03707b89aec | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | /*LoudSilence*/
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Solution {
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static FastScanner s = new FastScanner();
static FastWriter out = new FastWriter();
final static int mod = (int)1e9 + 7;
final static int INT_MAX = Integer.MAX_VALUE;
final static int INT_MIN = Integer.MIN_VALUE;
final static long LONG_MAX = Long.MAX_VALUE;
final static long LONG_MIN = Long.MIN_VALUE;
final static double DOUBLE_MAX = Double.MAX_VALUE;
final static double DOUBLE_MIN = Double.MIN_VALUE;
final static float FLOAT_MAX = Float.MAX_VALUE;
final static float FLOAT_MIN = Float.MIN_VALUE;
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static class FastScanner{BufferedReader br;StringTokenizer st;
public FastScanner() {if(System.getProperty("ONLINE_JUDGE") == null){try {br = new BufferedReader(new FileReader("E:\\Competitive Coding\\input.txt"));}
catch (FileNotFoundException e) {br = new BufferedReader(new InputStreamReader(System.in));}}else{br = new BufferedReader(new InputStreamReader(System.in));}}
String next(){while (st == null || !st.hasMoreElements()){try{st = new StringTokenizer(br.readLine());}catch (IOException e){e.printStackTrace();}}return st.nextToken();}
int nextInt(){return Integer.parseInt(next());}
long nextLong(){return Long.parseLong(next());}
double nextDouble(){return Double.parseDouble(next());}
List<Integer> readIntList(int n){List<Integer> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextInt()); return arr;}
List<Long> readLongList(int n){List<Long> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextLong()); return arr;}
int[] readIntArr(int n){int[] arr = new int[n]; for (int i = 0; i < n; i++) arr[i] = s.nextInt(); return arr;}
long[] readLongArr(int n){long[] arr = new long[n]; for (int i = 0; i < n; i++) arr[i] = s.nextLong(); return arr;}
String nextLine(){String str = "";try{str = br.readLine();}catch (IOException e){e.printStackTrace();}return str;}}
static class FastWriter{private BufferedWriter bw;public FastWriter(){if(System.getProperty("ONLINE_JUDGE") == null){try {this.bw = new BufferedWriter(new FileWriter("E:\\Competitive Coding\\output.txt"));}
catch (IOException e) {this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}else{this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}
public void print(Object object) throws IOException{bw.append(""+ object);}
public void println(Object object) throws IOException{print(object);bw.append("\n");}
public void debug(int object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void debug(long object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void close() throws IOException{bw.close();}}
public static void println(Object str) throws IOException{out.println(""+str);}
public static void println(Object str, int nextLine) throws IOException{out.print(""+str);}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static ArrayList<Integer> seive(int n){ArrayList<Integer> list = new ArrayList<>();int arr[] = new int[n+1];for(long i = 2; i <= n; i++) {if(arr[(int)i] == 1) {continue;}else {list.add((int)i);for(long j = i*i; j <= n; j = j + i) {arr[(int)j] = 1;}}}return list;}
public static long gcd(long a, long b){if(a > b) {a = (a+b)-(b=a);}if(a == 0L){return b;}return gcd(b%a, a);}
public static void swap(int[] arr, int i, int j) {arr[i] = arr[i] ^ arr[j]; arr[j] = arr[j] ^ arr[i]; arr[i] = arr[i] ^ arr[j];}
public static boolean isPrime(long n){if(n < 2){return false;}if(n == 2 || n == 3){return true;}if(n%2 == 0 || n%3 == 0){return false;}long sqrtN = (long)Math.sqrt(n)+1;for(long i = 6L; i <= sqrtN; i += 6) {if(n%(i-1) == 0 || n%(i+1) == 0) return false;}return true;}
public static long mod_add(long a, long b){ return (a%mod + b%mod)%mod;}
public static long mod_sub(long a, long b){ return (a%mod - b%mod + mod)%mod;}
public static long mod_mul(long a, long b){ return (a%mod * b%mod)%mod;}
public static long modInv(long a, long b){ return expo(a, b-2)%b;}
public static long mod_div(long a, long b){return mod_mul(a, modInv(b, mod));}
public static long expo (long a, long n){if(n == 0){return 1;}long recAns = expo(mod_mul(a,a), n/2);if(n % 2 == 0){return recAns;}else{return mod_mul(a, recAns);}}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Pair class
static class Pair implements Comparable<Pair>{
long f;
long s;
public Pair(long f, long s){
this.f = f;
this.s = s;
}
public String toString(){
return "("+f+", "+s+")";
}
@Override
public int compareTo(Pair o) {
if(f == o.f) return s-o.s < 0 ? -1 : 1;
else return f-o.f < 0 ? -1 : 1;
}
}
static class Triplet{
long f;
long s;
long t;
public Triplet(long f, long s, long t){
this.f = f;
this.s = s;
this.t = t;
}
public String toString(){
return "("+f+", "+s+", "+t+")";
}
}
// public static class Pair<X extends Comparable<X>,Y extends Comparable<Y>> implements Comparable<Pair<X,Y>>{
// X first;
// Y second;
// public Pair(X first, Y second){
// this.first = first;
// this.second = second;
// }
// public String toString(){
// return "( " + first+" , "+second+" )";
// }
// @Override
// public int compareTo(Pair<X, Y> o) {
// int t = first.compareTo(o.first);
// if(t == 0) return second.compareTo(o.second);
// return t;
// }
// }
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Code begins
public static void solve() throws IOException {
char arr[] = s.next().toCharArray();
if(arr.length==1){
println("NO");
return;
}
int a = 0;
int b = 0;
for(char ch : arr){
if(ch == 'A') a++;
else b++;
if(b > a){
println("NO");
return;
}
}
if(arr[arr.length-1] == 'B'){
println("YES");
}else{
println("NO");
}
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static void main(String[] args) throws IOException {
int test = s.nextInt();
for(int t = 1; t <= test; t++) {
solve();
}
out.close();
}
}
/*public static boolean allsame(int arr[]){
for(int i = 0; i < arr.length-1; i++){
if(arr[i] != arr[i+1]) return false;
}
return true;
}
public static List<List<Integer>> permutation(int arr[], int i){
if(i == 0){
List<List<Integer>> ans = new ArrayList<>();
List<Integer> li = new ArrayList<>();
li.add(arr[i]);
ans.add(li);
return ans;
}
int temp = arr[i];
List<List<Integer>> rec = permutation(arr, i-1);
List<List<Integer>> ans = new ArrayList<>();
for(List<Integer> li : rec){
for(int j = 0; j <= li.size(); j++){
List<Integer> list = new ArrayList<>();
for(int ele: li){
list.add(ele);
}
list.add(j, temp);
ans.add(list);
}
}
return ans;
}*/
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | b2044c7fe2597005b1260f6a72b045f1 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes |
import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
// System.out.println(sc.nextByte());
// foodanimal(sc);
goodstrings(sc);
}
private static void goodstrings(Scanner scanner){
int testcase=scanner.nextInt();
for(int i=0;i<testcase;i++){
String s=scanner.next();
if(s.charAt(0)=='B'||s.charAt(s.length()-1)=='A')System.out.println("NO");
else{
boolean possbile=true;
int acount=0;
int bcount=0;
for(int j=1;j<s.length();j++){
char f=s.charAt(j-1);
char sec=s.charAt(j);
if(f=='A')acount++;
if(sec=='B')bcount++;
if(f=='B'&&sec=='B' && acount<bcount){
possbile=false;
System.out.println("NO");
break;
}
}
if(possbile)System.out.println("YES");
}
}
}} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | f5be7df1f6c718963531894007f9ce70 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class B_I_love_AAAB {
static long mod = Long.MAX_VALUE;
public static void main(String[] args) {
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
FastReader f = new FastReader();
int t = f.nextInt();
while(t-- > 0){
solve(f, out);
}
out.close();
}
public static void solve(FastReader f, PrintWriter out) {
String s2 = f.nextLine();
int n = s2.length();
if(s2.charAt(n-1) != 'B') {
out.println("NO");
return;
}
int i = 0;
int flag = 0;
while(i < n) {
if(s2.charAt(i) == 'A') {
flag++;
} else {
flag--;
}
if(flag < 0) {
out.println("NO");
return;
}
i++;
}
out.println("YES");
}
// Sort an array
public static void sort(int arr[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i: arr) {
al.add(i);
}
Collections.sort(al);
for(int i = 0; i < arr.length; i++) {
arr[i] = al.get(i);
}
}
// Find all divisors of n
public static void allDivisors(int n) {
for(int i = 1; i*i <= n; i++) {
if(n%i == 0) {
System.out.println(i + " ");
if(i != n/i) {
System.out.println(n/i + " ");
}
}
}
}
// Check if n is prime or not
public static boolean isPrime(int n) {
if(n < 1) return false;
if(n == 2 || n == 3) return true;
if(n % 2 == 0 || n % 3 == 0) return false;
for(int i = 5; i*i <= n; i += 6) {
if(n % i == 0 || n % (i+2) == 0) {
return false;
}
}
return true;
}
// Find gcd of a and b
public static long gcd(long a, long b) {
long dividend = a > b ? a : b;
long divisor = a < b ? a : b;
while(divisor > 0) {
long reminder = dividend % divisor;
dividend = divisor;
divisor = reminder;
}
return dividend;
}
// Find lcm of a and b
public static long lcm(long a, long b) {
long lcm = gcd(a, b);
long hcf = (a * b) / lcm;
return hcf;
}
// Find factorial in O(n) time
public static long fact(int n) {
long res = 1;
for(int i = 2; i <= n; i++) {
res *= res * i;
}
return res;
}
// Find power in O(logb) time
public static long power(long a, long b) {
long res = 1;
while(b > 0) {
if((b&1) == 1) {
res = (res * a)%mod;
}
a = (a * a)%mod;
b >>= 1;
}
return res;
}
// Find nCr
public static long nCr(int n, int r) {
if(r < 0 || r > n) {
return 0;
}
long ans = fact(n) / (fact(r) * fact(n-r));
return ans;
}
// Find nPr
public static long nPr(int n, int r) {
if(r < 0 || r > n) {
return 0;
}
long ans = fact(n) / fact(r);
return ans;
}
// sort all characters of a string
public static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
// User defined class for fast I/O
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
boolean hasNext() {
if (st != null && st.hasMoreTokens()) {
return true;
}
String tmp;
try {
br.mark(1000);
tmp = br.readLine();
if (tmp == null) {
return false;
}
br.reset();
} catch (IOException e) {
return false;
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
float nextFloat() {
return Float.parseFloat(next());
}
boolean nextBoolean() {
return Boolean.parseBoolean(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] nextArray(int n) {
int[] a = new int[n];
for(int i=0; i<n; i++) {
a[i] = nextInt();
}
return a;
}
}
}
/**
Dec Char Dec Char Dec Char Dec Char
--------- --------- --------- ----------
0 NUL (null) 32 SPACE 64 @ 96 `
1 SOH (start of heading) 33 ! 65 A 97 a
2 STX (start of text) 34 " 66 B 98 b
3 ETX (end of text) 35 # 67 C 99 c
4 EOT (end of transmission) 36 $ 68 D 100 d
5 ENQ (enquiry) 37 % 69 E 101 e
6 ACK (acknowledge) 38 & 70 F 102 f
7 BEL (bell) 39 ' 71 G 103 g
8 BS (backspace) 40 ( 72 H 104 h
9 TAB (horizontal tab) 41 ) 73 I 105 i
10 LF (NL line feed, new line) 42 * 74 J 106 j
11 VT (vertical tab) 43 + 75 K 107 k
12 FF (NP form feed, new page) 44 , 76 L 108 l
13 CR (carriage return) 45 - 77 M 109 m
14 SO (shift out) 46 . 78 N 110 n
15 SI (shift in) 47 / 79 O 111 o
16 DLE (data link escape) 48 0 80 P 112 p
17 DC1 (device control 1) 49 1 81 Q 113 q
18 DC2 (device control 2) 50 2 82 R 114 r
19 DC3 (device control 3) 51 3 83 S 115 s
20 DC4 (device control 4) 52 4 84 T 116 t
21 NAK (negative acknowledge) 53 5 85 U 117 u
22 SYN (synchronous idle) 54 6 86 V 118 v
23 ETB (end of trans. block) 55 7 87 W 119 w
24 CAN (cancel) 56 8 88 X 120 x
25 EM (end of medium) 57 9 89 Y 121 y
26 SUB (substitute) 58 : 90 Z 122 z
27 ESC (escape) 59 ; 91 [ 123 {
28 FS (file separator) 60 < 92 \ 124 |
29 GS (group separator) 61 = 93 ] 125 }
30 RS (record separator) 62 > 94 ^ 126 ~
31 US (unit separator) 63 ? 95 _ 127 DEL
*/
// (a/b)%mod == (a * moduloInverse(b)) % mod;
// moduloInverse(b) = power(b, mod-2);
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 9a745d6240b1c9e3e94d9095155aaea4 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
Reader reader = new Reader();
int t = reader.nextInt();
while(t-- > 0){
String s = reader.next();
solve(s);
}
}
private static void solve(String s) {
var n = s.length();
if(n == 1 || s.charAt(0) == 'B' || s.charAt(n - 1) == 'A'){
System.out.println("No");
}else{
int a = 0;
int b = 0;
for(int i = 0;i<n;++i) {
var c = s.charAt(i);
a += c == 'A' ? 1 : 0;
b += c == 'B' ? 1 : 0;
if(b > a) {
System.out.println("no");
return;
}
}
System.out.println(b > a ? "no" : "yes");
}
}
static class Reader{
BufferedReader bufferedReader;
StringTokenizer stringTokenizer;
Reader(){
bufferedReader = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while(stringTokenizer == null || !stringTokenizer.hasMoreElements()){
try{
stringTokenizer = new StringTokenizer(bufferedReader.readLine());
}catch (Exception e){
System.out.println(e.getMessage());
}
}
return stringTokenizer.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String s = "";
try{
s = bufferedReader.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 8023441caeeea344958a9692f242b86f | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes |
import java.util.*;
public class codeFr6 {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-->0) {
String st = s.next();
boolean b=(st.charAt(st.length()-1)=='B');
int sum=0;
for (int i = 0;i<st.length(); i++){
if (st.charAt(i)=='A')
sum++;
else
sum--;
if (sum<0)
b=false;
}
if (b==true)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 9be41f5caee535f00debaa68029e9b90 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes |
import java.util.*;
import java.io.*;
import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.sqrt;
import static java.lang.Math.floor;
public class Solution {
static class ListNode {
int val;
ListNode next;
ListNode() {}
ListNode(int val) { this.val = val; }
ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}
static class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
public static int max(int []nums, int s, int e) {
int max = Integer.MIN_VALUE;
for(int i = s; i <= e; i++) {
max = Math.max(max, nums[i]);
}
return max;
}
static int depth(TreeNode root) {
if(root == null)return 0;
int a = 1+ depth(root.left);
int b = 1+ depth(root.right);
return Math.max(a,b);
}
static HashSet<Integer>set = new HashSet<>();
static void deepestLeaves(TreeNode root, int cur_depth, int depth) {
if(root == null)return;
if(cur_depth == depth)set.add(root.val);
deepestLeaves(root.left,cur_depth+1,depth);
deepestLeaves(root.right,cur_depth+1,depth);
}
public static void print(TreeNode root) {
if(root == null)return;
System.out.print(root.val+" ");
System.out.println("er");
print(root.left);
print(root.right);
}
public static HashSet<Integer>original(TreeNode root){
int d = depth(root);
deepestLeaves(root,0,d);
return set;
}
static HashSet<Integer>set1 = new HashSet<>();
static void leaves(TreeNode root) {
if(root == null)return;
if(root.left == null && root.right == null)set1.add(root.val);
leaves(root.left);
leaves(root.right);
}
public static boolean check(HashSet<Integer>s, HashSet<Integer>s1) {
if(s.size() != s1.size())return false;
for(int a : s) {
if(!s1.contains(a))return false;
}
return true;
}
static TreeNode subTree;
public static void smallest_subTree(TreeNode root) {
if(root == null)return;
smallest_subTree(root.left);
smallest_subTree(root.right);
set1 = new HashSet<>();
leaves(root);
boolean smallest = check(set,set1);
if(smallest) {
subTree = root;
return;
}
}
public static TreeNode answer(TreeNode root) {
smallest_subTree(root);
return subTree;
}
}
static class Key<K1, K2>
{
public K1 key1;
public K2 key2;
public Key(K1 key1, K2 key2)
{
this.key1 = key1;
this.key2 = key2;
}
@Override
public boolean equals(Object o)
{
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Key key = (Key) o;
if (key1 != null ? !key1.equals(key.key1) : key.key1 != null) {
return false;
}
if (key2 != null ? !key2.equals(key.key2) : key.key2 != null) {
return false;
}
return true;
}
@Override
public int hashCode()
{
int result = key1 != null ? key1.hashCode() : 0;
result = 31 * result + (key2 != null ? key2.hashCode() : 0);
return result;
}
@Override
public String toString() {
return "[" + key1 + ", " + key2 + "]";
}
}
public static int sumOfDigits (long n) {
int sum = 0;
while(n > 0) {
sum += n%10;
n /= 10;
}
return sum;
}
public static void swap(int []ar, int i, int j) {
for(int k= j; k >= i; k--) {
int temp = ar[k];
ar[k] = ar[k+1];
ar[k+1] = temp;
}
}
public static int findOr(int[]bits){
int or=0;
for(int i=0;i<32;i++){
or=or<<1;
if(bits[i]>0)
or=or+1;
}
return or;
}
public static boolean[] getSieve(int n) {
boolean[] isPrime = new boolean[n+1];
for (int i = 2; i <= n; i++) isPrime[i] = true;
for (int i = 2; i*i <= n; i++) if (isPrime[i])
for (int j = i; i*j <= n; j++) isPrime[i*j] = false;
return isPrime;
}
public static Long gcd(Long a, Long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static long nextPrime(long n) {
boolean found = false;
long prime = n;
while(!found) {
prime++;
if(isPrime(prime))
found = true;
}
return prime;
}
// method to return LCM of two numbers
static Long lcm(Long a, Long b)
{
return (a / gcd(a, b)) * b;
}
static boolean isPrime(long n) {
if(n <= 1)return false;
if(n <= 3)return true;
if(n%2 == 0 || n%3 == 0)return false;
for(int i = 5; i*i <= n; i+= 6) {
if(n%i == 0 || n%(i+2) == 0)
return false;
}
return true;
}
static int countGreater(int arr[], int n, int k){
int l = 0;
int r = n - 1;
// Stores the index of the left most element
// from the array which is greater than k
int leftGreater = n;
// Finds number of elements greater than k
while (l <= r) {
int m = l + (r - l) / 2;
// If mid element is greater than
// k update leftGreater and r
if (arr[m] > k) {
leftGreater = m;
r = m - 1;
}
// If mid element is less than
// or equal to k update l
else
l = m + 1;
}
// Return the count of elements greater than k
return (n - leftGreater);
}
static ArrayList<Integer>printDivisors(int n){
ArrayList<Integer>list = new ArrayList<>();
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
// If divisors are equal, print only one
if (n/i == i)
list.add(i);
else // Otherwise print both
list.add(i);
list.add(n/i);
}
}
return list;
}
static void leftRotate(int l, int r,int arr[], int d)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(l,r,arr);
}
static void leftRotatebyOne(int l, int r,int arr[])
{
int i, temp;
temp = arr[l];
for (i = l; i < r; i++)
arr[i] = arr[i + 1];
arr[r] = temp;
}
static class pairInPq<F extends Comparable<F>, S extends Comparable<S>>
implements Comparable<pairInPq<F, S>> {
private F first;
private S second;
public pairInPq(F first, S second){
this.first = first;
this.second = second;
}
public F getFirst(){return first;}
public S getSecond(){return second;}
// All the code you already have is fine
@Override
public int compareTo(pairInPq<F, S> o) {
int retVal = getSecond().compareTo(o.getSecond());
if (retVal != 0) {
return retVal;
}
return getFirst().compareTo(o.getFirst());
}
}
static long modInverse(long a, long m)
{
long m0 = m;
long y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1) {
// q is quotient
long q = a / m;
long t = m;
// m is remainder now, process
// same as Euclid's algo
m = a % m;
a = t;
t = y;
// Update x and y
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
static long power(long x, long y, long p)
{
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static TreeNode buildTree(TreeNode root,int []ar, int l, int r){
if(l > r)return null;
int len = l+r;
if(len%2 != 0)len++;
int mid = (len)/2;
int v = ar[mid];
TreeNode temp = new TreeNode(v);
root = temp;
root.left = buildTree(root.left,ar,l,mid-1);
root.right = buildTree(root.right,ar,mid+1,r);
return root;
}
static int LIS(int arr[], int n)
{
int lis[] = new int[n];
int i, j, max = 0;
/* Initialize LIS values for all indexes */
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute optimized LIS values in
bottom up manner */
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] >= arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Pick maximum of all LIS values */
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
static void permuteString(String s , String answer)
{
if (s.length() == 0)
{
System.out.print(answer + " ");
return;
}
for(int i = 0 ;i < s.length(); i++)
{
char ch = s.charAt(i);
String left_substr = s.substring(0, i);
String right_substr = s.substring(i + 1);
String rest = left_substr + right_substr;
permuteString(rest, answer + ch);
}
}
static boolean isPowerOfTwo(long n)
{
if(n==0)
return false;
return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==
(int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static class Pair1{
long x;
long y;
public Pair1(long x, long y) {
this.x = x;
this.y = y;
}
}
static class Pair {
int x;
int y;
// Constructor
public Pair(int x, int y)
{
this.x = x;
this.y = y;
}
}
static class Sorting implements Comparator<Pair>{
public int compare(Pair p1, Pair p2){
if(p1.x==p2.x){
return p1.y-p2.y;
}
return p1.x - p2.x;
}
}
static class Compare1{
static void compare(Pair arr[], int n)
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return p1.x - p2.x;
}
});
}
}
static int min;
static int max;
static LinkedList<Integer>[]adj;
static int n;
static void insertlist(int n) {
for(int i = 0; i <= n; i++) {
adj[i] = new LinkedList<>();
}
}
static int []ar;
static boolean []vis;
static HashMap<Key,Integer>map;
static ArrayList<Integer>list;
static int dfs(int parent, int child,LinkedList<Integer>[]adj) {
list.add(parent);
for(int a : adj[parent]) {
if(vis[a] == false) {
vis[a] = true;
return 1+dfs(a,parent,adj);
}
}
return 0;
}
static StringTokenizer st;
static BufferedReader ob;
static int [] readarrayInt( int n)throws IOException {
st = new StringTokenizer(ob.readLine());
int []ar = new int[n];
for(int i = 0; i < n; i++) {
ar[i] = Integer.parseInt(st.nextToken());
}
return ar;
}
static long [] readarrayLong( int n)throws IOException {
st = new StringTokenizer(ob.readLine());
long []ar = new long[n];
for(int i = 0; i < n; i++) {
ar[i] = Long.parseLong(st.nextToken());
}
return ar;
}
static int readInt()throws IOException {
return Integer.parseInt(ob.readLine());
}
static long readLong()throws IOException {
return Long.parseLong(ob.readLine());
}
static int nextTokenInt() {
return Integer.parseInt(st.nextToken());
}
static long nextTokenLong() {
return Long.parseLong(st.nextToken());
}
static int root(int u) {
if(ar[u] == -1)return -1;
if(u == ar[u]) {
return u;
}
return root(ar[u]);
}
static Pair []pairar;
static int numberOfDiv(long num) {
int c = 0;
for(int i = 1; i <= Math.sqrt(num ); i++) {
if(num%i == 0) {
long d = num/i;
if( d == i)c++; else c += 2;
}
}
return c;
}
static long ans;
static int count;
static boolean []computed;
static void NoOfWays(int n) {
if(n == 0 ) {
count++;
}
if(n <= 0)return;
for(int i = 1; i <= n; i++) {
if(n+i <= n)
NoOfWays(n-i);
}
}
static boolean binarylistsearch( List<Integer>list, int l, int r, int s, int e) {
if(s > e)return false;
int mid = (s+e)/2;
if(list.get(mid) >= l && list.get(mid) < r) {
return true;
}else if(list.get(mid) > r) {
return binarylistsearch(list,l,r,s,mid-1);
}else if(list.get(mid) < l) {
return binarylistsearch(list,l,r,mid+1,e);
}
return false;
}
static int [][] readmatrix(int r, int c)throws IOException{
int [][]mat = new int[r][c];
for(int i = 0; i < r; i++) {
st = new StringTokenizer(ob.readLine());
for(int j = 0; j < c; j++) {
mat[i][j] = nextTokenInt();
}
}
return mat;
}
static HashSet<Integer>set1;
static boolean possible;
static int c = 0;
static void isbeautiful(HashSet<Integer>s,int num, List<Integer>good, int i, int x, int count) {
if(c == 2) {
possible = true;
return;
}
if(num >x || i == good.size())return;
if(num > x)return;
if(i >= good.size())return;
for(int j = i; j < good.size(); j++) {
if(!map.containsKey(new Key(num,good.get(j))) &&
!map.containsKey(new Key(good.get(j),num))){
if(s.contains(num) && s.contains(good.get(j)))
map.put(new Key (num,good.get(j)),1);
isbeautiful(s,num*good.get(j),good,i,x,count);
}
}
}
static long sum;
static long mod;
static void recur(HashSet<Integer>set,HashMap<Integer,HashSet<Integer>>map, int n, int j) {
if(j > n)return;
int v = 0;
for(int a : set) {
if(map.get(j).contains(a)) {
v++;
}
}
long d = map.get(j).size()-v;
sum = (sum*d)%mod;
HashSet<Integer> temp = map.get(j);
for(int num : temp) {
if(!set.contains(num)) {
set.add(num);
recur(set,map,n,j+1);
set.remove(num);
}
}
}
static int key1;
static int key2;
static HashSet<Integer>set;
public static TreeNode lowestCommonAncestor(TreeNode root) {
if(root == null)return null;
TreeNode left = lowestCommonAncestor(root.left);
TreeNode right =lowestCommonAncestor(root.right);
if(left == null && right != null) {
System.out.println(right.val);
return right;
}else if(right == null && left != null) {
System.out.println(left.val);
return left;
}else if(left == null && right == null) {
return null;
}else {
System.out.println(root.val);
return root;
}
}
static ArrayList<Integer>res;
static boolean poss = false;
public static void recur1 (char []ar, int i) {
if(i >= ar.length) {
boolean isPalindrome = false;
for(int k = 0; k < ar.length; k++) {
for(int j = 0; j < ar.length; j++) {
if(j-k+1 >= 5) {
int x = k;
int y = j;
while(x < y && ar[x] == ar[y]) {
x++;
y--;
}
if(x == y || x > y) {
isPalindrome = true;
}
}
}
}
if(!isPalindrome) {
poss = true;
}
}
if(i < ar.length && ar[i] == '?' ) {
ar[i] = '0';
recur1(ar,i+1);
ar[i] = '?';
}
if(i < ar.length && ar[i] == '?') {
ar[i] = '1';
recur1(ar,i+1);
ar[i] = '?';
}
if(i < ar.length && ar[i] != '?') {
recur1(ar,i+1);
}
}
static int []theArray;
static int rotate(int []ar, int element, int i) {
int count = 0;
while(ar[i] != element) {
int last = ar[i];
count++;
int prev = ar[1];
for(int j = 2; j <= i; j++) {
int temp = ar[j];
ar[j] = prev;
prev = temp;
}
ar[1] = prev;
}
return count;
}
static int []A;
static long nth(long max, long min, int n){
long mod = 1000000007;
if(max%min == 0){
System.out.println(min*n);
return (min*n)%mod;
}else{
long d = max/min;
d++;
long e = n/d;
long ans = (max*e)%mod;
long r = n%d;
long div = (ans/min);
long temp = (div*min);
long f = (temp*r)%mod;
ans = f;
if(temp == ans){
ans += min;
}
System.out.println(ans);
return ans;
}
}
public static int index(int k, int l, int r) {
if(r-l == 1) {
if(ar[r] >= k && ar[l] < k) {
return l;
}
}else if(l >= r){
return l;
}
int mid = (l+r)/2;
if(ar[mid] >= k) {
return index(k,l,mid-1);
}else {
return index(k,mid+1,r);
}
}
public static int remSnakes(int start, int end, int k) {
int rem = k-ar[end];
if(start+rem > end ) {
return 0;
}else {
return 1+ remSnakes(start+rem,end-1,k);
}
}
static int N;
static int []tree = new int[2*N];
static void build(int []arr,boolean odd) {
for(int i = 0; i < N; i++) {
tree[N+i] = arr[i];
}
for(int i = N-1; i > 0; --i) {
int left = i << 1;
int right = i << 1|1;
if(odd)
tree[i] = tree[i<<1]|tree[i<<1|1];
else
tree[i] = tree[i<<1]^tree[i<<1|1];
}
}
// function to update a tree node
static void updateTreeNode(int p, int value, boolean odd)
{
// set value at position p
tree[p + N] = value;
p = p + N;
// move upward and update parents
for (int i = p; i > 1; i >>= 1)
if(odd)
tree[i >> 1] = tree[i] | tree[i^1];
else
tree[i >> 1] = tree[i]^tree[i^1];
}
static int query(int l ,int r) {
int res = 0;
//loop to build the sum in the range
for(l += N, r += N; l < r; l >>= 1, r >>= 1) {
if((l&1) > 0) {
res += tree[--r];
}
if((r&1) > 0) {
res += tree[l++];
}
}
return res;
}
static int sum1;
static void min_line(int u, int min) {
for(int i : adj[u]) {
min_line(i,min);
}
System.out.println(u);
}
static class Key1 {
public final int X;
public final int Y;
public Key1(final int X, final int Y) {
this.X = X;
this.Y = Y;
}
public boolean equals (final Object O) {
if (!(O instanceof Key1)) return false;
if (((Key1) O).X != X) return false;
if (((Key1) O).Y != Y) return false;
return true;
}
public int hashCode() {
return (X << 16) + Y;
}
}
static void solve() {
int []points = new int[26];
int []top_pos = new int[26];
String []ar = {"ABC"};
Arrays.fill(top_pos, Integer.MAX_VALUE);
for(int i = 0; i < ar.length; i++) {
String s = ar[i];
for(int j = 0; j < s.length(); j++) {
int pos = 97-(s.charAt(j)-'0');
points[pos] += (j);
if(j < top_pos[pos]) {
top_pos[pos] = j;
}
}
}
int i = 0;
boolean []vis = new boolean[26];
StringBuilder sb = new StringBuilder();
while(true) {
int min = Integer.MAX_VALUE;
for(int j = 0; j < 26; j++) {
if(vis[j] == false) {
min = Math.min(min, points[j]);
}
}
if(min == Integer.MAX_VALUE)break;
PriorityQueue<pairInPq>pq = new PriorityQueue<>();
for(int j = 0; j < 26; j++) {
if(points[j] == min) {
vis[j] = true;
char c = (char)(j + 'a');
pq.add(new pairInPq(c,top_pos[j]));
}
}
while(pq.size() > 0) {
pairInPq p = pq.poll();
char c = (char)p.first;
sb.append(c);
}
}
String res = sb.toString();
res = res.toUpperCase();
System.out.println(res);
}
static boolean al_sub(String s, String a) {
int k = 0;
for(int i = 0; i < s.length(); i++) {
if(k == a.length())break;
if(s.charAt(i) == a.charAt(k)) {
k++;
}
}
return k == a.length();
}
static String result(String s, String a) {
char []s_ar = s.toCharArray();
char []a_ar = a.toCharArray();
StringBuilder sb = new StringBuilder();
if(s.length() < a.length()) {
for(int i = 0; i < s.length(); i++) {
if(s_ar[i] == '?')s_ar[i] = 'a';
}
}else {
if(al_sub(s,a)) {
return "-1";
}else {
int k = 0;
char element = 'z';
for(char i = 'a'; i <= 'e'; i++) {
char []temp = s.toCharArray();
boolean pos = true;;
for(int j = 0; j < s.length(); j++) {
if(temp[j] == '?') {
temp[j] = i;
}
}
boolean sub = false;
k = 0;
for(int j = 0; j < s.length(); j++) {
if(k == a.length()) {
pos = false;
break;
}
if(a_ar[k] == temp[j]) {
k++;
}
}
if(pos && k < a.length()) {
element = i;
break;
}
}
if(element == 'z')return "-1";
for(int i = 0; i < s.length(); i++) {
if(s_ar[i] == '?') {
s_ar[i] = element;
}
sb.append(s_ar[i]);
}
return sb.toString();
}
}
return "-1";
}
static void addNodes(ListNode l, int k) {
if(k > 10 )return;
ListNode temp = new ListNode(k);
l.next = temp;
addNodes(l.next,k+1);
}
static ListNode head;
static int listnode_recur(ListNode tail) {
if(tail == null )return 0;
int n = listnode_recur(tail.next);
max = Math.max(max, tail.val+head.val);
head = head.next;
return max;
}
static void recursion(int []fill, int len, int []valid, int k) {
if(k == len) {
int num = 0;
for(int i = 0; i < fill.length; i++) {
num = (num*10)+fill[i];
}
System.out.println(num);
if(!set.contains(num)) {
set.add(num);
}
return;
}
for(int i = 0; i < valid.length; i++) {
if(fill[k] == 0) {
fill[k] = valid[i];
recursion(fill,len,valid,k+1);
fill[k] = 0;
}else {
recursion(fill,len,valid,k+1);
}
}
}
static long minans(Long []ar,Long max) {
Long one = 0l;
Long two = 0l;
for(int i = 0; i < ar.length; i++) {
Long d = (max-ar[i])/2;
two += d;
one += (max-ar[i])%2;
}
Long ans = 0l;
if(one >= two) {
ans += (two*2);
// System.out.println(one+" "+two+" "+ans);
one -= two;
if(one > 0)
ans += (one*2)-1;
}else {
ans += (one*2);
if(two > 0) {
two -= one;
Long d = two/3;
Long temp = d*3;
Long r = two%3;
ans += temp;
ans += d;
if(r == 2) {
ans += 3;
}else if(r == 1){
ans += 2;
}
}
}
if(ans < 0)ans = 0l;
// System.out.println(one+" "+two+" "+ans);
return ans;
}
static int binarySearch(long []dif,long left, int s, int e) {
if(s > e)return s;
int mid = (s+e)/2;
if(dif[mid] > left) {
return binarySearch(dif,left,s,mid-1);
}else {
return binarySearch(dif,left,mid+1,e);
}
}
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int size1 = getLength(l1);
int size2 = getLength(l2);
ListNode head = new ListNode(1);
// make sure length l1 >= length2
head.next = size1 < size2?helper(l2,l1,size2-size1):helper(l1,l2,size1-size2);
if(head.next.val > 9) {
head.next.val = head.next.val%10;
return head;
}
return head.next;
}
public static int getLength(ListNode l1) {
int count = 0;
while(l1 != null) {
l1 = l1.next;
count++;
}
return count;
}
public static ListNode helper(ListNode l1, ListNode l2, int offset) {
if(l1 == null)return null;
ListNode result = offset == 0?new ListNode(l1.val+l2.val):new ListNode(l1.val);
System.out.println(l1.val+" "+l2.val);
ListNode post = offset == 0?helper(l1.next,l2.next,0):helper(l1.next,l2,offset-1);
if(post != null && post.val > 9) {
result.val += 1;
post.val = post.val%10;
}
result.next = post;
return result;
}
public static ListNode arrayToLinkedList(int []ar, ListNode head) {
head = new ListNode(0);
head.next = null;
ListNode h = head;
for(int i = 0; i < ar.length; i++) {
ListNode cur = new ListNode(ar[i]);
if(h == null) {
h = cur;
}else{
h.next = cur;
h = h.next;
}
}
return head.next;
}
static Long numLessThanN(Long n, int s, int e,Long []powerful) {
if(s >= e) {
if(powerful[s] > n)return powerful[s-1];
return powerful[s];
}
int mid = (s+e)/2;
if(powerful[mid] == n) {
return powerful[mid];
}else if(powerful[mid] > n && powerful[mid-1] < n) {
return powerful[mid-1];
}else if(powerful[mid] > n){
return numLessThanN(n,0,mid-1,powerful);
}else {
return numLessThanN(n,mid+1,e,powerful);
}
}
static Long minimumX;
static void minX(Long x, Long c) {
Long p = 2l;
Long a = x;
if(x > minimumX)return;
if(x == 1)return;
while(p*p <= x) {
int count = 0;
Long num = (long)Math.pow(p, c);
Long minx = lcm(num,x);
minx = minx/(gcd(num,x));
minimumX = Math.min(minimumX, minx);
// System.out.println(minimumX+" "+x+" "+p+" "+num);
//System.out.println(minx+" "+"df"+" "+p);
if(minx < x) {
minX(minx,c);
}
p++;
}
}
static boolean isConsecutive (List<Integer>list) {
for(int i : list) {
if(i > 3)return false;
}
int x = 0;
int y = 0;
for(int i = 0; i < list.size(); i++) {
if(list.get(i) == 2) {
x++;
}else if(list.get(i) == 3)y++;
}
if(y >= 2 || x >= 3)return false;
if(x > 0 && y > 0)return false;
return true;
}
static int []ismean(int []ar, int one, int minusone, int []temp){
int n = ar.length;
for(int i = 1; i < n-1; i++) {
if(temp[i] == 1 && temp[i-1] == -1 && one > 0) {
one--;
temp[i+1] = 1;
}else if(temp[i-1] == -1 && temp[i] == -1 && one > 0) {
temp[i+1] = 1;
one--;
}else if(temp[i] == -1 && temp[i-1] == 1 && minusone > 0) {
minusone--;
temp[i+1] = -1;
}else if(temp[i] == 1 && temp[i-1] == 1 && minusone > 0) {
minusone--;
temp[i+1] = -1;
}else {
return temp;
}
}
return temp;
}
static boolean isnottwo(int []ar, int one, int minusone,int temp[]) {
int n = ar.length;
int []ans = ismean(ar,one,minusone,temp);
if(ans[n-1] != -2)return true;
return false;
}
public static long findClosest(long arr[], long target)
{
int n = arr.length;
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n-1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1], target);
i = mid + 1; // update i
}
}
// Only single element left after search
return arr[mid];
}
// Method to compare which one is the more close
// We find the closest by taking the difference
// between the target and both values. It assumes
// that val2 is greater than val1 and target lies
// between these two.
public static long getClosest(long val1, long val2,
long target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
static int gain(int i) {
if(i == 0 || i == n) {
return 0;
}
if(ar[i] > ar[i-1] && ar[i] > ar[i+1] || ar[i] < ar[i-1] && ar[i] < ar[i+1])return 1;
return 0;
}
static int number_of_theif(String s ) {
int c = 0;
int n = s.length();
int zero = 0;
for(char i : s.toCharArray()) {
if(i == '?')c++;
if(i == '0')zero++;
}
if(c == n)return n;
if(zero == n)return 1;
int index = -1;
for(int i = 0; i < n; i++) {
if(s.charAt(i) == '1') {
index = i;
}
}
if(index != -1) {
int zero_index = -1;
for(int j = index+1; j < n; j++) {
if(s.charAt(j) == '0') {
zero_index = j;
break;
}
}
if(zero_index == -1) {
return n-index;
}else {
return zero_index-index+1;
}
}else {
for(int i = 0; i < n; i++) {
if(s.charAt(i) == '0') {
return i+1;
}
}
}
return 1;
}
static List<List<Integer>>res0;
static int minpath;
static void recursion(List<Integer>l, int parent, LinkedList<Integer>[]adj) {
if(adj[parent].size() == 0) {
l.add(parent);
res0.add(new ArrayList(l));
minpath++;
return;
}else {
l.add(parent);
recursion(l,adj[parent].get(0),adj);
for(int i = 1; i < adj[parent].size(); i++) {
recursion(new ArrayList<>(), adj[parent].get(i),adj);
}
}
}
static class FastReader {
BufferedReader br; StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()) {
try {st = new StringTokenizer(br.readLine());}
catch (IOException e) {e.printStackTrace();}
}
return st.nextToken();
}
int nextInt(){ return Integer.parseInt(next()); }
long nextLong(){ return Long.parseLong(next()); }
double nextDouble(){ return Double.parseDouble(next()); }
String nextLine(){
String str = "";
try {str = br.readLine();}
catch (IOException e) {e.printStackTrace();}
return str;
}
}
static HashMap<Integer,Integer>Amap;
static HashMap<Integer,Integer>Bmap;
static void numberOfPerm(int index, int []A, int []B, int []C, long cur_ans, int n) {
if(index == n) {
sum += cur_ans%mod;
//System.out.println(cur_ans);
return;
}
Amap.remove(A[index]);
Bmap.remove(B[index]);
// System.out.println(set+" "+index);
if(C[index] != 0) {
numberOfPerm(index+1,A,B,C,cur_ans,n);
// System.out.println(index);
}else
if(!set.contains(Amap.get(A[index])) && !set.contains(Bmap.get(A[index])) &&
!set.contains(Amap.get(B[index])) && !set.contains(Bmap.get(B[index])) &&
!set.contains(A[index]) && !set.contains(B[index])){
set.add(A[index]);
set.add(B[index]);
numberOfPerm(index+1,A,B,C,(cur_ans*2)%mod,n);
}else {
if((!set.contains(Amap.get(A[index])) && !set.contains(Bmap.get(A[index]))
&& !set.contains(A[index]))) {
set.add(A[index]);
numberOfPerm(index+1,A,B,C,cur_ans,n);
}else if (!set.contains(Amap.get(B[index])) && !set.contains(Bmap.get(A[index]))
&& !set.contains(B[index])) {
set.add(B[index]);
// System.out.println("vampire");
numberOfPerm(index+1,A,B,C,cur_ans,n);
}else {
numberOfPerm(index+1,A,B,C,cur_ans,n);
}
}
}
static int [][]mat;
static int k;
public static void last_tree_down( int n, int m) {
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(mat[i][j] == k-1) {
if(i-1 >= 0 && mat[i-1][j] == 0) {
mat[i-1][j] = k;
}
if(i+1 < n && mat[i+1][j] == 0) {
mat[i+1][j] = k;
}
if(j-1 >= 0 && mat[i][j-1] == 0) {
mat[i][j-1] = k;
}
if(j+1 < n && mat[i][j+1] == 0) {
mat[i][j+1] = k;
}
}
}
}
boolean allTreesDown = true;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(mat[i][j] == 0) {
allTreesDown = false;
}
}
}
if(!allTreesDown) {
k++;
last_tree_down(n,m);
}
}
public static void main(String args[])throws IOException {
ob = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
BufferedOutputStream out = new BufferedOutputStream(System.out);
FastReader sc = new FastReader();
FileOutputStream out1 = new FileOutputStream("output.txt");
// BufferedReader ob1 = new BufferedReader(new FileReader("input.txt"));
PrintWriter writer = new PrintWriter("output.txt", "UTF-8");
int t = sc.nextInt();
while(t --> 0) {
String s = sc.next();
boolean pos = true;
boolean B = false;
int countA = 0;
int countB = 0;
for(int i = 0; i < s.length(); i++) {
if(s.charAt(i) == 'B') {
countB++;
}else {
countA++;
}
if(countB > countA) {
pos = false;
break;
}
}
if(countB == 0 || !pos || s.charAt(s.length()-1) == 'A') {
System.out.println("NO");
}else {
System.out.println("YES");
}
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | fb66b286a6af25515bfa9116f24c8c64 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.Scanner;
public class aa {
public static void main(String[] args) {
var in = new Scanner(System.in);
int n = in.nextInt();
for(int i = 0; i < n; i++) {
int nA = 0;
int nB = 0;
var word = in.next();
int m = word.length();
int j = 0;
if('B' != word.charAt(m -1)){
System.out.println("NO");
continue;
}
while(j < m){
if(word.charAt(j) != 'A') {
nB++;
if(nB > nA) {
System.out.println("NO");
break;
}
}
else nA++;
j++;
}
if(j == m)
System.out.println("YES");
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 0e3514def25a539f766cb6a228abaf24 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.security.cert.X509CRL;
import java.util.*;
import java.lang.*;
import java.util.stream.Collector;
import java.util.stream.Collectors;
@SuppressWarnings("unused")
public class Main {
static InputStream is;
static PrintWriter out;
static String INPUT = "";
static String OUTPUT = "";
//global
//private final static long BASE = 998244353L;
private final static int BASE = 1000000007;
private final static int INF_I = 1001001001;
private final static long INF_L = 1001001001001001001L;
private final static int MAXN = 100100;
private static List<Integer> palin = new ArrayList<>();
static void solve() {
int ntest = readInt();
for (int test=0;test<ntest;test++) {
char[] S = readString().toCharArray();
int cntA=0;
boolean isOk = true, hasB = false;
for (int i=0;i<S.length;i++) {
if (S[i]=='A') cntA+=1;
else {
hasB |= (i==S.length-1);
isOk &= (cntA>=1);
cntA-=1;
}
}
if (isOk && hasB) out.println("yes");
else out.println("no");
}
}
public static void main(String[] args) throws Exception
{
long S = System.currentTimeMillis();
if (INPUT=="") {
is = System.in;
} else {
File file = new File(INPUT);
is = new FileInputStream(file);
}
if (OUTPUT == "") out = new PrintWriter(System.out);
else out = new PrintWriter(OUTPUT);
solve();
out.flush();
long G = System.currentTimeMillis();
}
private static class Point<T extends Number & Comparable<T>> implements Comparable<Point<T>> {
private T x;
private T y;
public Point(T x, T y) {
this.x = x;
this.y = y;
}
public T getX() {return x;}
public T getY() {return y;}
@Override
public int compareTo(Point<T> o) {
int cmp = x.compareTo(o.getX());
if (cmp==0) return y.compareTo(o.getY());
return cmp;
}
}
private static class ClassComparator<T extends Comparable<T>> implements Comparator<T> {
public ClassComparator() {}
@Override
public int compare(T a, T b) {
return a.compareTo(b);
}
}
private static class ListComparator<T extends Comparable<T>> implements Comparator<List<T>> {
public ListComparator() {}
@Override
public int compare(List<T> o1, List<T> o2) {
for (int i = 0; i < Math.min(o1.size(), o2.size()); i++) {
int c = o1.get(i).compareTo(o2.get(i));
if (c != 0) {
return c;
}
}
return Integer.compare(o1.size(), o2.size());
}
}
private static boolean eof()
{
if(lenbuf == -1)return true;
int lptr = ptrbuf;
while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;
try {
is.mark(1000);
while(true){
int b = is.read();
if(b == -1){
is.reset();
return true;
}else if(!isSpaceChar(b)){
is.reset();
return false;
}
}
} catch (IOException e) {
return true;
}
}
private static byte[] inbuf = new byte[1024];
static int lenbuf = 0, ptrbuf = 0;
private static int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
// private static boolean isSpaceChar(int c) { return !(c >= 32 && c <= 126); }
private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private static double readDouble() { return Double.parseDouble(readString()); }
private static char readChar() { return (char)skip(); }
private static String readString()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private static char[] readChar(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private static char[][] readTable(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = readChar(m);
return map;
}
private static int[] readIntArray(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = readInt();
return a;
}
private static long[] readLongArray(int n) {
long[] a = new long[n];
for (int i=0;i<n;i++) a[i] = readLong();
return a;
}
private static int readInt()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static long readLong()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 4e3a84b3fefee03d15ff7794f14283ba | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | /*
Enjoy the code :>
Modular Arithematic
1. (a + b)%c = ((a%c)+(b%c))%c
2. (a - c)%c = ((a%c)-(b%c) + c)%c
3. (a * c)%c = ((a%c)*(b*c))%c
4. (a / b)%c = ((a%c)*(b-1%c))%c
Modular Exponetion --> time complexity O(logN)
*/
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
static int N = 1000000;
static int M = 1000000007;
public static void main(String[] args) throws Exception{
fileConnect();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
while(t-- > 0){
st = new StringTokenizer(br.readLine());
String s = st.nextToken();
boolean ans = s.endsWith("B");
boolean res = true;
int sum = 0;
for(int i=0; i<s.length(); i++){
if(s.charAt(i) == 'A') sum++;
else sum--;
if(sum < 0) {
res = false;
break;
}
}
if(res && ans) System.out.println("Yes");
else System.out.println("No");
}
}
static int stringcost(String s){
int cost = 0;
for(int i=0; i<s.length(); i++){
cost += (s.charAt(i) - '`');
}
return cost;
}
// read array
static int[] readArr(int N, BufferedReader br, StringTokenizer st) throws Exception
{
int[] arr = new int[N];
st = new StringTokenizer(br.readLine());
for(int i=0; i < N; i++)
arr[i] = Integer.parseInt(st.nextToken());
return arr;
}
// print array
static void printArr(int[] arr, int n){
for(int i=0; i<n; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
}
// gcd
static long gcd(long a, long b){
if(b == 0) return a;
return gcd(b, a%b);
}
// lcm
public static long lcm(long a, long b){
return (a * b)/gcd(a, b);
}
// binary exponentiation
static long binaryExp(long x, long n){
// calcute x ^ n;
if(n == 0){
return 1;
}
else if(n%2 == 0){
return binaryExp(x*x, n/2);
}else {
return x * binaryExp(x*x, (n-1)/2);
}
}
// modular exponentiation
static long modExp(long x, long n, long m){
// calcute (x ^ n)%m;
if(n == 0){
return 1;
}
else if(n%2 == 0){
long temp = (x * x)%m;
return modExp(temp, n/2, m)%m;
}else {
long temp = (x * x)%m;
return (x * modExp(temp, (n-1)/2, m))%m;
}
}
// number is prime of not
static boolean isPrime(int n){
if(n == 0 || n == 1) return false;
if(n == 2 || n == 3) return true;
for(int i=2; i*i<=n; i++){
if(n%i == 0){
return false;
}
}
return true;
}
/* seive of erotheneses
- mark all prime between 1 to maxN
*/
static void seiveOfEroth(){
int maxN = 1000000;
boolean[] prime = new boolean[maxN + 1];
// false for prime and true for non-prime
prime[0] = prime[1] = true;
for(int i=2; i<=maxN; i++){
if(!prime[i]){
for(int j=i*i; j<=maxN; j+=i){
prime[j] = true;
}
}
}
}
/* prime factorisation
optimise solution
time complexity = O(sqrt(N))
*/
static void primeFactor(int n){
if(n == 1) System.out.print(n);
for(int i=2; i*i<=n; i++){
if(n == 1) return;
int cnt = 0;
if(n%i == 0){
while(n%i == 0){
cnt++;
n = n/i;
}
if(n != 1)
System.out.print(i + " ^ " + cnt + " + ");
else {
System.out.print(i + " ^ " + cnt);
}
}
}
if(n > 1){
System.out.print(n + " ^ 1");
}
}
/*
prime factorisation using seive
-- time complexity - O(logN)
-- use when multiple testcases
*/
static void primeFactor2(int n){
if(n == 1) System.out.print(1);
int[] seive = new int[N+1];
seive[0] = 0;
seive[1] = 1;
// this loop mark all the position with first prime number divisor.
for(int i=2; i<=N; i++){
if(seive[i] == 0){
for(int j=i; j<=N; j+=i){
if(seive[j] == 0){
seive[j] = i;
}
}
}
} // end of for loop
while(n != 1){
if(n/seive[n] != 1)
System.out.print(seive[n] + "*");
else
System.out.print(seive[n]);
n = n/seive[n];
}// end of while
}
/*
Calculate the power of matrix
*/
static long[][] powerMatix(long[][] matrix, int dim, int n){
// create identity matrix
long[][] identity = new long[dim][dim];
for(int i=0; i<dim; i++){
identity[i][i] = 1;
}
// printMatrix(identity, dim);
if(n==0) return identity;
if(n==1) return matrix;
// time complexity O(n)
// overtime O(n*( n^3))
// for(int i=1; i<n; i++){
// matrix = matmat(matrix, matrix, dim);
// }
// now use binary exponenition concept
// new time complexity O(logN*(N^3))
long[][] temp = identity;
int cnt = 0;
while(n > 1){
if(n%2 == 1){
temp = multimat(matrix, temp, dim);
n--;
}else {
matrix = multimat(matrix, matrix, dim);
n = n/2;
}
}
matrix = multimat(matrix, temp, dim);
return matrix;
}
// print the matrix
static void printMatrix(long[][] matrix, int dim){
for(int i=0; i<dim; i++){
for(int j=0; j<dim; j++){
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
}
// matrix multiplication
// time complexity O(n^3)
// this is for square matrix.
static long[][] multimat(long[][] matrix, long[][] matrix2, int dim){
long[][] result = new long[dim][dim];
for(int i=0; i<dim; i++){
for(int j=0; j<dim; j++){
for(int k=0; k<dim; k++){
result[i][j] = (result[i][j] + ((matrix[i][k] % M)* (matrix2[k][j]%M))%M)%M;
}
}
}
// for(int i=0; i<dim; i++){
// for(int j=0; j<dim; j++){
// matrix[i][j] = result[i][j];
// }
// }
// printMatrix(result, dim);
return result;
}
// FileInputOutput
static void fileConnect(){
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
} catch (Exception e) {
System.err.println("Error");
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 0a12e5051099b09b0733fbe8bc82772d | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes |
import java.util.Scanner;
public class Main {
public static boolean function(String leitura){
boolean result = false;
int size = leitura.length();
int geral = 0;
int contador_A = 0;
int contador_B = 0;
for (int i = 0; i < size; i++) {
char letra = leitura.charAt(i);
if( letra == 'A'){
contador_A++;
}
else if(letra == 'B' ){
contador_B++;
}
if (contador_B > contador_A){
return false;
}
}
if (leitura.charAt(0) != 'B' && leitura.charAt(size-1) !='A'){
result = true;
}
return result;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for (int i = 0; i < n; i++) {
String leitura = sc.next();
boolean check = function(leitura);
if(check){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 67fae901eea648475f00c3fd355cc4ad | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
static int M = 1_000_000_007;
static Random rng = new Random();
private static boolean testCase(String s) {
int numA = 0;
if (s.charAt(s.length() - 1) == 'A') return false;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == 'A') {
numA++;
} else {
if (numA == 0) {
return false;
} else {
numA--;
}
}
}
return true;
}
public static void main(String[] args) {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc.
//in.nextLine();
for (int tt = 1; tt <= t; ++tt) {
String s = in.next();
out.println(testCase(s) ? "YES" : "NO");
}
out.close();
}
private static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
boolean hasNext() {
return st.hasMoreTokens();
}
char[] readCharArray(int n) {
char[] arr = new char[n];
try {
br.read(arr);
br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return arr;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
private static void sort(int[] arr) {
int temp, idx;
for (int i = arr.length - 1; i > 0; i--) {
idx = rng.nextInt(i + 1);
temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
}
Arrays.sort(arr);
}
private static void sort(long[] arr) {
long temp;
int idx;
for (int i = arr.length - 1; i > 0; i--) {
idx = rng.nextInt(i + 1);
temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
}
Arrays.sort(arr);
}
private static <T> void sort(T[] arr) {
T temp;
int idx;
for (int i = arr.length - 1; i > 0; i--) {
idx = rng.nextInt(i + 1);
temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
}
Arrays.sort(arr);
}
private static <T> void sort(T[] arr, Comparator<? super T> cmp) {
T temp;
int idx;
for (int i = arr.length - 1; i > 0; i--) {
idx = rng.nextInt(i + 1);
temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
}
Arrays.sort(arr, cmp);
}
private static <T extends Comparable<? super T>> void sort(List<T> list) {
T temp;
int idx;
for (int i = list.size() - 1; i > 0; i--) {
idx = rng.nextInt(i + 1);
temp = list.get(i);
list.set(i, list.get(idx));
list.set(idx, temp);
}
Collections.sort(list);
}
private static <T> void sort(List<T> list, Comparator<? super T> cmp) {
T temp;
int idx;
for (int i = list.size() - 1; i > 0; i--) {
idx = rng.nextInt(i + 1);
temp = list.get(i);
list.set(i, list.get(idx));
list.set(idx, temp);
}
Collections.sort(list, cmp);
}
static class DSU {
int[] parent, rank;
public DSU(int n) {
parent = new int[n];
rank = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
}
}
public int find(int i) {
if (parent[i] == i) {
return i;
} else {
int res = find(parent[i]);
parent[i] = res;
return res;
}
}
public boolean isSameSet(int i, int j) {
return find(i) == find(j);
}
public void union(int i, int j) {
int iParent = find(i), jParent = find(j);
if (iParent != jParent) {
if (rank[iParent] > rank[jParent]) {
parent[jParent] = iParent;
} else {
parent[iParent] = jParent;
if (rank[iParent] == rank[jParent]) {
rank[jParent]++;
}
}
}
}
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 68616a208eed49521709556e58dfb969 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
public class HelloWorld{
public static void main(String []args){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
String str=sc.next();
if(str.length()<2){
System.out.println("No");
continue;
}
if(str.charAt(str.length()-1)!='B'){
System.out.println("No");
continue;
}
if(str.charAt(0)!='A'){
System.out.println("No");
continue;
}
int A=0;
boolean flag=true;
for(int i=0;i<str.length();i++){
char ch=str.charAt(i);
if(ch=='A'){
A++;
}else{
A--;
}
if(A<0){
flag=false;
break;
}
}
if(flag){
System.out.println("Yes");
}else{
System.out.println("No");
}
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | b6f07579d8e452f79db164dbe5dc961b | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
//import java.util.*;
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-->0) {
String s = br.readLine();
System.out.println(findWays(s));
}
}
private static String findWays(String s) {
if(s.charAt(s.length()-1)=='A') return "NO";
if(s.charAt(0)=='B') return "NO";
int aCount = 0; int bCount = 0;
for(int i=0;i<s.length();i++){
if(s.charAt(i)=='A') aCount++;
else bCount++;
if(bCount> aCount) return "NO";
}
return "YES";
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 0e084fe365012c2d746a2e9da227fd21 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.Arrays;
public class Main {
public static void main(String[] args) throws IOException {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
//solve (in, out);
int tests = in.nextInt();
for (int i = 0;i < tests;i++) {
solve (in, out);
}
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
// main code
static void solve (FastScanner in, PrintWriter out) {
String s = in.next();
int count = 0;
boolean ans = s.length() < 2 || s.charAt(s.length()-1) == 'A' ? false : true;
for (int i = 0;i < s.length();++i) {
count += s.charAt(i) == 'A' ? 1 : -1;
if (count < 0) {
ans = false;
break;
}
}
out.println(ans ? "YES" : "NO");
out.flush();
}
}
| Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 2953b69e21287d2b1066f4cd84d6f790 | train_108.jsonl | 1650722700 | Let's call a string good if its length is at least $$$2$$$ and all of its characters are $$$\texttt{A}$$$ except for the last character which is $$$\texttt{B}$$$. The good strings are $$$\texttt{AB},\texttt{AAB},\texttt{AAAB},\ldots$$$. Note that $$$\texttt{B}$$$ is not a good string.You are given an initially empty string $$$s_1$$$.You can perform the following operation any number of times: Choose any position of $$$s_1$$$ and insert some good string in that position. Given a string $$$s_2$$$, can we turn $$$s_1$$$ into $$$s_2$$$ after some number of operations? | 256 megabytes | import java.util.*;
public class Code1672B {
//I love AAAB
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T=sc.nextInt();
for(int t=0; t<T; t++) {
String str = sc.next();
int countA = 0, countB = 0;
boolean res = true;
if(str.length() < 2 || str.charAt(0) == 'B' || str.charAt(str.length()-1) == 'A')
{
res = false;
}
else {
for(int i=0; i<str.length(); i++) {
if(str.charAt(i) == 'A') countA++;
else countB++;
if(countA < countB) {
res = false;
break;
}
}
}
if(res) System.out.println("YES");
else System.out.println("NO");
}
}
} | Java | ["4\n\nAABAB\n\nABB\n\nAAAAAAAAB\n\nA"] | 1 second | ["YES\nNO\nYES\nNO"] | NoteIn the first test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAB}} \to \texttt{A}\color{red}{\texttt{AB}}\texttt{AB}$$$.In the third test case, we transform $$$s_1$$$ as such: $$$\varnothing \to \color{red}{\texttt{AAAAAAAAB}}$$$.In the second and fourth test case, it can be shown that it is impossible to turn $$$s_1$$$ into $$$s_2$$$. | Java 11 | standard input | [
"constructive algorithms",
"implementation"
] | 0b9be2f076cfa13cdc76c489bf1ea416 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single string $$$s_2$$$ ($$$1 \leq |s_2| \leq 2 \cdot 10^5$$$). It is guaranteed that $$$s_2$$$ consists of only the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$. It is guaranteed that the sum of $$$|s_2|$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print "YES" (without quotes) if we can turn $$$s_1$$$ into $$$s_2$$$ after some number of operations, and "NO" (without quotes) otherwise. You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). | standard output | |
PASSED | 1324c875137cf5e5eef375df0948d1ef | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static ContestScanner sc = new ContestScanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
static StringBuilder sb = new StringBuilder();
static long mod = (long) 1e9 + 7;
public static void main(String[] args) throws Exception {
//int T = sc.nextInt();
//for(int i = 0; i < T; i++)solve();
solve();
pw.flush();
}
static int ask(int i){
if (i==0) return 0;
System.out.println("? " + i);
int temp = sc.nextInt();
if (temp==-1) System.exit(0);
return temp;
}
public static void solve() {
int n = sc.nextInt();
int left = -2;
int right = (int)5e6;
while (right - left > 1){
int mid = (right + left)/2;
if(ask(mid) == 1){
right = mid;
}else{
left = mid;
}
}
int ans = (int)1e9;
for(int i = 1; i <= n; i++){
int temp = ask(right / i);
if(temp >= 1){
ans = Math.min(ans,temp*(right/i));
}
}
System.out.println("! " + ans);
}
static class GeekInteger {
public static void save_sort(int[] array) {
shuffle(array);
Arrays.sort(array);
}
public static int[] shuffle(int[] array) {
int n = array.length;
Random random = new Random();
for (int i = 0, j; i < n; i++) {
j = i + random.nextInt(n - i);
int randomElement = array[j];
array[j] = array[i];
array[i] = randomElement;
}
return array;
}
public static void save_sort(long[] array) {
shuffle(array);
Arrays.sort(array);
}
public static long[] shuffle(long[] array) {
int n = array.length;
Random random = new Random();
for (int i = 0, j; i < n; i++) {
j = i + random.nextInt(n - i);
long randomElement = array[j];
array[j] = array[i];
array[i] = randomElement;
}
return array;
}
}
}
/**
* refercence : https://github.com/NASU41/AtCoderLibraryForJava/blob/master/ContestIO/ContestScanner.java
*/
class ContestScanner {
private final java.io.InputStream in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private static final long LONG_MAX_TENTHS = 922337203685477580L;
private static final int LONG_MAX_LAST_DIGIT = 7;
private static final int LONG_MIN_LAST_DIGIT = 8;
public ContestScanner(java.io.InputStream in){
this.in = in;
}
public ContestScanner(java.io.File file) throws java.io.FileNotFoundException {
this(new java.io.BufferedInputStream(new java.io.FileInputStream(file)));
}
public ContestScanner(){
this(System.in);
}
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
}else{
ptr = 0;
try {
buflen = in.read(buffer);
} catch (java.io.IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte()) return buffer[ptr++]; else return -1;
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
public boolean hasNext() {
while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
return hasNextByte();
}
public String next() {
if (!hasNext()) throw new java.util.NoSuchElementException();
StringBuilder sb = new StringBuilder();
int b = readByte();
while(isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext()) throw new java.util.NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
int digit = b - '0';
if (n >= LONG_MAX_TENTHS) {
if (n == LONG_MAX_TENTHS) {
if (minus) {
if (digit <= LONG_MIN_LAST_DIGIT) {
n = -n * 10 - digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("%d%s... is not number", n, Character.toString(b))
);
}
}
} else {
if (digit <= LONG_MAX_LAST_DIGIT) {
n = n * 10 + digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("%d%s... is not number", n, Character.toString(b))
);
}
}
}
}
throw new ArithmeticException(
String.format("%s%d%d... overflows long.", minus ? "-" : "", n, digit)
);
}
n = n * 10 + digit;
}else if(b == -1 || !isPrintableChar(b)){
return minus ? -n : n;
}else{
throw new NumberFormatException();
}
b = readByte();
}
}
public int nextInt() {
long nl = nextLong();
if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
return (int) nl;
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long[] nextLongArray(int length){
long[] array = new long[length];
for(int i=0; i<length; i++) array[i] = this.nextLong();
return array;
}
public long[] nextLongArray(int length, java.util.function.LongUnaryOperator map){
long[] array = new long[length];
for(int i=0; i<length; i++) array[i] = map.applyAsLong(this.nextLong());
return array;
}
public int[] nextIntArray(int length){
int[] array = new int[length];
for(int i=0; i<length; i++) array[i] = this.nextInt();
return array;
}
public int[] nextIntArray(int length, java.util.function.IntUnaryOperator map){
int[] array = new int[length];
for(int i=0; i<length; i++) array[i] = map.applyAsInt(this.nextInt());
return array;
}
public double[] nextDoubleArray(int length){
double[] array = new double[length];
for(int i=0; i<length; i++) array[i] = this.nextDouble();
return array;
}
public double[] nextDoubleArray(int length, java.util.function.DoubleUnaryOperator map){
double[] array = new double[length];
for(int i=0; i<length; i++) array[i] = map.applyAsDouble(this.nextDouble());
return array;
}
public long[][] nextLongMatrix(int height, int width){
long[][] mat = new long[height][width];
for(int h=0; h<height; h++) for(int w=0; w<width; w++){
mat[h][w] = this.nextLong();
}
return mat;
}
public int[][] nextIntMatrix(int height, int width){
int[][] mat = new int[height][width];
for(int h=0; h<height; h++) for(int w=0; w<width; w++){
mat[h][w] = this.nextInt();
}
return mat;
}
public double[][] nextDoubleMatrix(int height, int width){
double[][] mat = new double[height][width];
for(int h=0; h<height; h++) for(int w=0; w<width; w++){
mat[h][w] = this.nextDouble();
}
return mat;
}
public char[][] nextCharMatrix(int height, int width){
char[][] mat = new char[height][width];
for(int h=0; h<height; h++){
String s = this.next();
for(int w=0; w<width; w++){
mat[h][w] = s.charAt(w);
}
}
return mat;
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 4ec8898e0adc2f3e4faa2168c47be542 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.BufferedReader;
import java.io.Closeable;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.time.Clock;
import java.time.LocalDateTime;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.Map;
import java.util.Objects;
import java.util.Random;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.function.Supplier;
public class Solution {
private void solve() throws IOException {
int n = nextInt();
int maxWidth = (n + 1) * 2000 - 1;
int res;
{
Pii t = findWidth(1, maxWidth);
res = t.x * t.y;
maxWidth = t.x;
}
for (int i = 2; i <= n; i++) {
int w = maxWidth / i;
if (w > 0) {
int h = ask(w);
if (h <= n) {
res = Math.min(res, w*h);
}
}
}
result(res);
}
private Pii findWidth(int height, int maxWidth) throws IOException {
int left = 0;
int right = maxWidth;
while (left < right) {
int mid = (left + right) / 2;
if (ask(mid) > height) {
left = mid + 1;
} else {
right = mid;
}
}
return new Pii(left, ask(left));
}
private static class Pii implements Comparable<Pii> {
// not today, evil hacker
private static final int hashMultiplier = BigInteger.valueOf(new Random().nextInt(100000) + 100).nextProbablePrime().intValue();
private int x;
private int y;
public Pii(int x, int y) {
this.x = x;
this.y = y;
}
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pii pii = (Pii) o;
if (x != pii.x) return false;
return y == pii.y;
}
public int hashCode() {
return hashMultiplier * x + y;
}
public String toString() {
return "(" + x + ", " + y + ")";
}
public int compareTo(Pii o) {
int c = Integer.compare(x, o.x);
return c != 0 ? c : Integer.compare(y, o.y);
}
}
private int ask(int w) throws IOException {
if (w < 1) return Integer.MAX_VALUE-13;
out.println("? " + w);
out.flush();
int res = nextInt();
return res > 0 ? res: Integer.MAX_VALUE-13;
}
private void result(int sq) {
out.println("! " + sq);
out.flush();
}
private static final boolean runNTestsInProd = false;
private static final boolean printCaseNumber = false;
private static final boolean assertInProd = false;
private static final boolean logToFile = false;
private static final boolean readFromConsoleInDebug = true;
private static final boolean writeToConsoleInDebug = true;
private static final boolean testTimer = false;
private static Boolean isDebug = null;
private BufferedReader in;
private StringTokenizer line;
private PrintWriter out;
public static void main(String[] args) throws Exception {
isDebug = Arrays.asList(args).contains("DEBUG_MODE");
if (isDebug) {
log = logToFile ? new PrintWriter("logs/j_solution_" + System.currentTimeMillis() + ".log") : new PrintWriter(System.out);
clock = Clock.systemDefaultZone();
}
new Solution().run();
}
private void run() throws Exception {
in = new BufferedReader(new InputStreamReader(!isDebug || readFromConsoleInDebug ? System.in : new FileInputStream("input.txt")));
out = !isDebug || writeToConsoleInDebug ? new PrintWriter(System.out) : new PrintWriter("output.txt");
try (Timer totalTimer = new Timer("total")) {
int t = 1;//runNTestsInProd || isDebug ? nextInt() : 1;
for (int i = 0; i < t; i++) {
if (printCaseNumber) {
out.print("Case #" + (i + 1) + ": ");
}
if (testTimer) {
try (Timer testTimer = new Timer("test #" + (i + 1))) {
solve();
}
} else {
solve();
}
if (isDebug) {
out.flush();
}
}
}
in.close();
out.flush();
out.close();
}
private void println(Object... objects) {
boolean isFirst = true;
for (Object o : objects) {
if (!isFirst) {
out.print(" ");
} else {
isFirst = false;
}
out.print(o.toString());
}
out.println();
}
private int[] nextIntArray(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
private long[] nextLongArray(int n) throws IOException {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextLong();
}
return res;
}
private int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
private long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
private double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private char[] nextTokenChars() throws IOException {
return nextToken().toCharArray();
}
private String nextToken() throws IOException {
while (line == null || !line.hasMoreTokens()) {
line = new StringTokenizer(in.readLine());
}
return line.nextToken();
}
private static void assertPredicate(boolean p) {
if ((isDebug || assertInProd) && !p) {
throw new RuntimeException();
}
}
private static void assertPredicate(boolean p, String message) {
if ((isDebug || assertInProd) && !p) {
throw new RuntimeException(message);
}
}
private static <T> void assertNotEqual(T unexpected, T actual) {
if ((isDebug || assertInProd) && Objects.equals(actual, unexpected)) {
throw new RuntimeException("assertNotEqual: " + unexpected + " == " + actual);
}
}
private static <T> void assertEqual(T expected, T actual) {
if ((isDebug || assertInProd) && !Objects.equals(actual, expected)) {
throw new RuntimeException("assertEqual: " + expected + " != " + actual);
}
}
private static PrintWriter log = null;
private static Clock clock = null;
private static void log(Object... objects) {
log(true, objects);
}
private static void logNoDelimiter(Object... objects) {
log(false, objects);
}
private static void log(boolean printDelimiter, Object[] objects) {
if (isDebug) {
StringBuilder sb = new StringBuilder();
sb.append(LocalDateTime.now(clock)).append(" - ");
boolean isFirst = true;
for (Object o : objects) {
if (!isFirst && printDelimiter) {
sb.append(" ");
} else {
isFirst = false;
}
sb.append(o.toString());
}
log.println(sb);
log.flush();
}
}
private static class Timer implements Closeable {
private final String label;
private final long startTime = isDebug ? System.nanoTime() : 0;
public Timer(String label) {
this.label = label;
}
@Override
public void close() throws IOException {
if (isDebug) {
long executionTime = System.nanoTime() - startTime;
String fraction = Long.toString(executionTime / 1000 % 1_000_000);
logNoDelimiter("Timer[", label, "]: ", executionTime / 1_000_000_000, '.', "00000".substring(0, 6 - fraction.length()), fraction, 's');
}
}
}
private static <T> T timer(String label, Supplier<T> f) throws Exception {
if (isDebug) {
try (Timer timer = new Timer(label)) {
return f.get();
}
} else {
return f.get();
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | c068da0ac227ccdd66ffe523edb2133e | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.BufferedReader;
import java.io.Closeable;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.time.Clock;
import java.time.LocalDateTime;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.Map;
import java.util.Objects;
import java.util.Random;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.function.Supplier;
public class Solution {
private void solve() throws IOException {
int n = nextInt();
int maxWidth = (n + 1) * 2000 - 1;
int res;
{
Pii t = findWidth(1, maxWidth);
res = t.x * t.y;
maxWidth = t.x;
}
for (int i = 2; i <= n; i++) {
int w = (maxWidth - 1) / i;
if (w > 0) {
int h = ask(w);
if (h <= n) {
res = Math.min(res, w*h);
}
}
}
result(res);
}
private Pii findWidth(int height, int maxWidth) throws IOException {
int left = 0;
int right = maxWidth;
while (left < right) {
int mid = (left + right) / 2;
if (ask(mid) > height) {
left = mid + 1;
} else {
right = mid;
}
}
return new Pii(left, ask(left));
}
private static class Pii implements Comparable<Pii> {
// not today, evil hacker
private static final int hashMultiplier = BigInteger.valueOf(new Random().nextInt(100000) + 100).nextProbablePrime().intValue();
private int x;
private int y;
public Pii(int x, int y) {
this.x = x;
this.y = y;
}
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pii pii = (Pii) o;
if (x != pii.x) return false;
return y == pii.y;
}
public int hashCode() {
return hashMultiplier * x + y;
}
public String toString() {
return "(" + x + ", " + y + ")";
}
public int compareTo(Pii o) {
int c = Integer.compare(x, o.x);
return c != 0 ? c : Integer.compare(y, o.y);
}
}
private int ask(int w) throws IOException {
if (w < 1) return Integer.MAX_VALUE-13;
out.println("? " + w);
out.flush();
int res = nextInt();
return res > 0 ? res: Integer.MAX_VALUE-13;
}
private void result(int sq) {
out.println("! " + sq);
out.flush();
}
private static final boolean runNTestsInProd = false;
private static final boolean printCaseNumber = false;
private static final boolean assertInProd = false;
private static final boolean logToFile = false;
private static final boolean readFromConsoleInDebug = true;
private static final boolean writeToConsoleInDebug = true;
private static final boolean testTimer = false;
private static Boolean isDebug = null;
private BufferedReader in;
private StringTokenizer line;
private PrintWriter out;
public static void main(String[] args) throws Exception {
isDebug = Arrays.asList(args).contains("DEBUG_MODE");
if (isDebug) {
log = logToFile ? new PrintWriter("logs/j_solution_" + System.currentTimeMillis() + ".log") : new PrintWriter(System.out);
clock = Clock.systemDefaultZone();
}
new Solution().run();
}
private void run() throws Exception {
in = new BufferedReader(new InputStreamReader(!isDebug || readFromConsoleInDebug ? System.in : new FileInputStream("input.txt")));
out = !isDebug || writeToConsoleInDebug ? new PrintWriter(System.out) : new PrintWriter("output.txt");
try (Timer totalTimer = new Timer("total")) {
int t = 1;//runNTestsInProd || isDebug ? nextInt() : 1;
for (int i = 0; i < t; i++) {
if (printCaseNumber) {
out.print("Case #" + (i + 1) + ": ");
}
if (testTimer) {
try (Timer testTimer = new Timer("test #" + (i + 1))) {
solve();
}
} else {
solve();
}
if (isDebug) {
out.flush();
}
}
}
in.close();
out.flush();
out.close();
}
private void println(Object... objects) {
boolean isFirst = true;
for (Object o : objects) {
if (!isFirst) {
out.print(" ");
} else {
isFirst = false;
}
out.print(o.toString());
}
out.println();
}
private int[] nextIntArray(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
private long[] nextLongArray(int n) throws IOException {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextLong();
}
return res;
}
private int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
private long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
private double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private char[] nextTokenChars() throws IOException {
return nextToken().toCharArray();
}
private String nextToken() throws IOException {
while (line == null || !line.hasMoreTokens()) {
line = new StringTokenizer(in.readLine());
}
return line.nextToken();
}
private static void assertPredicate(boolean p) {
if ((isDebug || assertInProd) && !p) {
throw new RuntimeException();
}
}
private static void assertPredicate(boolean p, String message) {
if ((isDebug || assertInProd) && !p) {
throw new RuntimeException(message);
}
}
private static <T> void assertNotEqual(T unexpected, T actual) {
if ((isDebug || assertInProd) && Objects.equals(actual, unexpected)) {
throw new RuntimeException("assertNotEqual: " + unexpected + " == " + actual);
}
}
private static <T> void assertEqual(T expected, T actual) {
if ((isDebug || assertInProd) && !Objects.equals(actual, expected)) {
throw new RuntimeException("assertEqual: " + expected + " != " + actual);
}
}
private static PrintWriter log = null;
private static Clock clock = null;
private static void log(Object... objects) {
log(true, objects);
}
private static void logNoDelimiter(Object... objects) {
log(false, objects);
}
private static void log(boolean printDelimiter, Object[] objects) {
if (isDebug) {
StringBuilder sb = new StringBuilder();
sb.append(LocalDateTime.now(clock)).append(" - ");
boolean isFirst = true;
for (Object o : objects) {
if (!isFirst && printDelimiter) {
sb.append(" ");
} else {
isFirst = false;
}
sb.append(o.toString());
}
log.println(sb);
log.flush();
}
}
private static class Timer implements Closeable {
private final String label;
private final long startTime = isDebug ? System.nanoTime() : 0;
public Timer(String label) {
this.label = label;
}
@Override
public void close() throws IOException {
if (isDebug) {
long executionTime = System.nanoTime() - startTime;
String fraction = Long.toString(executionTime / 1000 % 1_000_000);
logNoDelimiter("Timer[", label, "]: ", executionTime / 1_000_000_000, '.', "00000".substring(0, 6 - fraction.length()), fraction, 's');
}
}
}
private static <T> T timer(String label, Supplier<T> f) throws Exception {
if (isDebug) {
try (Timer timer = new Timer(label)) {
return f.get();
}
} else {
return f.get();
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 286b73d989e88fea46646f3732767095 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | // https://codeforces.com/blog/entry/102172
import java.io.*;
import java.util.*;
public class CF1672E extends PrintWriter {
CF1672E() { super(System.out, true); }
Scanner sc = new Scanner(System.in);
public static void main(String[] $) {
CF1672E o = new CF1672E(); o.main(); o.flush();
}
int query(int w) {
println("? " + w);
return sc.nextInt();
}
void main() {
int n = sc.nextInt();
int lower = n * 1 + n - 1 - 1;
int upper = n * 2000 + n - 1;
while (upper - lower > 1) {
int w = (lower + upper) / 2;
int h = query(w);
if (h == 1)
upper = w;
else
lower = w;
}
long ans = upper;
for (int h = 2; h <= n; h++) {
int w = upper / h;
int h_ = query(w);
if (h_ > 0)
ans = Math.min(ans, (long) w * h_);
}
println("! " + ans);
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 218a054b5f8a7e95e4156bf051bce885 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | //Utilities
import java.io.*;
import java.util.*;
public class a {
static int t;
static int n;
static int s;
static int minArea;
public static void main(String[] args) throws IOException {
t = 1;
while (t-- > 0) {
n = in.iscan();
minArea = Integer.MAX_VALUE;
int l = 1, r = 2000 * n + n - 1, mid;
s = 0;
while (l <= r) {
mid = (l+r)/2;
int h = query(mid);
if (h == 0 || h > 1) {
l = mid+1;
}
else { // h == 1
s = mid;
r = mid-1;
}
}
minArea = s;
for (int h = 2; h <= n; h++) {
int w = s / h;
int newH = query(s / h);
if (newH != 0) {
minArea = Math.min(minArea, w * newH);
}
}
System.out.println("! " + minArea);
System.out.flush();
}
out.close();
}
static int query(int w) throws IOException {
System.out.println("? " + w);
System.out.flush();
return in.iscan();
}
static INPUT in = new INPUT(System.in);
static PrintWriter out = new PrintWriter(System.out);
private static class INPUT {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar, numChars;
public INPUT (InputStream stream) {
this.stream = stream;
}
public INPUT (String file) throws IOException {
this.stream = new FileInputStream (file);
}
public int cscan () throws IOException {
if (curChar >= numChars) {
curChar = 0;
numChars = stream.read (buf);
}
if (numChars == -1)
return numChars;
return buf[curChar++];
}
public int iscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
int res = 0;
do {
res = (res << 1) + (res << 3);
res += c - '0';
c = cscan ();
}
while (!space (c));
return res * sgn;
}
public String sscan () throws IOException {
int c = cscan ();
while (space (c))
c = cscan ();
StringBuilder res = new StringBuilder ();
do {
res.appendCodePoint (c);
c = cscan ();
}
while (!space (c));
return res.toString ();
}
public double dscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
double res = 0;
while (!space (c) && c != '.') {
if (c == 'e' || c == 'E')
return res * UTILITIES.fast_pow (10, iscan ());
res *= 10;
res += c - '0';
c = cscan ();
}
if (c == '.') {
c = cscan ();
double m = 1;
while (!space (c)) {
if (c == 'e' || c == 'E')
return res * UTILITIES.fast_pow (10, iscan ());
m /= 10;
res += (c - '0') * m;
c = cscan ();
}
}
return res * sgn;
}
public long lscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
long res = 0;
do {
res = (res << 1) + (res << 3);
res += c - '0';
c = cscan ();
}
while (!space (c));
return res * sgn;
}
public boolean space (int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
public static class UTILITIES {
static final double EPS = 10e-6;
public static void sort(int[] a, boolean increasing) {
ArrayList<Integer> arr = new ArrayList<Integer>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static void sort(long[] a, boolean increasing) {
ArrayList<Long> arr = new ArrayList<Long>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static void sort(double[] a, boolean increasing) {
ArrayList<Double> arr = new ArrayList<Double>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static int lower_bound (int[] arr, int x) {
int low = 0, high = arr.length, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x)
high = mid;
else
low = mid + 1;
}
return low;
}
public static int upper_bound (int[] arr, int x) {
int low = 0, high = arr.length, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] > x)
high = mid;
else
low = mid + 1;
}
return low;
}
public static void updateMap(HashMap<Integer, Integer> map, int key, int v) {
if (!map.containsKey(key)) {
map.put(key, v);
}
else {
map.put(key, map.get(key) + v);
}
if (map.get(key) == 0) {
map.remove(key);
}
}
public static long gcd (long a, long b) {
return b == 0 ? a : gcd (b, a % b);
}
public static long lcm (long a, long b) {
return a * b / gcd (a, b);
}
public static long fast_pow_mod (long b, long x, int mod) {
if (x == 0) return 1;
if (x == 1) return b;
if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
}
public static long fast_pow (long b, long x) {
if (x == 0) return 1;
if (x == 1) return b;
if (x % 2 == 0) return fast_pow (b * b, x / 2);
return b * fast_pow (b * b, x / 2);
}
public static long choose (long n, long k) {
k = Math.min (k, n - k);
long val = 1;
for (int i = 0; i < k; ++i)
val = val * (n - i) / (i + 1);
return val;
}
public static long permute (int n, int k) {
if (n < k) return 0;
long val = 1;
for (int i = 0; i < k; ++i)
val = (val * (n - i));
return val;
}
// start of permutation and lower/upper bound template
public static void nextPermutation(int[] nums) {
//find first decreasing digit
int mark = -1;
for (int i = nums.length - 1; i > 0; i--) {
if (nums[i] > nums[i - 1]) {
mark = i - 1;
break;
}
}
if (mark == -1) {
reverse(nums, 0, nums.length - 1);
return;
}
int idx = nums.length-1;
for (int i = nums.length-1; i >= mark+1; i--) {
if (nums[i] > nums[mark]) {
idx = i;
break;
}
}
swap(nums, mark, idx);
reverse(nums, mark + 1, nums.length - 1);
}
public static void swap(int[] nums, int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
public static void reverse(int[] nums, int i, int j) {
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
static int lower_bound (int[] arr, int hi, int cmp) {
int low = 0, high = hi, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= cmp) high = mid;
else low = mid + 1;
}
return low;
}
static int upper_bound (int[] arr, int hi, int cmp) {
int low = 0, high = hi, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] > cmp) high = mid;
else low = mid + 1;
}
return low;
}
// end of permutation and lower/upper bound template
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | b7a143042043e2f9f2505274de713464 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
Grader grader = new Grader() {
@Override
public int test(int w) {
System.out.println("? " + w);
return scanner.nextInt();
}
@Override
public int getN() {
return n;
}
@Override
public void checkAns(int s) {
System.out.println("! " + s);
}
};
Solver solver = new Solver(grader);
solver.solve();
}
public interface Grader {
int test(int w);
int getN();
void checkAns(int s);
}
public static class Solver {
private static final int MAX_LEN = 2000;
private final Grader grader;
public Solver(Grader grader) {
this.grader = grader;
}
public void solve() {
int n = grader.getN();
int lo = 2 * n - 1, hi = n * (MAX_LEN + 1) - 1;
// possible length of the longest string
int currentMaxLengthMin = 1, currentMaxLengthMax = MAX_LEN;
Map<Integer, Integer> answers = new TreeMap<>();
int minArea = hi;
while (lo < hi) {
int w = (lo + hi) / 2;
int h = grader.test(w);
if (h != 1) lo = w + 1;
else hi = w;
if (h != 0) {
answers.put(w, h);
currentMaxLengthMax = Math.min(currentMaxLengthMax, w);
minArea = Math.min(minArea, w * h);
} else currentMaxLengthMin = Math.max(currentMaxLengthMin, w + 1);
}
int[] heights = new int[n];
for (Map.Entry<Integer, Integer> entry : answers.entrySet()) {
int w = entry.getKey();
int h = entry.getValue();
if (heights[h - 1] == 0) heights[h - 1] = w;
else heights[h - 1] = Math.min(heights[h - 1], w);
}
int s = lo;
heights[0] = s;
for (int h = 2; h <= n; h++) {
if (heights[h - 1] == 0) heights[h - 1] = heights[h - 2];
else heights[h - 1] = Math.min(heights[h - 1], heights[h - 2]);
int length = s / h;
length = Math.min(length, heights[h - 2] - 1);
length = Math.max(length, currentMaxLengthMin);
if (heights[h - 1] != 0 && heights[h - 1] <= length) continue;
if (length * h >= minArea) continue;
int actualH;
if (answers.containsKey(length)) actualH = answers.get(length);
else actualH = grader.test(length);
if (actualH != 0) {
minArea = Math.min(minArea, length * actualH);
answers.put(length, actualH);
currentMaxLengthMax = Math.min(currentMaxLengthMax, length);
if (actualH == h) {
heights[h - 1] = length;
} else {
if (heights[actualH - 1] == 0 || heights[actualH - 1] > length) heights[actualH - 1] = length;
}
} else currentMaxLengthMin = Math.max(currentMaxLengthMin, length + 1);
}
grader.checkAns(minArea);
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | a47f3ae09f641e028e91f3bbd57a5515 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
static PrintWriter pw;
static Scanner sc;
static int ask(int mid) throws IOException {
pw.println("? " + mid);
pw.flush();
return sc.nextInt();
}
public static void main(String[] args) throws IOException {
sc = new Scanner(System.in);
pw = new PrintWriter(System.out);
int n = sc.nextInt();
int start = 1, end = (int) 5e6;
while (start <= end) {
int mid = start + end >> 1;
int v = ask(Math.max(1, mid));
if (v == 1) {
end = mid - 1;
} else {
start = mid + 1;
}
}
int ans = start;
for (int i = 1; i < n; i++) {
int cur = (start ) / (i + 1);
int v = ask(Math.max(1, cur));
if (v != 0) {
ans = Math.min(ans, v * cur);
}
}
pw.println("! " + ans);
pw.flush();
pw.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String readAllLines(BufferedReader reader) throws IOException {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line);
content.append(System.lineSeparator());
}
return content.toString();
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | fff6684b82b9322ed3603d35fbca3187 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.FileNotFoundException;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastScanner in = new FastScanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
EBloknotexe solver = new EBloknotexe();
solver.solve(1, in, out);
out.close();
}
static class EBloknotexe {
FastScanner in;
PrintWriter out;
int n;
int qCount = 0;
public void solve(int testNumber, FastScanner in, PrintWriter out) {
this.in = in;
this.out = out;
n = in.nextInt();
int left = (1 + 1) * n - 2, right = (2000 + 1) * n - 1;
while (left < right - 1) {
int mid = (left + right) >>> 1;
if (ask(mid) == 1) {
right = mid;
} else {
left = mid;
}
}
int sumL = right - (n - 1);
long ans = right;
for (int h = 1; h <= n; h++) {
int w = (sumL + n - h + h - 1) / h;
ans = Math.min(ans, (long) w * ask(w));
}
out.println("! " + ans);
out.flush();
}
int ask(int w) {
qCount++;
out.println("? " + w);
out.flush();
int h = in.nextInt();
if (h == 0) {
h = Integer.MAX_VALUE;
}
return h;
}
}
static class FastScanner {
public BufferedReader br;
public StringTokenizer st;
public FastScanner(InputStream in) {
br = new BufferedReader(new InputStreamReader(in));
}
public FastScanner(String fileName) {
try {
br = new BufferedReader(new FileReader(fileName));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public int nextInt() {
return Integer.parseInt(next());
}
public String next() {
while (st == null || !st.hasMoreElements()) {
String line = null;
try {
line = br.readLine();
} catch (IOException e) {
throw new UnknownError();
}
if (line == null) {
throw new UnknownError();
}
st = new StringTokenizer(line);
}
return st.nextToken();
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 8e9d70f43849dc8ac39cc744e3649527 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.*;
import java.io.*;
import java.text.*;
public class CF_1672_E{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int N = ni();
int lo = 1, hi = 4000000+N;
while (lo < hi){
int mid = lo + (hi-lo)/2;
int h = query(mid);
if(h == 1)hi = mid;
else lo = mid+1;
}
int ans = hi;
for(int h = 2; h <= N; h++){
int w = hi/h, H = query(w);
ans = Math.min(ans, w*H);
}
pni("!", ans);
}
final int INF = 1000000;
int query(int w) throws Exception{
pni("?", w);
int h = ni();
if(h == 0)return INF;
return h;
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
void exit(boolean b){if(!b)System.exit(0);}
static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
DecimalFormat df = new DecimalFormat("0.00000000000");
double PI = 3.141592653589793238462643383279502884197169399;
static boolean multipleTC = false, memory = true, fileIO = false;
FastReader in;PrintWriter out;
void run() throws Exception{
long ct = System.currentTimeMillis();
if (fileIO) {
in = new FastReader("");
out = new PrintWriter("");
} else {
in = new FastReader();
out = new PrintWriter(System.out);
}
//Solution Credits: Taranpreet Singh
int T = multipleTC? ni():1;
pre();
for (int t = 1; t <= T; t++) solve(t);
out.flush();
out.close();
System.err.println("Runtime: "+(System.currentTimeMillis() - ct));
}
public static void main(String[] args) throws Exception{
if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1672_E().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
else new CF_1672_E().run();
}
int[][] make(int n, int e, int[] from, int[] to, boolean f){
int[][] g = new int[n][];int[]cnt = new int[n];
for(int i = 0; i< e; i++){
cnt[from[i]]++;
if(f)cnt[to[i]]++;
}
for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
for(int i = 0; i< e; i++){
g[from[i]][--cnt[from[i]]] = to[i];
if(f)g[to[i]][--cnt[to[i]]] = from[i];
}
return g;
}
int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
int[][][] g = new int[n][][];int[]cnt = new int[n];
for(int i = 0; i< e; i++){
cnt[from[i]]++;
if(f)cnt[to[i]]++;
}
for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
for(int i = 0; i< e; i++){
g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
}
return g;
}
int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object... o){for(Object oo:o)out.print(oo+" ");}
void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str;
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 26ddf639a33fd096424bbecfa0b3f4a2 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.*;
import java.io.*;
// you can compare with output.txt and expected out
public class RoundGlobal20E {
MyPrintWriter out;
MyScanner in;
// final static long FIXED_RANDOM;
// static {
// FIXED_RANDOM = System.currentTimeMillis();
// }
final static String IMPOSSIBLE = "IMPOSSIBLE";
final static String POSSIBLE = "POSSIBLE";
final static String YES = "YES";
final static String NO = "NO";
private void initIO(boolean isFileIO) {
if (System.getProperty("ONLINE_JUDGE") == null && isFileIO) {
try{
in = new MyScanner(new FileInputStream("input.txt"));
out = new MyPrintWriter(new FileOutputStream("output.txt"));
}
catch(FileNotFoundException e){
e.printStackTrace();
}
}
else{
in = new MyScanner(System.in);
out = new MyPrintWriter(new BufferedOutputStream(System.out));
}
}
public static void main(String[] args){
// Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
RoundGlobal20E sol = new RoundGlobal20E();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = false;
boolean hasMultipleTests = false;
initIO(isFileIO);
solve(in, out);
in.close();
out.close();
}
public void solve(MyScanner in, MyPrintWriter out) {
n = in.nextInt();
// takes lg(1998n) queries
// 22 queries at n = 2000
int l = findSumLengths();
// line i: [ai... a(i+1))
// sum of length <= w
// lg(2000n) + lg(1000n) + lg(500n) + ....
// n lgn +
// n = 1
// easy
// n = 2
// need to find max(ai) or min(ai), or search for h=2
// lg(4000) + lg(2000) < 30
// n = 3
// search for h = 1, 2, 3
// n = 5 -- have 35 queries
// h = 1 -> log(10000)
// h = 2 -> log(6000)
// h = 3 -> log(4000)
// h = 4 -> log(4000)
// h = 5 -> log(2000)
// unless ai's have very even structure, we are unlikely to save a lot from high h
// at h=k, the saving is at most k-1
// h = 1 -> w = l+n-1
// for h=2 to be better than that,
// w must be (l+n-2)/2
// maintain the best area
// h = k, ask for w = (area-1)/h
// w*h <= area-1 -> w <= (area-1)/h
int area = l+n-1;
for(int h=2; h<=n; h++) {
int w = (area-1)/h;
if(w == 0)
continue;
int ans = askQuery(w);
if(ans != 0)
area = Math.min(area, w*ans);
}
giveAnswer(area);
}
int n;
private int findSumLengths() {
int left = n+n-1;
int right = 2000*n;
while(left < right) {
int mid = (left+right)/2;
int h = askQuery(mid);
if(h == 1) { // mid is incl.
right = mid;
}
else { // mid is not incl.
left = mid+1;
}
}
return left-(n-1);
}
private void giveAnswer(int area) {
out.println("! " + area);
out.flush();
}
private int askQuery(int width) {
out.println("? " + width);
out.flush();
return in.nextInt();
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
// 32768?
public MyScanner(InputStream is, int bufferSize) {
br = new BufferedReader(new InputStreamReader(is), bufferSize);
}
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
// br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
}
public void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[][] nextTreeEdges(int n, int offset){
int[][] e = new int[n-1][2];
for(int i=0; i<n-1; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[][] nextMatrix(int n, int m) {
return nextMatrix(n, m, 0);
}
int[][] nextMatrix(int n, int m, int offset) {
int[][] mat = new int[n][m];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[i][j] = nextInt()+offset;
}
}
return mat;
}
int[][] nextPairs(int n){
return nextPairs(n, 0);
}
int[][] nextPairs(int n, int offset) {
int[][] xy = new int[2][n];
for(int i=0; i<n; i++) {
xy[0][i] = nextInt() + offset;
xy[1][i] = nextInt() + offset;
}
return xy;
}
int[][] nextGraphEdges(){
return nextGraphEdges(0);
}
int[][] nextGraphEdges(int offset) {
int m = nextInt();
int[][] e = new int[m][2];
for(int i=0; i<m; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[] nextIntArray(int len) {
return nextIntArray(len, 0);
}
int[] nextIntArray(int len, int offset){
int[] a = new int[len];
for(int j=0; j<len; j++)
a[j] = nextInt()+offset;
return a;
}
long[] nextLongArray(int len) {
return nextLongArray(len, 0);
}
long[] nextLongArray(int len, int offset){
long[] a = new long[len];
for(int j=0; j<len; j++)
a[j] = nextLong()+offset;
return a;
}
}
public static class MyPrintWriter extends PrintWriter{
public MyPrintWriter(OutputStream os) {
super(os);
}
public void printlnAns(boolean ans) {
if(ans)
println(YES);
else
println(NO);
}
public void printAns(long[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(long[] arr){
printAns(arr);
println();
}
public void printAns(int[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(int[] arr){
printAns(arr);
println();
}
public <T> void printAns(ArrayList<T> arr){
if(arr != null && arr.size() > 0){
print(arr.get(0));
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i));
}
}
}
public <T> void printlnAns(ArrayList<T> arr){
printAns(arr);
println();
}
public void printAns(int[] arr, int add){
if(arr != null && arr.length > 0){
print(arr[0]+add);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]+add);
}
}
}
public void printlnAns(int[] arr, int add){
printAns(arr, add);
println();
}
public void printAns(ArrayList<Integer> arr, int add) {
if(arr != null && arr.size() > 0){
print(arr.get(0)+add);
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i)+add);
}
}
}
public void printlnAns(ArrayList<Integer> arr, int add){
printAns(arr, add);
println();
}
public void printlnAnsSplit(long[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public void printlnAnsSplit(int[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public <T> void printlnAnsSplit(ArrayList<T> arr, int split){
if(arr != null && !arr.isEmpty()){
for(int i=0; i<arr.size(); i+=split){
print(arr.get(i));
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr.get(j));
}
println();
}
}
}
}
static private void permutateAndSort(int[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
int temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private int[][] constructChildren(int n, int[] parent, int parentRoot){
int[][] childrens = new int[n][];
int[] numChildren = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
numChildren[parent[i]]++;
}
for(int i=0; i<n; i++) {
childrens[i] = new int[numChildren[i]];
}
int[] idx = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
childrens[parent[i]][idx[parent[i]]++] = i;
}
return childrens;
}
static private int[][][] constructDirectedNeighborhood(int n, int[][] e){
int[] inDegree = new int[n];
int[] outDegree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outDegree[u]++;
inDegree[v]++;
}
int[][] inNeighbors = new int[n][];
int[][] outNeighbors = new int[n][];
for(int i=0; i<n; i++) {
inNeighbors[i] = new int[inDegree[i]];
outNeighbors[i] = new int[outDegree[i]];
}
int[] inIdx = new int[n];
int[] outIdx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outNeighbors[u][outIdx[u]++] = v;
inNeighbors[v][inIdx[v]++] = u;
}
return new int[][][] {inNeighbors, outNeighbors};
}
static private int[][] constructNeighborhood(int n, int[][] e) {
int[] degree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
degree[u]++;
degree[v]++;
}
int[][] neighbors = new int[n][];
for(int i=0; i<n; i++)
neighbors[i] = new int[degree[i]];
int[] idx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
neighbors[u][idx[u]++] = v;
neighbors[v][idx[v]++] = u;
}
return neighbors;
}
static private void makeDotUndirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict graph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "--" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
static private void makeDotDirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict digraph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "->" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 70b1cfef6c45a00b21d05ba9aad3ec4d | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | // JAI SHREE RAM, HAR HAR MAHADEV, HARE KRISHNA
import java.util.*;
import java.util.Map.Entry;
import java.util.stream.*;
import java.lang.*;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.io.*;
public class CodeForces {
static private final String INPUT = "input.txt";
static private final String OUTPUT = "output.txt";
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st;
static PrintWriter out = new PrintWriter(System.out);
static DecimalFormat df = new DecimalFormat("0.00000");
final static int mod = (int) (1e9 + 7);
final static int MAX = Integer.MAX_VALUE;
final static int MIN = Integer.MIN_VALUE;
final static long INF = Long.MAX_VALUE;
final static long NEG_INF = Long.MIN_VALUE;
static Random rand = new Random();
// ======================= MAIN ==================================
public static void main(String[] args) throws IOException {
long time = System.currentTimeMillis();
boolean oj = System.getProperty("ONLINE_JUDGE") != null;
// ==== start ====
input();
preprocess();
int t = 1;
// t = readInt();
while (t-- > 0) {
solve();
}
out.flush();
// ==== end ====
if (!oj)
System.out.println(Arrays.deepToString(new Object[] { System.currentTimeMillis() - time + " ms" }));
}
private static void solve() throws IOException {
int n = readInt();
int lo = n + n - 1, hi = 2000 * n + n - 1, maxL = hi;
while (lo <= hi) {
int mid = (lo + hi) / 2;
out.println("? " + mid);
out.flush();
int x = readInt();
if (x == 1) {
maxL = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
int ans = maxL;
for (int i = 1; i <= n; i++) {
int w = maxL / i;
out.println("? " + w);
out.flush();
int l = readInt();
if (l != 0)
ans = Math.min(ans, l * w);
}
out.println("! " + ans);
}
private static void preprocess() throws IOException {
}
// cd C:\Users\Eshan Bhatt\Visual Studio Code\Competitive Programming\CodeForces
// javac CodeForces.java
// java CodeForces
// javac CodeForces.java && java CodeForces
// ==================== CUSTOM CLASSES ================================
static class Pair {
int first, second;
Pair(int f, int s) {
first = f;
second = s;
}
public int compareTo(Pair o) {
if (this.first == o.first)
return this.second - o.second;
return this.first - o.first;
}
@Override
public boolean equals(Object obj) {
if (obj == this)
return true;
if (obj == null)
return false;
if (this.getClass() != obj.getClass())
return false;
Pair other = (Pair) (obj);
if (this.first != other.first)
return false;
if (this.second != other.second)
return false;
return true;
}
@Override
public int hashCode() {
return this.first ^ this.second;
}
@Override
public String toString() {
return this.first + " " + this.second;
}
}
static class DequeNode {
DequeNode prev, next;
int val;
DequeNode(int val) {
this.val = val;
}
DequeNode(int val, DequeNode prev, DequeNode next) {
this.val = val;
this.prev = prev;
this.next = next;
}
}
// ======================= FOR INPUT ==================================
private static void input() {
FileInputStream instream = null;
PrintStream outstream = null;
try {
instream = new FileInputStream(INPUT);
outstream = new PrintStream(new FileOutputStream(OUTPUT));
System.setIn(instream);
System.setOut(outstream);
} catch (Exception e) {
System.err.println("Error Occurred.");
}
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
static String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(readLine());
return st.nextToken();
}
static long readLong() throws IOException {
return Long.parseLong(next());
}
static int readInt() throws IOException {
return Integer.parseInt(next());
}
static double readDouble() throws IOException {
return Double.parseDouble(next());
}
static char readCharacter() throws IOException {
return next().charAt(0);
}
static String readString() throws IOException {
return next();
}
static String readLine() throws IOException {
return br.readLine().trim();
}
static int[] readIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = readInt();
return arr;
}
static int[][] read2DIntArray(int n, int m) throws IOException {
int[][] arr = new int[n][m];
for (int i = 0; i < n; i++)
arr[i] = readIntArray(m);
return arr;
}
static List<Integer> readIntList(int n) throws IOException {
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(readInt());
return list;
}
static long[] readLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = readLong();
return arr;
}
static long[][] read2DLongArray(int n, int m) throws IOException {
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++)
arr[i] = readLongArray(m);
return arr;
}
static List<Long> readLongList(int n) throws IOException {
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(readLong());
return list;
}
static char[] readCharArray(int n) throws IOException {
return readString().toCharArray();
}
static char[][] readMatrix(int n, int m) throws IOException {
char[][] mat = new char[n][m];
for (int i = 0; i < n; i++)
mat[i] = readCharArray(m);
return mat;
}
// ========================= FOR OUTPUT ==================================
private static void printIList(List<Integer> list) {
for (int i = 0; i < list.size(); i++)
out.print(list.get(i) + " ");
out.println(" ");
}
private static void printLList(List<Long> list) {
for (int i = 0; i < list.size(); i++)
out.print(list.get(i) + " ");
out.println(" ");
}
private static void printIArray(int[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println(" ");
}
private static void print2DIArray(int[][] arr) {
for (int i = 0; i < arr.length; i++)
printIArray(arr[i]);
}
private static void printLArray(long[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println(" ");
}
private static void print2DLArray(long[][] arr) {
for (int i = 0; i < arr.length; i++)
printLArray(arr[i]);
}
// ====================== TO CHECK IF STRING IS NUMBER ========================
private static boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
private static boolean isLong(String s) {
try {
Long.parseLong(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
// ==================== FASTER SORT ================================
private static void sort(int[] arr) {
int n = arr.length;
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void reverseSort(int[] arr) {
int n = arr.length;
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void sort(long[] arr) {
int n = arr.length;
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void reverseSort(long[] arr) {
int n = arr.length;
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
// ==================== MATHEMATICAL FUNCTIONS ===========================
private static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
private static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static int mod_pow(int a, int b, int mod) {
if (b == 0)
return 1;
int temp = mod_pow(a, b >> 1, mod);
temp %= mod;
temp = (int) ((1L * temp * temp) % mod);
if ((b & 1) == 1)
temp = (int) ((1L * temp * a) % mod);
return temp;
}
private static int multiply(int a, int b) {
return (int) ((((1L * a) % mod) * ((1L * b) % mod)) % mod);
}
private static int divide(int a, int b) {
return multiply(a, mod_pow(b, mod - 2, mod));
}
private static boolean isPrime(long n) {
for (long i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return true;
}
private static long nCr(long n, long r) {
if (n - r > r)
r = n - r;
long ans = 1L;
for (long i = r + 1; i <= n; i++)
ans *= i;
for (long i = 2; i <= n - r; i++)
ans /= i;
return ans;
}
private static List<Integer> factors(int n) {
List<Integer> list = new ArrayList<>();
for (int i = 1; 1L * i * i <= n; i++)
if (n % i == 0) {
list.add(i);
if (i != n / i)
list.add(n / i);
}
return list;
}
private static List<Long> factors(long n) {
List<Long> list = new ArrayList<>();
for (long i = 1; i * i <= n; i++)
if (n % i == 0) {
list.add(i);
if (i != n / i)
list.add(n / i);
}
return list;
}
// ==================== Primes using Seive =====================
private static List<Integer> getPrimes(int n) {
boolean[] prime = new boolean[n + 1];
Arrays.fill(prime, true);
for (int i = 2; 1L * i * i <= n; i++)
if (prime[i])
for (int j = i * i; j <= n; j += i)
prime[j] = false;
// return prime;
List<Integer> list = new ArrayList<>();
for (int i = 2; i <= n; i++)
if (prime[i])
list.add(i);
return list;
}
private static int[] SeivePrime(int n) {
int[] primes = new int[n];
for (int i = 0; i < n; i++)
primes[i] = i;
for (int i = 2; 1L * i * i < n; i++) {
if (primes[i] != i)
continue;
for (int j = i * i; j < n; j += i)
if (primes[j] == j)
primes[j] = i;
}
return primes;
}
// ==================== STRING FUNCTIONS ================================
private static boolean isPalindrome(String str) {
int i = 0, j = str.length() - 1;
while (i < j)
if (str.charAt(i++) != str.charAt(j--))
return false;
return true;
}
// check if a is subsequence of b
private static boolean isSubsequence(String a, String b) {
int idx = 0;
for (int i = 0; i < b.length() && idx < a.length(); i++)
if (a.charAt(idx) == b.charAt(i))
idx++;
return idx == a.length();
}
private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
private static String sortString(String str) {
int[] arr = new int[256];
for (char ch : str.toCharArray())
arr[ch]++;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < 256; i++)
while (arr[i]-- > 0)
sb.append((char) i);
return sb.toString();
}
// ==================== LIS & LNDS ================================
private static int LIS(int arr[], int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find1(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find1(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) >= val) {
ret = mid;
j = mid - 1;
} else {
i = mid + 1;
}
}
return ret;
}
private static int LNDS(int[] arr, int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find2(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find2(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) <= val) {
i = mid + 1;
} else {
ret = mid;
j = mid - 1;
}
}
return ret;
}
// =============== Lower Bound & Upper Bound ===========
// less than or equal
private static int lower_bound(List<Integer> list, int val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(List<Long> list, long val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(int[] arr, int val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(long[] arr, long val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
// greater than or equal
private static int upper_bound(List<Integer> list, int val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(List<Long> list, long val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(int[] arr, int val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(long[] arr, long val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
// ==================== UNION FIND =====================
private static int find(int x, int[] parent) {
if (parent[x] == x)
return x;
return parent[x] = find(parent[x], parent);
}
private static boolean union(int x, int y, int[] parent, int[] rank) {
int lx = find(x, parent), ly = find(y, parent);
if (lx == ly)
return false;
if (rank[lx] > rank[ly])
parent[ly] = lx;
else if (rank[lx] < rank[ly])
parent[lx] = ly;
else {
parent[lx] = ly;
rank[ly]++;
}
return true;
}
// ================== SEGMENT TREE (RANGE SUM & RANGE UPDATE) ==================
public static class SegmentTree {
int n;
long[] arr, tree, lazy;
SegmentTree(long arr[]) {
this.arr = arr;
this.n = arr.length;
this.tree = new long[(n << 2)];
this.lazy = new long[(n << 2)];
build(1, 0, n - 1);
}
void build(int id, int start, int end) {
if (start == end)
tree[id] = arr[start];
else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
build(left, start, mid);
build(right, mid + 1, end);
tree[id] = tree[left] + tree[right];
}
}
void update(int l, int r, long val) {
update(1, 0, n - 1, l, r, val);
}
void update(int id, int start, int end, int l, int r, long val) {
distribute(id, start, end);
if (end < l || r < start)
return;
if (start == end)
tree[id] += val;
else if (l <= start && end <= r) {
lazy[id] += val;
distribute(id, start, end);
} else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
update(left, start, mid, l, r, val);
update(right, mid + 1, end, l, r, val);
tree[id] = tree[left] + tree[right];
}
}
long query(int l, int r) {
return query(1, 0, n - 1, l, r);
}
long query(int id, int start, int end, int l, int r) {
if (end < l || r < start)
return 0L;
distribute(id, start, end);
if (start == end)
return tree[id];
else if (l <= start && end <= r)
return tree[id];
else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
return query(left, start, mid, l, r) + query(right, mid + 1, end, l, r);
}
}
void distribute(int id, int start, int end) {
if (start == end)
tree[id] += lazy[id];
else {
tree[id] += lazy[id] * (end - start + 1);
lazy[(id << 1)] += lazy[id];
lazy[(id << 1) + 1] += lazy[id];
}
lazy[id] = 0;
}
}
// ==================== TRIE ================================
static class Trie {
class Node {
Node[] children;
boolean isEnd;
Node() {
children = new Node[26];
}
}
Node root;
Trie() {
root = new Node();
}
void insert(String word) {
Node curr = root;
for (char ch : word.toCharArray()) {
if (curr.children[ch - 'a'] == null)
curr.children[ch - 'a'] = new Node();
curr = curr.children[ch - 'a'];
}
curr.isEnd = true;
}
boolean find(String word) {
Node curr = root;
for (char ch : word.toCharArray()) {
if (curr.children[ch - 'a'] == null)
return false;
curr = curr.children[ch - 'a'];
}
return curr.isEnd;
}
}
// ==================== FENWICK TREE ================================
static class FT {
long[] tree;
int n;
FT(int[] arr, int n) {
this.n = n;
this.tree = new long[n + 1];
for (int i = 1; i <= n; i++) {
update(i, arr[i - 1]);
}
}
void update(int idx, int val) {
while (idx <= n) {
tree[idx] += val;
idx += idx & -idx;
}
}
long query(int l, int r) {
return getSum(r) - getSum(l - 1);
}
long getSum(int idx) {
long ans = 0L;
while (idx > 0) {
ans += tree[idx];
idx -= idx & -idx;
}
return ans;
}
}
// ==================== BINARY INDEX TREE ================================
static class BIT {
long[][] tree;
int n, m;
BIT(int[][] mat, int n, int m) {
this.n = n;
this.m = m;
tree = new long[n + 1][m + 1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
update(i, j, mat[i - 1][j - 1]);
}
}
}
void update(int x, int y, int val) {
while (x <= n) {
int t = y;
while (t <= m) {
tree[x][t] += val;
t += t & -t;
}
x += x & -x;
}
}
long query(int x1, int y1, int x2, int y2) {
return getSum(x2, y2) - getSum(x1 - 1, y2) - getSum(x2, y1 - 1) + getSum(x1 - 1, y1 - 1);
}
long getSum(int x, int y) {
long ans = 0L;
while (x > 0) {
int t = y;
while (t > 0) {
ans += tree[x][t];
t -= t & -t;
}
x -= x & -x;
}
return ans;
}
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 6df4c99e02704e263dd235a3080f33db | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
public class codeforces_G20_E {
private static void solve(FastIOAdapter in, PrintWriter out) {
int n = in.nextInt();
int l = 1, r = 5_000_000;
while (l != r) {
int mid = (l + r) / 2;
int ans = query(mid, in, out);
if (ans == 1) r = mid;
else l = mid + 1;
}
int ans = Integer.MAX_VALUE;
for (int i = 1; i <= n; i++) {
int resp = query(l / i, in, out);
if (resp != 0) ans = Math.min(ans, resp * (l / i));
}
out.println("! " + ans);
out.flush();
}
private static int query(int width, FastIOAdapter in, PrintWriter out) {
out.println("? " + width);
out.flush();
return in.nextInt();
}
public static void main(String[] args) throws Exception {
try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
int count = 1;
//count = ioAdapter.nextInt();
while (count-- > 0) {
solve(ioAdapter, ioAdapter.out);
}
}
}
static void ruffleSort(int[] arr) {
int n = arr.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
int tmp = arr[i];
int randomPos = i + rnd.nextInt(n - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
Arrays.sort(arr);
}
static class FastIOAdapter implements AutoCloseable {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
@Override
public void close() throws Exception {
out.flush();
out.close();
br.close();
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 8b37a4eca09ae96f4b605df7f249e93a | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.*;
public class E {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
// TODO Auto-generated method stub
solve(0);
}
private static void solve(int t) {
int n = sc.nextInt();
int space = n - 1;
int low = n + space;
int high = 2000*n + space + 100;
int maxArea = findMinWidth(low, high);
int widthAtHeightOne = maxArea;
int target, height, width;
for (int i = 2; i <= n; ++i) {
target = maxArea / i;
target *= i;
if (target == maxArea)
target -= i;
width = target / i;
if (width > 0) {
height = query(width);
if (height != 0 && height*width < maxArea) {
maxArea = height*width;
}
}
}
System.out.println("! " + maxArea);
System.out.flush();
}
private static int findMinWidth(int l, int h) {
if (query(l) == 1)
return l;
int mid;
while (h - l > 1) {
mid = (h + l)/2;
if (query(mid) == 1)
h = mid;
else
l = mid;
}
return h;
}
private static int query (int val) {
System.out.println("? " + val);
System.out.flush();
int height = sc.nextInt();
return height;
}
public static void print(int test, long result) {
System.out.println("Case #" + test + ": " + result);
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | fc9fba582946718606ddaf8bada1c8bb | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.*;
public class Q1672E {
public static void main(String args[])
{
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int l = 2*n - 1, r = 2001*n - 1;
while(l!=r)
{
int m = (l+r)/2;
System.out.println("? " + m);
System.out.flush();
int h = s.nextInt();
if(h==1) r = m;
else l = m + 1;
}
int ans = l;
for(int i=2; i<=n; i++)
{
int w = l/i;
System.out.println("? " + w);
System.out.flush();
int h = s.nextInt();
if(h!=0&&w*h<ans) ans = w*h;
}
System.out.println("! " + ans);
System.out.flush();
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | aab0b7f6163d2297c58ab75a7a64065e | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author lucasr
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
MyScanner in = new MyScanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
ENotepadexe solver = new ENotepadexe();
solver.solve(1, in, out);
out.close();
}
static class ENotepadexe {
public static MyScanner sc;
public static PrintWriter out;
public void solve(int testNumber, MyScanner sc, PrintWriter out) {
ENotepadexe.sc = sc;
ENotepadexe.out = out;
int n = sc.nextInt();
int left = 2 * n - 2, right = 2001 * n - 1;
while (left + 1 < right) {
int med = (left + right) / 2;
if (query(med) == 1) {
right = med;
} else {
left = med;
}
}
int bestArea = right;
for (int i = 2; i <= n; i++) {
if (query(bestArea / i) == i) {
bestArea = (bestArea / i) * i;
}
}
guess(bestArea);
}
int query(int w) {
out.println("? " + w);
out.flush();
return sc.nextInt();
}
void guess(int area) {
out.println("! " + area);
out.flush();
}
}
static class MyScanner {
private BufferedReader br;
private StringTokenizer tokenizer;
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 4440c21c49464b764be894d243549412 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.min;
import static java.lang.System.exit;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class E {
static void solve() throws Exception {
int n = scanInt();
int l = 2 * n - 1, r = 2001 * n - 1;
while (l < r) {
int m = (l + r) >> 1;
out.println("? " + m);
out.flush();
if (scanInt() == 1) {
r = m;
} else {
l = m + 1;
}
}
int ans = l;
for (int i = 2; i <= n; i++) {
if (ans >= i + 1) {
int w = (ans - 1) / i;
out.println("? " + w);
out.flush();
int h = scanInt();
if (h == 0) {
break;
}
ans = min(ans, w * h);
}
}
out.println("! " + ans);
}
static int scanInt() throws IOException {
return parseInt(scanString());
}
static long scanLong() throws IOException {
return parseLong(scanString());
}
static String scanString() throws IOException {
while (tok == null || !tok.hasMoreTokens()) {
tok = new StringTokenizer(in.readLine());
}
return tok.nextToken();
}
static BufferedReader in;
static PrintWriter out;
static StringTokenizer tok;
public static void main(String[] args) {
try {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
solve();
in.close();
out.close();
} catch (Throwable e) {
e.printStackTrace();
exit(1);
}
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 11 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 03ebdee51c6ed6141b43a88d3b5648e2 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.net.InetSocketAddress;
import java.util.*;
public class Main {
//--------------------------INPUT READER---------------------------------//
static class fs {
public BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public fs() { this(System.in); }
public fs(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String ns() { return next(); }
int[] na(long nn) {
int n = (int) nn;
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
long[] nal(long nn) {
int n = (int) nn;
long[] l = new long[n];
for(int i = 0; i < n; i++) l[i] = nl();
return l;
}
}
//-----------------------------------------------------------------------//
//---------------------------PRINTER-------------------------------------//
static class Printer {
static PrintWriter w;
public Printer() {this(System.out);}
public Printer(OutputStream os) {
w = new PrintWriter(os, true);
}
public void p(int i) {w.println(i);}
public void p(long l) {w.println(l);}
public void p(double d) {w.println(d);}
public void p(String s) { w.println(s);}
public void pr(int i) {w.print(i);}
public void pr(long l) {w.print(l);}
public void pr(double d) {w.print(d);}
public void pr(String s) { w.print(s);}
public void pl() {w.println();}
public void close() {w.close();}
}
//-----------------------------------------------------------------------//
//--------------------------VARIABLES------------------------------------//
static fs sc = new fs();
static OutputStream outputStream = System.out;
static Printer w = new Printer(outputStream);
static long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
static int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
static long mod = 1000000007;
//-----------------------------------------------------------------------//
//--------------------------ADMIN_MODE-----------------------------------//
private static void ADMIN_MODE() throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
w = new Printer(new FileOutputStream("output.txt"));
sc = new fs(new FileInputStream("input.txt"));
}
}
//-----------------------------------------------------------------------//
//----------------------------START--------------------------------------//
public static void main(String[] args)
throws IOException {
// ADMIN_MODE();
//int t = sc.ni();while(t-->0)
solve();
w.close();
}
static void solve() {
int n = sc.ni();
// min width for ht = 1
int l = 0; // ht(l) > 1
int r = (2000*n) + (n-1); // ht(r) <= 1
while(l < r-1) {
int mid = (r+l)/2;
w.p("? "+mid);
int ht = sc.ni();
if(ht == 0) {
l = mid;
continue;
}
if(ht > 1) l = mid;
else {
r = mid;
}
}
int minWidthForHeight1 = r;
long minArea = minWidthForHeight1;
int nextHeight = 2;
while(nextHeight <= n) {
long minAreaForNextHeight = minWidthForHeight1 - nextHeight + 1;
long maxAreaForNextHeight = minWidthForHeight1;
for(long i = minAreaForNextHeight; i < maxAreaForNextHeight; i++) {
if(i%nextHeight==0) {
long width = i/nextHeight;
w.p("? "+width);
int ht = sc.ni();
if(ht == 0) break;
long area = ht*width;
if(area < minArea) minArea = area;
break;
}
}
nextHeight++;
}
w.p("! "+minArea);
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 3fdcd2d90c319fe6e185d22168fd474d | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
public class NotepadExe {
private static final int W_MAX = 40022222;
public static void solve(FastIO io) {
final int N = io.nextInt();
int allTextWidth = BinarySearch.firstThat(1, W_MAX, new BinarySearch.IntCheck() {
@Override
public boolean valid(int value) {
int h = query(io, value);
return h == 1;
}
});
int best = allTextWidth;
for (int h = 2; h <= N; ++h) {
int w = allTextWidth / h;
int hActual = query(io, w);
if (hActual > 0) {
best = Math.min(best, w * hActual);
}
}
io.println("! " + best);
}
private static int query(FastIO io, int w) {
io.println("? " + w);
io.flush();
final int H = io.nextInt();
return H;
}
public static class BinarySearch {
// Finds the left-most value that satisfies the IntCheck in the range [L, R).
// It will return R if the nothing in the range satisfies the check.
public static int firstThat(int L, int R, IntCheck check) {
while (L < R) {
int M = (L >> 1) + (R >> 1) + (L & R & 1);
if (check.valid(M)) {
R = M;
} else {
L = M + 1;
}
}
return L;
}
// Finds the right-most value that satisfies the IntCheck in the range [L, R).
// It will return L - 1 if nothing in the range satisfies the check.
public static int lastThat(int L, int R, IntCheck check) {
int firstValue = firstThat(L, R, new IntCheck() {
@Override
public boolean valid(int value) {
return !check.valid(value);
}
});
return firstValue - 1;
}
// Finds the left-most value that satisfies the LongCheck in the range [L, R).
public static long firstThat(long L, long R, LongCheck check) {
while (L < R) {
long M = (L >> 1) + (R >> 1) + (L & R & 1);
if (check.valid(M)) {
R = M;
} else {
L = M + 1;
}
}
return L;
}
// Finds the right-most value that satisfies the IntCheck in the range [L, R).
// It will return L - 1 if nothing in the range satisfies the check.
public static long lastThat(long L, long R, LongCheck check) {
long firstValue = firstThat(L, R, new LongCheck() {
@Override
public boolean valid(long value) {
return !check.valid(value);
}
});
return firstValue - 1;
}
public static interface LongCheck {
public boolean valid(long value);
}
public static interface IntCheck {
public boolean valid(int value);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | e5726b450d3a12dacf6cde5bf3e5567e | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
public class NotepadExe {
private static final int W_MAX = 40022222;
public static void solve(FastIO io) {
final int N = io.nextInt();
int allTextWidth = BinarySearch.firstThat(1, W_MAX, new BinarySearch.IntCheck() {
@Override
public boolean valid(int value) {
int h = query(io, value);
return h == 1;
}
});
int best = allTextWidth;
for (int h = 2; h <= N; ++h) {
int w = best / h;
int hActual = query(io, w);
if (hActual > 0) {
best = Math.min(best, w * hActual);
}
}
io.println("! " + best);
}
private static int query(FastIO io, int w) {
io.println("? " + w);
io.flush();
final int H = io.nextInt();
return H;
}
public static class BinarySearch {
// Finds the left-most value that satisfies the IntCheck in the range [L, R).
// It will return R if the nothing in the range satisfies the check.
public static int firstThat(int L, int R, IntCheck check) {
while (L < R) {
int M = (L >> 1) + (R >> 1) + (L & R & 1);
if (check.valid(M)) {
R = M;
} else {
L = M + 1;
}
}
return L;
}
// Finds the right-most value that satisfies the IntCheck in the range [L, R).
// It will return L - 1 if nothing in the range satisfies the check.
public static int lastThat(int L, int R, IntCheck check) {
int firstValue = firstThat(L, R, new IntCheck() {
@Override
public boolean valid(int value) {
return !check.valid(value);
}
});
return firstValue - 1;
}
// Finds the left-most value that satisfies the LongCheck in the range [L, R).
public static long firstThat(long L, long R, LongCheck check) {
while (L < R) {
long M = (L >> 1) + (R >> 1) + (L & R & 1);
if (check.valid(M)) {
R = M;
} else {
L = M + 1;
}
}
return L;
}
// Finds the right-most value that satisfies the IntCheck in the range [L, R).
// It will return L - 1 if nothing in the range satisfies the check.
public static long lastThat(long L, long R, LongCheck check) {
long firstValue = firstThat(L, R, new LongCheck() {
@Override
public boolean valid(long value) {
return !check.valid(value);
}
});
return firstValue - 1;
}
public static interface LongCheck {
public boolean valid(long value);
}
public static interface IntCheck {
public boolean valid(int value);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | ccb42f05bde155a0dcf97e6fc3cb0cfd | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | //make sure to make new file!
import java.io.*;
import java.util.*;
public class EG20{
public static BufferedReader f;
public static PrintWriter out;
public static void main(String[] args)throws IOException{
f = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
int n = Integer.parseInt(f.readLine());
int l = 1;
int r = 4003000;
int ans = -1;
while(l <= r){
int mid = l + (r-l)/2;
int response = query(mid);
if(response != 1){
l = mid+1;
} else {
r = mid-1;
ans = mid;
}
}
long answer = (long)ans;
ArrayList<Integer> alist = new ArrayList<Integer>();
for(int sub = 2; sub <= n; sub++){
int i = ans-sub+1;
for(int k = sub; k <= n; k++){
if(i%k == 0){
alist.add(i/k);
}
}
}
for(int i : alist){
int res = query(i);
if(res == 0) continue;
answer = Math.min(answer,(long)i*(long)res);
}
out.println("! " + answer);
out.close();
}
public static int query(int x) throws IOException{
out.println("? " + x);
out.flush();
return Integer.parseInt(f.readLine());
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 5286a33016178e444984a32438f44a08 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class NotepadExe
{
public static void main(String omkar[]) throws Exception
{
BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(infile.readLine());
int N = Integer.parseInt(st.nextToken());
int low = 1;
int high = 1<<22;
while(low != high)
{
int mid = (low+high)>>1;
if(query(mid, infile) == 1)
high = mid;
else
low = mid+1;
}
int total = low;
int res = total;
for(int h=2; h <= N; h++)
{
int actualHeight = query(total/h, infile);
if(actualHeight != 0)
res = min(res, actualHeight*(total/h));
}
System.out.println("! "+res);
}
public static int query(int k, BufferedReader infile) throws Exception
{
System.out.println("? "+k);
return Integer.parseInt(infile.readLine());
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 2b861ebed35914fea73a33efe8039484 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
public class problemE {
public static void main(String[] args)throws IOException {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int n = Integer.parseInt(br.readLine());
int joyce_qu_so_cute = 1;
int r = 2001*n;
while(joyce_qu_so_cute<r) {
int m = (joyce_qu_so_cute+r)/2;
System.out.println("? "+m);
if(Integer.parseInt(br.readLine())==1)
r = m;
else
joyce_qu_so_cute = m+1;
}
int ans = joyce_qu_so_cute;
for(int i=2; i<=n; i++) {
double val = joyce_qu_so_cute/i;
if(Math.ceil(val)-val<0.000001)val = Math.ceil(val);
else val = Math.floor(val);
System.out.println("? "+(int)val);
int thing = Integer.parseInt(br.readLine());
if(thing!=0)
ans = Math.min(ans,thing*(int)val);
}
System.out.println("! "+ans);
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | b4373c54c54ab9500652a66365144644 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
public class problemE {
public static void main(String[] args)throws IOException {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int n = Integer.parseInt(br.readLine());
int joyce_qu_so_cute = 1;
int r = 2001*n;
while(joyce_qu_so_cute<r) {
int m = (joyce_qu_so_cute+r)/2;
System.out.println("? "+m);
if(Integer.parseInt(br.readLine())==1)
r = m;
else
joyce_qu_so_cute = m+1;
}
int len = joyce_qu_so_cute;
int ans = len;
for(int i=2; i<=n; i++) {
double val = len/i;
if(Math.ceil(val)-val<0.000001)val = Math.ceil(val);
else val = Math.floor(val);
System.out.println("? "+(int)val);
int thing = Integer.parseInt(br.readLine());
if(thing!=0)
ans = Math.min(ans,thing*(int)val);
}
System.out.println("! "+ans);
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 7aaca86e423d80dd77626f8c620e01f4 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) {new Main().run();}
FastReader in = new FastReader(false);
PrintWriter out = new PrintWriter(System.out);
void run() {
work();
out.flush();
}
long mod=998244353;
long gcd(long a,long b) {
return a==0?b:gcd(b%a,a);
}
final long inf=Long.MAX_VALUE/3;
void work(){
int n=ni();
int l=1,r=1000000000;
while(l<r){
int m=(l+r)/2;
System.out.println("? "+m);
int h=ni();
if(h!=1){
l=m+1;
}else{
r=m;
}
}
long ret=l;
for(int i=2;i<=n;i++){
int w=l/i;
System.out.println("? "+w);
long h=nl();
if(h>0){
ret=Math.min(ret,h*w);
}
}
out.println("! "+ret);
}
@SuppressWarnings("unused")
private ArrayList<Integer>[] ng(int n, int m) {
ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
for(int i=0;i<n;i++) {
graph[i]=new ArrayList<>();
}
for(int i=1;i<=m;i++) {
int s=in.nextInt()-1,e=in.nextInt()-1;
graph[s].add(e);
graph[e].add(s);
}
return graph;
}
private ArrayList<long[]>[] ngw(int n, int m) {
ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
for(int i=0;i<n;i++) {
graph[i]=new ArrayList<>();
}
for(int i=1;i<=m;i++) {
long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
graph[(int)s].add(new long[] {e,w});
graph[(int)e].add(new long[] {s,w});
}
return graph;
}
private int ni() {
return in.nextInt();
}
private long nl() {
return in.nextLong();
}
private double nd() {
return in.nextDouble();
}
private String ns() {
return in.next();
}
private long[] na(int n) {
long[] A=new long[n];
for(int i=0;i<n;i++) {
A[i]=in.nextLong();
}
return A;
}
private int[] nia(int n) {
int[] A=new int[n];
for(int i=0;i<n;i++) {
A[i]=in.nextInt();
}
return A;
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
InputStreamReader input;//no buffer
public FastReader()
{
br=new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(boolean isBuffer)
{
if(!isBuffer){
input=new InputStreamReader(System.in);
}else{
br=new BufferedReader(new InputStreamReader(System.in));
}
}
public boolean hasNext(){
try{
String s=br.readLine();
if(s==null){
return false;
}
st=new StringTokenizer(s);
}catch(IOException e){
e.printStackTrace();
}
return true;
}
public String next()
{
if(input!=null){
try {
StringBuilder sb=new StringBuilder();
int ch=input.read();
while(ch=='\n'||ch=='\r'||ch==32){
ch=input.read();
}
while(ch!='\n'&&ch!='\r'&&ch!=32){
sb.append((char)ch);
ch=input.read();
}
return sb.toString();
}catch (Exception e){
e.printStackTrace();
}
}
while(st==null || !st.hasMoreElements())//回车，空行情况
{
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt()
{
return (int)nextLong();
}
public long nextLong() {
try {
if(input!=null){
long ret=0;
int b=input.read();
while(b<'0'||b>'9'){
b=input.read();
}
while(b>='0'&&b<='9'){
ret=ret*10+(b-'0');
b=input.read();
}
return ret;
}
}catch (Exception e){
e.printStackTrace();
}
return Long.parseLong(next());
}
public double nextDouble()
{
return Double.parseDouble(next());
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 0f3b254b49d905ad137a07e58cb317fa | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Locale;
import java.util.StringTokenizer;
public class Solution implements Runnable {
private static final int MAXL = 2000;
private PrintStream out;
private BufferedReader in;
private StringTokenizer st;
public void solve() throws IOException {
long time0 = System.currentTimeMillis();
int n = nextInt();
int l = 0;
int r = n * (MAXL + 1);
while (l + 1 < r) {
int m = (l + r) / 2;
int res = query(m);
if (res != 1) {
l = m;
} else {
r = m;
}
}
int answer = r;
for (int i = 2; i <= n; i++) {
int opt_w = (r - 1) / i;
int opt_h = query(opt_w);
if (opt_h > 0 && answer > opt_w * opt_h) {
answer = opt_w * opt_h;
}
}
out.println("! " + answer);
System.err.println("time: " + (System.currentTimeMillis() - time0));
}
private int query(int w) throws IOException {
out.println("? " + w);
out.flush();
int result = nextInt();
return result;
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public String next() throws IOException {
while (!st.hasMoreTokens()) {
String line = in.readLine();
if (line == null) {
return null;
}
st = new StringTokenizer(line);
}
return st.nextToken();
}
@Override
public void run() {
try {
solve();
out.close();
} catch (Throwable e) {
throw new RuntimeException(e);
}
}
public Solution(String name) throws IOException {
Locale.setDefault(Locale.US);
if (name == null) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintStream(new BufferedOutputStream(System.out));
} else {
in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
}
st = new StringTokenizer("");
}
public static void main(String[] args) throws IOException {
new Thread(new Solution(null)).start();
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | 0235a5eced308e3c42e95592c28329a8 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Random;
import java.util.Scanner;
import java.util.StringTokenizer;
/*
*/
public class E {
static Scanner fs=new Scanner(System.in);
public static void main(String[] args) {
int n=fs.nextInt();
int minTotal=1, maxTotal=2001*2001;
while (minTotal!=maxTotal) {
int mid=(minTotal+maxTotal)/2;
if (query(mid)==1) {
maxTotal=mid;
}
else {
minTotal=mid+1;
}
}
// int totalNonSpace=minTotal-(n-1);
long ans=minTotal;
for (int toSave=1; toSave<n; toSave++) {
int divideInto=(toSave+1);
int remaining=minTotal-toSave;
int perLine=(divideInto-1+remaining)/divideInto;
long curHeight=query(perLine);
ans=Math.min(ans, curHeight*perLine);
}
System.out.println("! "+ans);
}
static final Random random=new Random();
static final int mod=1_000_000_007;
static int query(int w) {
System.out.println("? "+w);
int returned = fs.nextInt();
if (returned ==0) {
return Integer.MAX_VALUE/2;
}
return returned;
}
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static long add(long a, long b) {
return (a+b)%mod;
}
static long sub(long a, long b) {
return ((a-b)%mod+mod)%mod;
}
static long mul(long a, long b) {
return (a*b)%mod;
}
static long exp(long base, long exp) {
if (exp==0) return 1;
long half=exp(base, exp/2);
if (exp%2==0) return mul(half, half);
return mul(half, mul(half, base));
}
static long[] factorials=new long[2_000_001];
static long[] invFactorials=new long[2_000_001];
static void precompFacts() {
factorials[0]=invFactorials[0]=1;
for (int i=1; i<factorials.length; i++) factorials[i]=mul(factorials[i-1], i);
invFactorials[factorials.length-1]=exp(factorials[factorials.length-1], mod-2);
for (int i=invFactorials.length-2; i>=0; i--)
invFactorials[i]=mul(invFactorials[i+1], i+1);
}
static long nCk(int n, int k) {
return mul(factorials[n], mul(invFactorials[k], invFactorials[n-k]));
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | feb439f8148b08587c4f1d1d68d38867 | train_108.jsonl | 1650722700 | This is an interactive problem.There are $$$n$$$ words in a text editor. The $$$i$$$-th word has length $$$l_i$$$ ($$$1 \leq l_i \leq 2000$$$). The array $$$l$$$ is hidden and only known by the grader. The text editor displays words in lines, splitting each two words in a line with at least one space. Note that a line does not have to end with a space. Let the height of the text editor refer to the number of lines used. For the given width, the text editor will display words in such a way that the height is minimized.More formally, suppose that the text editor has width $$$w$$$. Let $$$a$$$ be an array of length $$$k+1$$$ where $$$1=a_1 < a_2 < \ldots < a_{k+1}=n+1$$$. $$$a$$$ is a valid array if for all $$$1 \leq i \leq k$$$, $$$l_{a_i}+1+l_{a_i+1}+1+\ldots+1+l_{a_{i+1}-1} \leq w$$$. Then the height of the text editor is the minimum $$$k$$$ over all valid arrays.Note that if $$$w < \max(l_i)$$$, the text editor cannot display all the words properly and will crash, and the height of the text editor will be $$$0$$$ instead.You can ask $$$n+30$$$ queries. In one query, you provide a width $$$w$$$. Then, the grader will return the height $$$h_w$$$ of the text editor when its width is $$$w$$$.Find the minimum area of the text editor, which is the minimum value of $$$w \cdot h_w$$$ over all $$$w$$$ for which $$$h_w \neq 0$$$.The lengths are fixed in advance. In other words, the interactor is not adaptive. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
public class E {
public static void main(String[] args) throws IOException {
/**/
Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
/*/
Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(new FileInputStream("src/e.in"))));
/**/
int n = sc.nextInt();
int min = n;
int max = n*2001-1;
while (min<max) {
int w = (min+max)/2;
System.out.println("? "+w);
System.out.flush();
int h = sc.nextInt();
if (h==1)
max = w;
else
min = w+1;
}
min = max-(n-1);
for (int i = 2; i <= n; ++i) {
while ((max-1)/i*i>=min) {
int w = (max-1)/i;
System.out.println("? "+w);
System.out.flush();
int h = sc.nextInt();
if (h==i)
max = w*h;
else
break;
}
}
System.out.println("! "+max);
System.out.flush();
}
} | Java | ["6\n\n0\n\n4\n\n2"] | 1 second | ["? 1\n\n? 9\n\n? 16\n\n! 32"] | NoteIn the first test case, the words are $$$\{\texttt{glory},\texttt{to},\texttt{ukraine},\texttt{and},\texttt{anton},\texttt{trygub}\}$$$, so $$$l=\{5,2,7,3,5,6\}$$$. If $$$w=1$$$, then the text editor is not able to display all words properly and will crash. The height of the text editor is $$$h_1=0$$$, so the grader will return $$$0$$$.If $$$w=9$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory__to}$$$ $$$\texttt{ukraine__}$$$ $$$\texttt{and_anton}$$$ $$$\texttt{__trygub_}$$$ The height of the text editor is $$$h_{9}=4$$$, so the grader will return $$$4$$$.If $$$w=16$$$, then a possible way that the words will be displayed on the text editor is: $$$\texttt{glory_to_ukraine}$$$ $$$\texttt{and_anton_trygub}$$$ The height of the text editor is $$$h_{16}=2$$$, so the grader will return $$$2$$$.We have somehow figured out that the minimum area of the text editor is $$$32$$$, so we answer it. | Java 8 | standard input | [
"binary search",
"constructive algorithms",
"greedy",
"interactive"
] | 8eb4ce3fb9f6220ab6dbc12819680c1e | The first and only line of input contains a single integer $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of words on the text editor. It is guaranteed that the hidden lengths $$$l_i$$$ satisfy $$$1 \leq l_i \leq 2000$$$. | 2,200 | null | standard output | |
PASSED | edd368e7659eacd5d2ab0669aa296fef | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.TreeMap;
public class sol2 {
public static void main (String[] args) throws Exception
{
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int []f=new int [n];
for(int i=0;i<n;i++)
f[i]=sc.nextInt();
int min=-1,max=-1;
for(int i=1;i<n;i++) {
if(f[i-1]==f[i]) {
if(min==-1)
min=i;
max=i;
}
}
// System.out.println("max="+max+" min="+min);
if(min==max)System.out.println(0);
else System.out.println(Math.max(1,max-min-1));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 75a171685b614ce7378296cd2b3b6423 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static int gcd(int a, int b) {
if (a == 0) return b;
if (b == 0) return a;
if (a == b) return a;
if (a > b) return gcd(a-b, b);
return gcd(a, b-a);
}
static void print(long []arr) {
for(long a : arr) System.out.print(a + " ");
System.out.println();
}
static void print(int []arr) {
for(int a : arr) System.out.print(a + " ");
System.out.println();
}
static void print(long a) {
System.out.print(a);
}
static void printl(long a) {
System.out.println(a);
}
static void printl(char ch) {
System.out.println(ch);
}
static void print(char ch) {
System.out.print(ch);
}
static void printl(String s) {
System.out.println(s);
}
static void print(String s) {
System.out.print(s);
}
static int[] intArray(int n) {
int arr[] = new int[n];
for(int i = 0; i < n; i++) arr[i] = sc.nextInt();
return arr;
}
static long[] longArray(int n) {
long arr[] = new long[n];
for(int i = 0; i < n; i++) arr[i] = sc.nextLong();
return arr;
}
static double[] doubleArray(int n) {
double arr[] = new double[n];
for(int i = 0; i < n; i++) arr[i] = sc.nextDouble();
return arr;
}
static int[][] intMatrix(int n, int m) {
int mat[][] = new int[n][m];
for(int i = 0; i < mat.length; i++)
for(int j = 0; j < mat[0].length; j++) mat[i][j] = sc.nextInt();
return mat;
}
static long[][] longMatrix(int n, int m) {
long mat[][] = new long[n][m];
for(int i = 0; i < mat.length; i++)
for(int j = 0; j < mat[0].length; j++) mat[i][j] = sc.nextLong();
return mat;
}
static class IntPair {
int v1, v2;
IntPair(int v1, int v2) {
this.v1 = v1;
this.v2 = v2;
}
}
static class LongPair {
long v1, v2;
LongPair(long v1, long v2) {
this.v1 = v1;
this.v2 = v2;
}
}
static FastReader sc = new FastReader();
public static void main(String[] args) {
int tc = sc.nextInt();
while(tc-- > 0) {
int n = sc.nextInt();
long a[] = longArray(n);
int left = -1, right = -1;
for(int i = 0; i < n; i++) {
if(i + 1 < n && a[i] == a[i + 1]) {
right = i;
if(left == -1) left = i;
}
}
if(left == -1 || right == -1 || left == right) printl(0);
else printl(Math.max(1, right - left - 1));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | ae4b0a7949b71b479f9162272748d49d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.util.spi.AbstractResourceBundleProvider;
import static java.lang.Integer.MAX_VALUE;
public class codeforces {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
int min = -1;
int max =-1;
for(int i=1;i<n;i++){
if(a[i]==a[i-1]){
if(min==-1){
min = i;
}
max=i;
}
}
if(max==min) System.out.println(0);
else System.out.println(Math.max(1,max-min-1));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 8667553dc5c8542dcd5069dc66ffeb9d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.io.*;
import java.util.*;
public class Solution {
static int mod=(int)998244353;
public static void main(String[] args) {
Copied io = new Copied(System.in, System.out);
// int k=1;
int t = 1;
t = io.nextInt();
for (int i = 0; i < t; i++) {
// io.print("Case #" + k + ": ");
solve(io);
// k++;
}
io.close();
}
public static void solve(Copied io) {
int n = io.nextInt();
int a[] = io.readArray(n);
int st = -1, en = -1;
int ans = 0;
for(int i = 0; i < n - 1; i++) {
if(a[i] == a[i + 1]) {
if(st == -1) {
st = i;
}else {
en = i + 1;
}
}
}
// io.println(st + " " + en);
if(en != -1) {
ans = en - st + 1 == 3 ? 1 : en - st - 2;
}
io.println(ans);
}
static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
static void binaryOutput(boolean ans, Copied io){
String a=new String("YES"), b=new String("NO");
if(ans){
io.println(a);
}
else{
io.println(b);
}
}
static int power(int x, int y, int p)
{
int res = 1;
x = x % p;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % p;
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>();
for (long i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
static void printArr(int[] arr,Copied io) {
for (int x : arr)
io.print(x + " ");
io.println();
}
static void printArr(long[] arr,Copied io) {
for (long x : arr)
io.print(x + " ");
io.println();
}
static int[] listToInt(ArrayList<Integer> a){
int n=a.size();
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=a.get(i);
}
return arr;
}
static int mod_mul(int a, int b){ a = a % mod; b = b % mod; return (((a * b) % mod) + mod) % mod;}
static int mod_add(int a, int b){ a = a % mod; b = b % mod; return (((a + b) % mod) + mod) % mod;}
static int mod_sub(int a, int b){ a = a % mod; b = b % mod; return (((a - b) % mod) + mod) % mod;}
}
class Pair implements Cloneable, Comparable<Pair>
{
int x,y;
Pair(int a,int b)
{
this.x=a;
this.y=b;
}
@Override
public boolean equals(Object obj)
{
if(obj instanceof Pair)
{
Pair p=(Pair)obj;
return p.x==this.x && p.y==this.y;
}
return false;
}
@Override
public int hashCode()
{
return Math.abs(x)+101*Math.abs(y);
}
@Override
public String toString()
{
return "("+x+" "+y+")";
}
@Override
protected Pair clone() throws CloneNotSupportedException {
return new Pair(this.x,this.y);
}
@Override
public int compareTo(Pair a)
{
int t= this.x-a.x;
if(t!=0)
return t;
else
return this.y-a.y;
}
}
class byY implements Comparator<Pair>{
// @Override
public int compare(Pair o1,Pair o2){
// -1 if want to swap and (1,0) otherwise.
int t= o1.y-o2.y;
if(t!=0)
return t;
else
return o1.x-o2.x;
}
}
class byX implements Comparator<Pair>{
// @Override
public int compare(Pair o1,Pair o2){
// -1 if want to swap and (1,0) otherwise.
int t= o1.x-o2.x;
if(t!=0)
return t;
else
return o1.y-o2.y;
}
}
class DescendingOrder<T> implements Comparator<T>{
@Override
public int compare(Object o1,Object o2){
// -1 if want to swap and (1,0) otherwise.
int addingNumber=(Integer) o1,existingNumber=(Integer) o2;
if(addingNumber>existingNumber) return -1;
else if(addingNumber<existingNumber) return 1;
else return 0;
}
}
class Copied extends PrintWriter {
public Copied(InputStream i) {
super(new BufferedOutputStream(System.out));
r = new BufferedReader(new InputStreamReader(i));
}
public Copied(InputStream i, OutputStream o) {
super(new BufferedOutputStream(o));
r = new BufferedReader(new InputStreamReader(i));
}
public boolean hasMoreTokens() {
return peekToken() != null;
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public String next() {
return nextToken();
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
private BufferedReader r;
private String line;
private StringTokenizer st;
private String token;
private String peekToken() {
if (token == null)
try {
while (st == null || !st.hasMoreTokens()) {
line = r.readLine();
if (line == null) return null;
st = new StringTokenizer(line);
}
token = st.nextToken();
} catch (IOException e) { }
return token;
}
private String nextToken() {
String ans = peekToken();
token = null;
return ans;
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 8a00cfee5702b29cb165b7ec0688ec33 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class text1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int cs = sc.nextInt();
for(int dz = 0;dz < cs;dz ++) {
int length = sc.nextInt();
long[] number = new long[length];
Set<Integer> set = new TreeSet<>();
for(int dz1 = 0;dz1 < length;dz1 ++){
number[dz1] = sc.nextLong();
if(dz1 > 0 && number[dz1 - 1] == number[dz1])
set.add(dz1 - 1);
}
if(set.size() <= 1)
System.out.println(0);
else
System.out.println(find(set,0));
}
return ;
}
public static int find(Set<Integer> set, int time){
for(int a : set){
set.remove(a);
set.add(a + 1);
if(set.contains(a + 2))
set.remove(a + 2);
time ++;
break;
}
if(set.size() == 1)
return time;
else
return find(set,time);
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 7aba9c1af7389998257bb03fcdd5ac8b | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.StringTokenizer;
public class UnequalArray {
public static void main(String[] args) {
Kattio in = new Kattio();
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = in.nextInt();
int left = 0, right = 0;
for (int i = 0; i < n - 1; i++)
if (a[i] == a[i + 1]) {
left = i + 1;
break;
}
for (int i = n - 1; i >= 1; i--) {
if (a[i] == a[i - 1]) {
right = i;
break;
}
}
if (a[right] == a[left]&&right-left==1)
System.out.println(1);
else
System.out.println(Math.max(right - left-1, 0));
}
}
}
class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// 标准 IO
public Kattio() {
this(System.in, System.out);
}
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// 文件 IO
public Kattio(String intput, String output) throws IOException {
super(output);
r = new BufferedReader(new FileReader(intput));
}
// 在没有其他输入时返回 null
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) {
}
return null;
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long nextLong() {
return Long.parseLong(next());
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 0a8277f7518aef293ecf6392353e326e | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.util.stream.Collectors;
public class coding {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(st == null || !st.hasMoreElements()) {
try {
String str = br.readLine();
if(str == null)
throw new InputMismatchException();
st = new StringTokenizer(str);
} catch(IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch(IOException e) {
e.printStackTrace();
}
if(str == null)
throw new InputMismatchException();
return str;
}
public BigInteger nextBigInteger() {
// TODO Auto-generated method stub
return null;
}
public void close() {
// TODO Auto-generated method stub
}
}
public static <K, V> K getKey(Map<K, V> map, V value)
{
for (Map.Entry<K, V> entry: map.entrySet())
{
if (value.equals(entry.getValue())) {
return entry.getKey();
}
}
return null;
}
public static int median(ArrayList<Integer> x)
{
int p=0;
if(x.size()==1)
{
p=x.get(0);
}
else
{
p=x.get((x.size()-1)/2 );
}
return p;
}
// public static boolean palindrome(long x)
// {
// String s="",s1="";
// while(x!=BigInteger.valueOf(0))
// {
// s=s+x%10;
// x=x/10;
// }
// for(int i=0;i<s.length();i++)
// {
// s1=s1+s.charAt(s.length()-i-1);
// }
// if(s1.contentEquals(s))
// {
// return true;
// }
// else
// {
// return false;
// }
// }
static void reverse(String a[], int n)
{
String k="", t="";
for (int i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static int fact(int n)
{
int fact=1;
int i=1;
while(i<=n)
{
fact=fact*i;
i++;
}
return fact;
}
static int gcd(int y, int x)
{
while(x!=y)
{
if(x>y)
{
x=x-y;
}
else
{
y=y-x;
}
}
return y;
}
public static String generatePermutation(String str, int start, int end,ArrayList<String> x)
{
if (start == end-1)
System.out.println(x);
else
{
for (int i = start; i < end; i++)
{
str = swapString(str,start,i);
generatePermutation(str,start+1,end,x);
String s1=generatePermutation(str,start+1,end,x);
x.add(s1);
str = swapString(str,start,i);
}
}
return str;
}
public static String swapString(String a, int i, int j) {
char[] b =a.toCharArray();
char ch;
ch = b[i];
b[i] = b[j];
b[j] = ch;
return String.valueOf(b);
}
public static List<String> findPermutations(String str)
{
// base case
if (str == null || str.length() == 0) {
return null;
}
// create an empty ArrayList to store (partial) permutations
List<String> partial = new ArrayList<>();
// initialize the list with the first character of the string
partial.add(String.valueOf(str.charAt(0)));
// do for every character of the specified string
for (int i = 1; i < str.length(); i++)
{
// consider previously constructed partial permutation one by one
// (iterate backward to avoid ConcurrentModificationException)
for (int j = partial.size() - 1; j >= 0 ; j--)
{
// remove current partial permutation from the ArrayList
String s = partial.remove(j);
// Insert the next character of the specified string at all
// possible positions of current partial permutation. Then
// insert each of these newly constructed strings in the list
for (int k = 0; k <= s.length(); k++)
{
// Advice: use StringBuilder for concatenation
if(!partial.contains(s.substring(0, k) + str.charAt(i) + s.substring(k)))
{
partial.add(s.substring(0, k) + str.charAt(i) + s.substring(k));
}
}
}
}
return partial;
}
public static void main(String args[] ) throws Exception {
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
StringBuilder sb = new StringBuilder();
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextLong();
}
int max=-1,min=Integer.MAX_VALUE;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
min=Math.min(min, i);
max=Math.max(max, i);
}
}
if(min==max || max==-1)
{
System.out.println(0);
}
else
{
if(max-min==1)
{
System.out.println(max-min);
}
else
{
System.out.println(max-min-1);
}
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1f6d596748ae40a9ea5120a4a846e319 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.*;
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
FastReader sc = new FastReader(new BufferedInputStream(System.in));
/* For Multiple Test Cases */
int times = sc.nextInt();
// int times = 1;
while (times-- > 0){
solve(sc);
}
}
public static void solve(FastReader sc) {
int n = sc.nextInt();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextLong();
}
int min = -1;
int max = -1;
for (int i = 1; i < n; i++) {
if (arr[i]==arr[i-1]){
if (min==-1){
min = i;
}
max = i;
}
}
if (min==max){
out.println(0);
}
else {
out.println(max(1,(max-min-1)));
}
}
/*---------------------------------------------------------------------------------------
Template Credit - MagentaCobra
--------------------------------------------------------------------------------------- */
public static void sort(int[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void sort(long[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Long> ls = new ArrayList<Long>();
for(long x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void print(int[] arr)
{
//for debugging only
for(int x: arr)
out.print(x+" ");
out.println();
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/*-----------------------------------------------------------------------------------------------
End for MagentaCobra's Template
------------------------------------------------------------------------------------------------ */
static long maxSubArraySum(Deque<Long> a)
{
long size = a.size();
long tempMax = Integer.MIN_VALUE, maxEnd
= 0;
for (int i = 0; i < size; i++) {
maxEnd = maxEnd + a.pollFirst();
if (tempMax < maxEnd)
tempMax = maxEnd;
if (maxEnd < 0)
maxEnd = 0;
}
return tempMax;
}
/*-----------------------------------------------------------------------------------------------
Class For Fast I/O
------------------------------------------------------------------------------------------------ */
private static class FastReader {
public BufferedReader br;
public StringTokenizer st;
public FastReader(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
st = null;
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 9c3804f69ea053996e04d7b8e3b4cd0d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.*;
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
FastReader sc = new FastReader(new BufferedInputStream(System.in));
/* For Multiple Test Cases */
int times = sc.nextInt();
// For single Test Case
// int times = 1;
while (times-- > 0){
solve(sc);
}
}
public static void solve(FastReader sc) {
int n = sc.nextInt();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextLong();
}
int min = -1;
int max = -1;
for (int i = 1; i < n; i++) {
if (arr[i]==arr[i-1]){
if (min==-1){
min = i;
}
max = i;
}
}
if (min==max){
out.println(0);
}
else {
out.println(max(1,(max-min-1)));
}
}
/*---------------------------------------------------------------------------------------
Template Credit - MagentaCobra
--------------------------------------------------------------------------------------- */
public static void sort(int[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void sort(long[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Long> ls = new ArrayList<Long>();
for(long x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void print(int[] arr)
{
//for debugging only
for(int x: arr)
out.print(x+" ");
out.println();
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/*-----------------------------------------------------------------------------------------------
End for MagentaCobra's Template
------------------------------------------------------------------------------------------------ */
static long maxSubArraySum(Deque<Long> a)
{
long size = a.size();
long tempMax = Integer.MIN_VALUE, maxEnd
= 0;
for (int i = 0; i < size; i++) {
maxEnd = maxEnd + a.pollFirst();
if (tempMax < maxEnd)
tempMax = maxEnd;
if (maxEnd < 0)
maxEnd = 0;
}
return tempMax;
}
/*-----------------------------------------------------------------------------------------------
Class For Fast I/O
------------------------------------------------------------------------------------------------ */
private static class FastReader {
public BufferedReader br;
public StringTokenizer st;
public FastReader(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
st = null;
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 6453ceaab32d151e49af8df823cd1b14 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class practice {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new BufferedWriter(new PrintWriter(System.out)));
StringBuilder sb = new StringBuilder();
int t = Integer.parseInt(br.readLine());
while (t --> 0) {
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
int arr[] = new int[n];
for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(st.nextToken());
int l = -1;
int r = -1;
for (int i = 0; i < n - 1 && l == -1; i++) {
if (arr[i] == arr[i + 1]) l = i;
}
for (int i = n - 2; i >= 0 && r == -1 && l != -1; i--) {
if (arr[i] == arr[i + 1]) r = i;
}
int ans = 0;
if (l != r) ans = Math.max(1, r - l - 1);
sb.append(ans).append("\n");
}
pw.println(sb.toString().trim());
pw.close();
br.close();
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | deae83bf65a020086b2f44eaff769475 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.lang.*;
import java.util.*;
public class ComdeFormces {
public static int cc2;
public static pair pr;
public static long sum;
public static void main(String[] args) throws Exception{
// TODO Auto-generated method stub
// Reader.init(System.in);
FastReader sc=new FastReader();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
// OutputStream out = new BufferedOutputStream ( System.out );
int t=sc.nextInt();
int tc=1;
while(t--!=0) {
int n=sc.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
}
int cnt=0;
int vl=0;
int s=-1;
int e=-1;
for(int i=0;i<n-1;i++) {
if(a[i]==a[i+1]) {
if(s==-1)s=i;
cnt++;
}
else {
cnt++;
if(cnt>1) {
e=i;
vl++;
}
cnt=0;
}
}
cnt++;
if(cnt>1) {
e=n-1;
vl++;
}
if(vl==0)log.write("0\n");
else if(vl==1) {
int sz=e-s+1;
if(sz>2) {
if(sz%2==0) {
if(sz==4)log.write("1\n");
else log.write((sz-3)+"\n");
}
else {
if(sz==3)log.write("1\n");
else log.write((sz-3)+"\n");
}
}
else log.write("0\n");
}
else log.write((e-s-2)+"\n");
log.flush();
}
}
static long evv(int a[],int mx,int mn,int n,long ans) {
int ind=-1;
int min=Integer.MAX_VALUE;
int min2=Integer.MAX_VALUE;
for(int i=0;i<n;i++) {
if(i==0 || i==n-1) {
if(mx-a[i]<min) {
min=mx-a[i];
ind=i;
}
if(ind==i) {
if(min2>mx-1) {
min2=mx-1;
}
}
else if(ind!=i) {
if(min2>a[i]-1)min2=a[i]-1;
}
}
else {
if(Math.abs(mx-a[i])+Math.abs(mx-a[i+1])<min) {
min=Math.abs(mx-a[i])+Math.abs(mx-a[i+1]);
ind=i;
}
if(ind==i) {
if(min2>mx+a[i]-2) {
min2=mx+a[i]-2;
}
}
else if(ind!=i) {
if(min2>a[i]+a[i+1]-2)min2=a[i]+a[i+1]-2;
}
}
// upd[i]=min;
}
return ans+=min+min2;
}
static long eval(ArrayList<ArrayList<Integer>> ar,int src,long f[], boolean vis[]) {
long mn=Integer.MAX_VALUE;
vis[src]=true;
for(int p:ar.get(src)) {
if(!vis[p]) {
long s=eval(ar,p,f,vis);
mn=Math.min(mn,s);
sum+=s;
}
}
if(src==0)return 0;
if(mn==Integer.MAX_VALUE)return f[src];
sum-=mn;
return Math.max(f[src], mn);
}
public static void radixSort(int a[]) {
int n=a.length;
int res[]=new int[n];
int p=1;
for(int i=0;i<=8;i++) {
int cnt[]=new int[10];
for(int j=0;j<n;j++) {
a[j]=res[j];
cnt[(a[j]/p)%10]++;
}
for(int j=1;j<=9;j++) {
cnt[j]+=cnt[j-1];
}
for(int j=n-1;j>=0;j--) {
res[cnt[(a[j]/p)%10]-1]=a[j];
cnt[(a[j]/p)%10]--;
}
p*=10;
}
}
static int bits(long n) {
int ans=0;
while(n!=0) {
if((n&1)==1)ans++;
n>>=1;
}
return ans;
}
static long flor(ArrayList<Long> ar,long el) {
int s=0;
int e=ar.size()-1;
while(s<=e) {
int m=s+(e-s)/2;
if(ar.get(m)==el)return ar.get(m);
else if(ar.get(m)<el)s=m+1;
else e=m-1;
}
return e>=0?e:-1;
}
public static int kadane(int a[]) {
int sum=0,mx=Integer.MIN_VALUE;
for(int i=0;i<a.length;i++) {
sum+=a[i];
mx=Math.max(mx, sum);
if(sum<0) sum=0;
}
return mx;
}
public static int m=1000000007;
public static int mul(int a, int b) {
return ((a%m)*(b%m))%m;
}
public static long mul(long a, long b) {
return ((a%m)*(b%m))%m;
}
public static int add(int a, int b) {
return ((a%m)+(b%m))%m;
}
public static long add(long a, long b) {
return ((a%m)+(b%m))%m;
}
//debug
public static <E> void p(E[][] a,String s) {
System.out.println(s);
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
public static void p(int[] a,String s) {
System.out.print(s+"=");
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
public static void p(long[] a,String s) {
System.out.print(s+"=");
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
public static <E> void p(E a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(ArrayList<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(LinkedList<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(HashSet<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(Stack<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(Queue<E> a,String s){
System.out.println(s+"="+a);
}
//utils
static ArrayList<Integer> divisors(int n){
ArrayList<Integer> ar=new ArrayList<>();
for (int i=2; i<=Math.sqrt(n); i++){
if (n%i == 0){
if (n/i == i) {
ar.add(i);
}
else {
ar.add(i);
ar.add(n/i);
}
}
}
return ar;
}
static ArrayList<Integer> prime(int n){
ArrayList<Integer> ar=new ArrayList<>();
int cnt=0;
boolean pr=false;
while(n%2==0) {
ar.add(2);
n/=2;
}
for(int i=3;i*i<=n;i+=2) {
pr=false;
while(n%i==0) {
n/=i;
ar.add(i);
pr=true;
}
}
if(n>2) ar.add(n);
return ar;
}
static long gcd(long a,long b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static int gcd(int a,int b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static long factmod(long n,long mod,long img) {
if(n==0)return 1;
long ans=1;
long temp=1;
while(n--!=0) {
if(temp!=img) {
ans=((ans%mod)*((temp)%mod))%mod;
}
temp++;
}
return ans%mod;
}
static int ncr(int n, int r){
if(r>n-r)r=n-r;
int ans=1;
for(int i=0;i<r;i++){
ans*=(n-i);
ans/=(i+1);
}
return ans;
}
public static class trip{
int a,b;
int c;
public trip(int a,int b,int c) {
this.a=a;
this.b=b;
this.c=c;
}
public int compareTo(trip q) {
return this.b-q.b;
}
}
static void mergesort(long[] a,int start,int end) {
if(start>=end)return ;
int mid=start+(end-start)/2;
mergesort(a,start,mid);
mergesort(a,mid+1,end);
merge(a,start,mid,end);
}
static void merge(long[] a, int start,int mid,int end) {
int ptr1=start;
int ptr2=mid+1;
long b[]=new long[end-start+1];
int i=0;
while(ptr1<=mid && ptr2<=end) {
if(a[ptr1]<=a[ptr2]) {
b[i]=a[ptr1];
ptr1++;
i++;
}
else {
b[i]=a[ptr2];
ptr2++;
i++;
}
}
while(ptr1<=mid) {
b[i]=a[ptr1];
ptr1++;
i++;
}
while(ptr2<=end) {
b[i]=a[ptr2];
ptr2++;
i++;
}
for(int j=start;j<=end;j++) {
a[j]=b[j-start];
}
}
public static class FastReader {
BufferedReader b;
StringTokenizer s;
public FastReader() {
b=new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(s==null ||!s.hasMoreElements()) {
try {
s=new StringTokenizer(b.readLine());
}
catch(IOException e) {
e.printStackTrace();
}
}
return s.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str="";
try {
str=b.readLine();
}
catch(IOException e) {
e.printStackTrace();
}
return str;
}
boolean hasNext() {
if (s != null && s.hasMoreTokens()) {
return true;
}
String tmp;
try {
b.mark(1000);
tmp = b.readLine();
if (tmp == null) {
return false;
}
b.reset();
} catch (IOException e) {
return false;
}
return true;
}
}
public static class pair{
int a;
int b;
public pair(int a,int b) {
this.a=a;
this.b=b;
}
public int compareTo(pair b) {
return this.a-b.a;
}
// public int compareToo(pair b) {
// return this.b-b.b;
// }
@Override
public String toString() {
return "{"+this.a+" "+this.b+"}";
}
}
static long pow(long a, long pw) {
long temp;
if(pw==0)return 1;
temp=pow(a,pw/2);
if(pw%2==0)return temp*temp;
return a*temp*temp;
}
static int pow(int a, int pw) {
int temp;
if(pw==0)return 1;
temp=pow(a,pw/2);
if(pw%2==0)return temp*temp;
return a*temp*temp;
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 5f882ef76d69fe871d29ef06e310657d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
/**
*
* @author Ahmad1423
*/
public class trial {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
Scanner read = new Scanner(System.in);
int t = read.nextInt();
for (int i = 0; i < t; i++) {
Colorful_Stamp(read);
}
}
public static void Colorful_Stamp(Scanner read) {
int arr[] = new int[read.nextInt()];
for (int i = 0; i < arr.length; i++) {
arr[i] = read.nextInt();
}
int firstPair = -1;
for (int i = 0; i < arr.length - 1; i++) {
if(arr[i] == arr[i + 1]){
firstPair = i + 1;
break;
}
}
if(firstPair == -1){
System.out.println(0);
return;
}
int secondPair = -1;
for (int i = arr.length - 1; i > firstPair; i--) {
if (arr[i] == arr[i - 1]){
secondPair = i - 1;
break;
}
}
if(secondPair == -1){
System.out.println(0);
return;
}
if (secondPair - firstPair == 0){
System.out.println(1);
return;
}
// System.out.println(" first pair index = " + firstPair);
// System.out.println("second pair index = " + secondPair);
if (secondPair - firstPair == 1){
System.out.println(1);
return;
}
if ((secondPair - firstPair) % 2 == 0){
System.out.println(secondPair - firstPair );
return;
} else {
System.out.println(secondPair - firstPair );
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 86d902868b58e91e5a8f67381ef9022b | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | // import static java.lang.System.out;
import static java.lang.Math.*;
import static java.lang.System.*;
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
static FastReader sc;
static long mod = ((long) 1e9) + 7;
static long lmax=Long.MAX_VALUE;
static long lmin=Long.MIN_VALUE;
static int imax=Integer.MAX_VALUE;
static int imin=Integer.MIN_VALUE;
// static FastWriter out;
static FastWriter out;
public static void main(String hi[]) throws IOException {
initializeIO();
out = new FastWriter();
sc = new FastReader();
long startTimeProg = System.currentTimeMillis();
long endTimeProg = Long.MAX_VALUE;
int t = sc.nextInt();
// int t=1;
// boolean[] seave=sieveOfEratosthenes((int)(1e5));
// int[] seave2=sieveOfEratosthenesInt((long)sqrt(1e9));
// debug(seave2);
while (t-- != 0) {
int n=sc.nextInt();
int[] arr=readIntArray(n);
int i=1,j=n-2;
int c=0;
debug(arr);
while(i<n&&arr[i]!=arr[i-1])i++;
if(i>=n){
print(0);
continue;
}
while (j>=0&&arr[j]!=arr[j+1])j--;
if(i==j){
c=1;
}else if(i<j){
c=j-i;
}
debug(i+" "+j);
print(c);
}
endTimeProg = System.currentTimeMillis();
debug("[finished : " + (endTimeProg - startTimeProg) + ".ms ]");
// System.out.println(String.format("%.9f", max));
}
private static long findchanges(int[] arr, int ind, long avg) {
long v=0;
long changes=0;
int n=arr.length;
for(int i=0;i<n;i++){
if(ind==i)continue;
v+=avg-arr[i]+v;
changes+=abs(avg-arr[i]+v);
}
return changes;
}
static int c=0;
private static int find(int[] arr, int i, int j) {
List<Integer> li=new ArrayList<>();
while (i<=j)li.add(arr[i++]);
c=0;
return chk(li);
}
private static int chk(List<Integer> li) {
if(li.size()<2)return c;
List<Integer> res=new ArrayList<>();
int n=li.size();
int v=0;
for(int i=0;i<n-1;i=i+2){
int l=li.get(i);
int r=li.get(i+1);
if(l+r>=10)c++;
res.add((l+r)%10);
}
return chk(res);
}
private static boolean isPeak(int[] arr,int i,int l,int r){
if(i==l||i==r)return false;
return (arr[i+1]<arr[i]&&arr[i]>arr[i-1]);
}
private static boolean isPeak(long[] arr,int i,int l,int r){
if(i==l||i==r)return false;
return (arr[i+1]<arr[i]&&arr[i]>arr[i-1]);
}
private static long kadens(List<Long> li, int l, int r) {
long max = Long.MIN_VALUE;
long ans = 0;
for (int i = l; i <= r; i++) {
ans += li.get(i);
max = max(ans, max);
if (ans < 0) ans = 0;
}
return max;
}
public static boolean isNumeric(String strNum) {
if (strNum == null) {
return false;
}
try {
double d = Double.parseDouble(strNum);
} catch (NumberFormatException nfe) {
return false;
}
return true;
}
private static boolean isSorted(List<Integer> li) {
int n = li.size();
if (n <= 1) return true;
for (int i = 0; i < n - 1; i++) {
if (li.get(i) > li.get(i + 1)) return false;
}
return true;
}
static boolean isPowerOfTwo(long x) {
return x != 0 && ((x & (x - 1)) == 0);
}
private static boolean ispallindromeList(List<Integer> res) {
int l = 0, r = res.size() - 1;
while (l < r) {
if (res.get(l) != res.get(r)) return false;
l++;
r--;
}
return true;
}
private static class Pair {
int first = 0;
int sec = 0;
int[] arr;
char ch;
String s;
Map<Integer, Integer> map;
Pair(int first, int sec) {
this.first = first;
this.sec = sec;
}
Pair(int[] arr) {
this.map = new HashMap<>();
for (int x : arr) this.map.put(x, map.getOrDefault(x, 0) + 1);
this.arr = arr;
}
Pair(char ch, int first) {
this.ch = ch;
this.first = first;
}
Pair(String s, int first) {
this.s = s;
this.first = first;
}
}
private static int sizeOfSubstring(int st, int e) {
int s = e - st + 1;
return (s * (s + 1)) / 2;
}
private static Set<Long> factors(long n) {
Set<Long> res = new HashSet<>();
// res.add(n);
for (long i = 1; i * i <= (n); i++) {
if (n % i == 0) {
res.add(i);
if (n / i != i) {
res.add(n / i);
}
}
}
return res;
}
private static long fact(long n) {
if (n <= 2) return n;
return n * fact(n - 1);
}
private static long ncr(long n, long r) {
return fact(n) / (fact(r) * fact(n - r));
}
private static int lessThen(long[] nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums[mid] <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int lessThen(List<Long> nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int lessThen(List<Integer> nums, int val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int greaterThen(List<Long> nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static int greaterThen(List<Integer> nums, int val, boolean work) {
int i = -1, l = 0, r = nums.size() - 1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static int greaterThen(long[] nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums[mid] >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static long gcd(long[] arr) {
long ans = 0;
for (long x : arr) {
ans = gcd(x, ans);
}
return ans;
}
private static int gcd(int[] arr) {
int ans = 0;
for (int x : arr) {
ans = gcd(x, ans);
}
return ans;
}
private static long sumOfAp(long a, long n, long d) {
long val = (n * (2 * a + ((n - 1) * d)));
return val / 2;
}
//geometrics
private static double areaOftriangle(double x1, double y1, double x2, double y2, double x3, double y3) {
double[] mid_point = midOfaLine(x1, y1, x2, y2);
// debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" "+y3);
double height = distanceBetweenPoints(mid_point[0], mid_point[1], x3, y3);
double wight = distanceBetweenPoints(x1, y1, x2, y2);
// debug(height+" "+wight);
return (height * wight) / 2;
}
private static double distanceBetweenPoints(double x1, double y1, double x2, double y2) {
double x = x2 - x1;
double y = y2 - y1;
return (Math.pow(x, 2) + Math.pow(y, 2));
}
public static boolean isPerfectSquareByUsingSqrt(long n) {
if (n <= 0) {
return false;
}
double squareRoot = Math.sqrt(n);
long tst = (long) (squareRoot + 0.5);
return tst * tst == n;
}
private static double[] midOfaLine(double x1, double y1, double x2, double y2) {
double[] mid = new double[2];
mid[0] = (x1 + x2) / 2;
mid[1] = (y1 + y2) / 2;
return mid;
}
private static long sumOfN(long n) {
return (n * (n + 1)) / 2;
}
private static long power(long x, long y, long p) {
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
/* Function to calculate x raised to the power y in O(logn)*/
static long power(long x, long y) {
long temp;
if (y == 0)
return 1l;
temp = power(x, y / 2);
if (y % 2 == 0)
return (temp * temp);
else
return (x * temp * temp);
}
private static StringBuilder reverseString(String s) {
StringBuilder sb = new StringBuilder(s);
int l = 0, r = sb.length() - 1;
while (l <= r) {
char ch = sb.charAt(l);
sb.setCharAt(l, sb.charAt(r));
sb.setCharAt(r, ch);
l++;
r--;
}
return sb;
}
private static void swap(List<Integer> li, int i, int j) {
int t = li.get(i);
li.set(i, li.get(j));
li.set(j, t);
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static String decimalToString(int x) {
return Integer.toBinaryString(x);
}
private static String decimalToString(long x) {
return Long.toBinaryString(x);
}
private static boolean isPallindrome(String s, int l, int r) {
while (l < r) {
if (s.charAt(l) != s.charAt(r)) return false;
l++;
r--;
}
return true;
}
private static boolean isSubsequence(String s, String t) {
if (s == null || t == null) return false;
Map<Character, List<Integer>> map = new HashMap<>(); //<character, index>
//preprocess t
for (int i = 0; i < t.length(); i++) {
char curr = t.charAt(i);
if (!map.containsKey(curr)) {
map.put(curr, new ArrayList<Integer>());
}
map.get(curr).add(i);
}
int prev = -1; //index of previous character
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (map.get(c) == null) {
return false;
} else {
List<Integer> list = map.get(c);
prev = lessThen(list, prev, false, 0, list.size() - 1);
if (prev == -1) {
return false;
}
prev++;
}
}
return true;
}
private static StringBuilder removeLeadingZero(StringBuilder sb) {
int i = 0;
while (i < sb.length() && sb.charAt(i) == '0') i++;
// debug("remove "+i);
if (i == sb.length()) return new StringBuilder();
return new StringBuilder(sb.substring(i, sb.length()));
}
private static StringBuilder removeEndingZero(StringBuilder sb) {
int i = sb.length() - 1;
while (i >= 0 && sb.charAt(i) == '0') i--;
// debug("remove "+i);
if (i < 0) return new StringBuilder();
return new StringBuilder(sb.substring(0, i + 1));
}
private static int stringToDecimal(String binaryString) {
// debug(decimalToString(n<<1));
return Integer.parseInt(binaryString, 2);
}
private static int stringToInt(String s) {
return Integer.parseInt(s);
}
private static String toString(long val) {
return String.valueOf(val);
}
private static void debug(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(long[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(List<int[]> arr) {
for (int[] a : arr) {
err.println(Arrays.toString(a));
}
}
private static void debug(float[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(double[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(boolean[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void print(String s) throws IOException {
out.println(s);
}
private static void debug(String s) throws IOException {
err.println(s);
}
private static int charToIntS(char c) {
return ((((int) (c - '0')) % 48));
}
private static void print(double s) throws IOException {
out.println(s);
}
private static void debug(char[] s1) throws IOException {
debug(Arrays.toString(s1));
}
private static void print(float s) throws IOException {
out.println(s);
}
private static void print(long s) throws IOException {
out.println(s);
}
private static void print(int s) throws IOException {
out.println(s);
}
private static void debug(double s) throws IOException {
err.println(s);
}
private static void debug(float s) throws IOException {
err.println(s);
}
private static void debug(long s) {
err.println(s);
}
private static void debug(int s) {
err.println(s);
}
private static boolean isPrime(long n) {
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return false;
}
return true;
}
private static List<List<Integer>> readUndirectedGraph(int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readUndirectedGraph(int[][] intervals, int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int[][] intervals, int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
static String[] readStringArray(int n) {
String[] arr = new String[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.next();
}
return arr;
}
private static Map<Character, Integer> freq(String s) {
Map<Character, Integer> map = new HashMap<>();
for (char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}
return map;
}
private static Map<Long, Integer> freq(long[] arr) {
Map<Long, Integer> map = new HashMap<>();
for (long x : arr) {
map.put(x, map.getOrDefault(x, 0) + 1);
}
return map;
}
private static Map<Integer, Integer> freq(int[] arr) {
Map<Integer, Integer> map = new HashMap<>();
for (int x : arr) {
map.put(x, map.getOrDefault(x, 0) + 1);
}
return map;
}
static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int) n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int[] sieveOfEratosthenesInt(long n) {
boolean prime[] = new boolean[(int) n + 1];
Set<Integer> li = new HashSet<>();
for (int i = 1; i <= n; i++) {
prime[i] = true;
li.add(i);
}
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) {
li.remove(i);
prime[i] = false;
}
}
}
int[] arr = new int[li.size()];
int i = 0;
for (int x : li) {
arr[i++] = x;
}
return arr;
}
static boolean isMemberAC(int a, int d, int x) {
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
static boolean isMemberAC(long a, long d, long x) {
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
private static void sort(int[] arr) {
int n = arr.length;
List<Integer> li = new ArrayList<>();
for (int x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(int[] arr) {
int n = arr.length;
List<Integer> li = new ArrayList<>();
for (int x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sort(double[] arr) {
int n = arr.length;
List<Double> li = new ArrayList<>();
for (double x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(double[] arr) {
int n = arr.length;
List<Double> li = new ArrayList<>();
for (double x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(long[] arr) {
int n = arr.length;
List<Long> li = new ArrayList<>();
for (long x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sort(long[] arr) {
int n = arr.length;
List<Long> li = new ArrayList<>();
for (long x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static long sum(int[] arr) {
long sum = 0;
for (int x : arr) {
sum += x;
}
return sum;
}
private static long sum(long[] arr) {
long sum = 0;
for (long x : arr) {
sum += x;
}
return sum;
}
private static void initializeIO() {
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output")));
System.setErr(new PrintStream(new FileOutputStream("error.txt")));
} catch (Exception e) {
System.err.println(e.getMessage());
}
}
private static int maxOfArray(int[] arr) {
int max = Integer.MIN_VALUE;
for (int x : arr) {
max = Math.max(max, x);
}
return max;
}
private static long maxOfArray(long[] arr) {
long max = Long.MIN_VALUE;
for (long x : arr) {
max = Math.max(max, x);
}
return max;
}
private static int[][] readIntIntervals(int n, int m) {
int[][] arr = new int[n][m];
for (int j = 0; j < n; j++) {
for (int i = 0; i < m; i++) {
arr[j][i] = sc.nextInt();
}
}
return arr;
}
private static long gcd(long a, long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
private static int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
private static int[] readIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
return arr;
}
private static double[] readDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextDouble();
}
return arr;
}
private static long[] readLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextLong();
}
return arr;
}
private static void debug(String[] arr) {
err.println(Arrays.toString(arr));
}
private static void debug(Boolean[][] arr) {
for (int i = 0; i < arr.length; i++) err.println(Arrays.toString(arr[i]));
}
private static void debug(int[] arr) {
err.println(Arrays.toString(arr));
}
private static void debug(long[] arr) {
err.println(Arrays.toString(arr));
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
BufferedWriter bw;
List<String> list = new ArrayList<>();
Set<String> set = new HashSet<>();
FastWriter() {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
<T> void print(T obj) throws IOException {
bw.write(obj.toString());
bw.flush();
}
void println() throws IOException {
print("\n");
}
<T> void println(T obj) throws IOException {
print(obj.toString() + "\n");
}
void print(int[] arr) throws IOException {
for (int x : arr) {
print(x + " ");
}
println();
}
void print(long[] arr) throws IOException {
for (long x : arr) {
print(x + " ");
}
println();
}
void print(double[] arr) throws IOException {
for (double x : arr) {
print(x + " ");
}
println();
}
void printCharN(char c, int n) throws IOException {
for (int i = 0; i < n; i++) {
print(c);
}
}
}
static class Dsu {
int[] parent, size;
Dsu(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
private int findParent(int u) {
if (parent[u] == u) return u;
return parent[u] = findParent(parent[u]);
}
private boolean union(int u, int v) {
// System.out.println("uf "+u+" "+v);
int pu = findParent(u);
// System.out.println("uf2 "+pu+" "+v);
int pv = findParent(v);
// System.out.println("uf3 " + u + " " + pv);
if (pu == pv) return false;
if (size[pu] <= size[pv]) {
parent[pu] = pv;
size[pv] += size[pu];
} else {
parent[pv] = pu;
size[pu] += size[pv];
}
return true;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 08315321ecc73690a3130b2b7da6c519 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | // import static java.lang.System.out;
import static java.lang.Math.*;
import static java.lang.System.*;
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
static FastReader sc;
static long mod = ((long) 1e9) + 7;
static long lmax=Long.MAX_VALUE;
static long lmin=Long.MIN_VALUE;
static int imax=Integer.MAX_VALUE;
static int imin=Integer.MIN_VALUE;
// static FastWriter out;
static FastWriter out;
public static void main(String hi[]) throws IOException {
initializeIO();
out = new FastWriter();
sc = new FastReader();
long startTimeProg = System.currentTimeMillis();
long endTimeProg = Long.MAX_VALUE;
int t = sc.nextInt();
// int t=1;
// boolean[] seave=sieveOfEratosthenes((int)(1e5));
// int[] seave2=sieveOfEratosthenesInt((long)sqrt(1e9));
// debug(seave2);
while (t-- != 0) {
int n=sc.nextInt();
int[] arr=readIntArray(n);
// debug("done");
int i=0,j=n-1;
int c=0;
while(i<n-1){
if(arr[i]==arr[i+1]){
break;
}
i++;
}
if(i==n-1){
print(0);
continue;
}
while(j>0){
if(arr[j]==arr[j-1])break;
j--;
}
int values=j-i+1;
if(values==2){
print(0);
}else {
print(max(1,values-3));
}
}
endTimeProg = System.currentTimeMillis();
debug("[finished : " + (endTimeProg - startTimeProg) + ".ms ]");
// System.out.println(String.format("%.9f", max));
}
private static long findchanges(int[] arr, int ind, long avg) {
long v=0;
long changes=0;
int n=arr.length;
for(int i=0;i<n;i++){
if(ind==i)continue;
v+=avg-arr[i]+v;
changes+=abs(avg-arr[i]+v);
}
return changes;
}
static int c=0;
private static int find(int[] arr, int i, int j) {
List<Integer> li=new ArrayList<>();
while (i<=j)li.add(arr[i++]);
c=0;
return chk(li);
}
private static int chk(List<Integer> li) {
if(li.size()<2)return c;
List<Integer> res=new ArrayList<>();
int n=li.size();
int v=0;
for(int i=0;i<n-1;i=i+2){
int l=li.get(i);
int r=li.get(i+1);
if(l+r>=10)c++;
res.add((l+r)%10);
}
return chk(res);
}
private static boolean isPeak(int[] arr,int i,int l,int r){
if(i==l||i==r)return false;
return (arr[i+1]<arr[i]&&arr[i]>arr[i-1]);
}
private static boolean isPeak(long[] arr,int i,int l,int r){
if(i==l||i==r)return false;
return (arr[i+1]<arr[i]&&arr[i]>arr[i-1]);
}
private static long kadens(List<Long> li, int l, int r) {
long max = Long.MIN_VALUE;
long ans = 0;
for (int i = l; i <= r; i++) {
ans += li.get(i);
max = max(ans, max);
if (ans < 0) ans = 0;
}
return max;
}
public static boolean isNumeric(String strNum) {
if (strNum == null) {
return false;
}
try {
double d = Double.parseDouble(strNum);
} catch (NumberFormatException nfe) {
return false;
}
return true;
}
private static boolean isSorted(List<Integer> li) {
int n = li.size();
if (n <= 1) return true;
for (int i = 0; i < n - 1; i++) {
if (li.get(i) > li.get(i + 1)) return false;
}
return true;
}
static boolean isPowerOfTwo(long x) {
return x != 0 && ((x & (x - 1)) == 0);
}
private static boolean ispallindromeList(List<Integer> res) {
int l = 0, r = res.size() - 1;
while (l < r) {
if (res.get(l) != res.get(r)) return false;
l++;
r--;
}
return true;
}
private static class Pair {
int first = 0;
int sec = 0;
int[] arr;
char ch;
String s;
Map<Integer, Integer> map;
Pair(int first, int sec) {
this.first = first;
this.sec = sec;
}
Pair(int[] arr) {
this.map = new HashMap<>();
for (int x : arr) this.map.put(x, map.getOrDefault(x, 0) + 1);
this.arr = arr;
}
Pair(char ch, int first) {
this.ch = ch;
this.first = first;
}
Pair(String s, int first) {
this.s = s;
this.first = first;
}
}
private static int sizeOfSubstring(int st, int e) {
int s = e - st + 1;
return (s * (s + 1)) / 2;
}
private static Set<Long> factors(long n) {
Set<Long> res = new HashSet<>();
// res.add(n);
for (long i = 1; i * i <= (n); i++) {
if (n % i == 0) {
res.add(i);
if (n / i != i) {
res.add(n / i);
}
}
}
return res;
}
private static long fact(long n) {
if (n <= 2) return n;
return n * fact(n - 1);
}
private static long ncr(long n, long r) {
return fact(n) / (fact(r) * fact(n - r));
}
private static int lessThen(long[] nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums[mid] <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int lessThen(List<Long> nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int lessThen(List<Integer> nums, int val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int greaterThen(List<Long> nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static int greaterThen(List<Integer> nums, int val, boolean work) {
int i = -1, l = 0, r = nums.size() - 1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static int greaterThen(long[] nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums[mid] >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static long gcd(long[] arr) {
long ans = 0;
for (long x : arr) {
ans = gcd(x, ans);
}
return ans;
}
private static int gcd(int[] arr) {
int ans = 0;
for (int x : arr) {
ans = gcd(x, ans);
}
return ans;
}
private static long sumOfAp(long a, long n, long d) {
long val = (n * (2 * a + ((n - 1) * d)));
return val / 2;
}
//geometrics
private static double areaOftriangle(double x1, double y1, double x2, double y2, double x3, double y3) {
double[] mid_point = midOfaLine(x1, y1, x2, y2);
// debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" "+y3);
double height = distanceBetweenPoints(mid_point[0], mid_point[1], x3, y3);
double wight = distanceBetweenPoints(x1, y1, x2, y2);
// debug(height+" "+wight);
return (height * wight) / 2;
}
private static double distanceBetweenPoints(double x1, double y1, double x2, double y2) {
double x = x2 - x1;
double y = y2 - y1;
return (Math.pow(x, 2) + Math.pow(y, 2));
}
public static boolean isPerfectSquareByUsingSqrt(long n) {
if (n <= 0) {
return false;
}
double squareRoot = Math.sqrt(n);
long tst = (long) (squareRoot + 0.5);
return tst * tst == n;
}
private static double[] midOfaLine(double x1, double y1, double x2, double y2) {
double[] mid = new double[2];
mid[0] = (x1 + x2) / 2;
mid[1] = (y1 + y2) / 2;
return mid;
}
private static long sumOfN(long n) {
return (n * (n + 1)) / 2;
}
private static long power(long x, long y, long p) {
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
/* Function to calculate x raised to the power y in O(logn)*/
static long power(long x, long y) {
long temp;
if (y == 0)
return 1l;
temp = power(x, y / 2);
if (y % 2 == 0)
return (temp * temp);
else
return (x * temp * temp);
}
private static StringBuilder reverseString(String s) {
StringBuilder sb = new StringBuilder(s);
int l = 0, r = sb.length() - 1;
while (l <= r) {
char ch = sb.charAt(l);
sb.setCharAt(l, sb.charAt(r));
sb.setCharAt(r, ch);
l++;
r--;
}
return sb;
}
private static void swap(List<Integer> li, int i, int j) {
int t = li.get(i);
li.set(i, li.get(j));
li.set(j, t);
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static String decimalToString(int x) {
return Integer.toBinaryString(x);
}
private static String decimalToString(long x) {
return Long.toBinaryString(x);
}
private static boolean isPallindrome(String s, int l, int r) {
while (l < r) {
if (s.charAt(l) != s.charAt(r)) return false;
l++;
r--;
}
return true;
}
private static boolean isSubsequence(String s, String t) {
if (s == null || t == null) return false;
Map<Character, List<Integer>> map = new HashMap<>(); //<character, index>
//preprocess t
for (int i = 0; i < t.length(); i++) {
char curr = t.charAt(i);
if (!map.containsKey(curr)) {
map.put(curr, new ArrayList<Integer>());
}
map.get(curr).add(i);
}
int prev = -1; //index of previous character
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (map.get(c) == null) {
return false;
} else {
List<Integer> list = map.get(c);
prev = lessThen(list, prev, false, 0, list.size() - 1);
if (prev == -1) {
return false;
}
prev++;
}
}
return true;
}
private static StringBuilder removeLeadingZero(StringBuilder sb) {
int i = 0;
while (i < sb.length() && sb.charAt(i) == '0') i++;
// debug("remove "+i);
if (i == sb.length()) return new StringBuilder();
return new StringBuilder(sb.substring(i, sb.length()));
}
private static StringBuilder removeEndingZero(StringBuilder sb) {
int i = sb.length() - 1;
while (i >= 0 && sb.charAt(i) == '0') i--;
// debug("remove "+i);
if (i < 0) return new StringBuilder();
return new StringBuilder(sb.substring(0, i + 1));
}
private static int stringToDecimal(String binaryString) {
// debug(decimalToString(n<<1));
return Integer.parseInt(binaryString, 2);
}
private static int stringToInt(String s) {
return Integer.parseInt(s);
}
private static String toString(long val) {
return String.valueOf(val);
}
private static void debug(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(long[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(List<int[]> arr) {
for (int[] a : arr) {
err.println(Arrays.toString(a));
}
}
private static void debug(float[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(double[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(boolean[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void print(String s) throws IOException {
out.println(s);
}
private static void debug(String s) throws IOException {
err.println(s);
}
private static int charToIntS(char c) {
return ((((int) (c - '0')) % 48));
}
private static void print(double s) throws IOException {
out.println(s);
}
private static void debug(char[] s1) throws IOException {
debug(Arrays.toString(s1));
}
private static void print(float s) throws IOException {
out.println(s);
}
private static void print(long s) throws IOException {
out.println(s);
}
private static void print(int s) throws IOException {
out.println(s);
}
private static void debug(double s) throws IOException {
err.println(s);
}
private static void debug(float s) throws IOException {
err.println(s);
}
private static void debug(long s) {
err.println(s);
}
private static void debug(int s) {
err.println(s);
}
private static boolean isPrime(long n) {
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return false;
}
return true;
}
private static List<List<Integer>> readUndirectedGraph(int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readUndirectedGraph(int[][] intervals, int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int[][] intervals, int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
static String[] readStringArray(int n) {
String[] arr = new String[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.next();
}
return arr;
}
private static Map<Character, Integer> freq(String s) {
Map<Character, Integer> map = new HashMap<>();
for (char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}
return map;
}
private static Map<Long, Integer> freq(long[] arr) {
Map<Long, Integer> map = new HashMap<>();
for (long x : arr) {
map.put(x, map.getOrDefault(x, 0) + 1);
}
return map;
}
private static Map<Integer, Integer> freq(int[] arr) {
Map<Integer, Integer> map = new HashMap<>();
for (int x : arr) {
map.put(x, map.getOrDefault(x, 0) + 1);
}
return map;
}
static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int) n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int[] sieveOfEratosthenesInt(long n) {
boolean prime[] = new boolean[(int) n + 1];
Set<Integer> li = new HashSet<>();
for (int i = 1; i <= n; i++) {
prime[i] = true;
li.add(i);
}
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) {
li.remove(i);
prime[i] = false;
}
}
}
int[] arr = new int[li.size()];
int i = 0;
for (int x : li) {
arr[i++] = x;
}
return arr;
}
static boolean isMemberAC(int a, int d, int x) {
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
static boolean isMemberAC(long a, long d, long x) {
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
private static void sort(int[] arr) {
int n = arr.length;
List<Integer> li = new ArrayList<>();
for (int x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(int[] arr) {
int n = arr.length;
List<Integer> li = new ArrayList<>();
for (int x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sort(double[] arr) {
int n = arr.length;
List<Double> li = new ArrayList<>();
for (double x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(double[] arr) {
int n = arr.length;
List<Double> li = new ArrayList<>();
for (double x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(long[] arr) {
int n = arr.length;
List<Long> li = new ArrayList<>();
for (long x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sort(long[] arr) {
int n = arr.length;
List<Long> li = new ArrayList<>();
for (long x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static long sum(int[] arr) {
long sum = 0;
for (int x : arr) {
sum += x;
}
return sum;
}
private static long sum(long[] arr) {
long sum = 0;
for (long x : arr) {
sum += x;
}
return sum;
}
private static void initializeIO() {
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output")));
System.setErr(new PrintStream(new FileOutputStream("error.txt")));
} catch (Exception e) {
System.err.println(e.getMessage());
}
}
private static int maxOfArray(int[] arr) {
int max = Integer.MIN_VALUE;
for (int x : arr) {
max = Math.max(max, x);
}
return max;
}
private static long maxOfArray(long[] arr) {
long max = Long.MIN_VALUE;
for (long x : arr) {
max = Math.max(max, x);
}
return max;
}
private static int[][] readIntIntervals(int n, int m) {
int[][] arr = new int[n][m];
for (int j = 0; j < n; j++) {
for (int i = 0; i < m; i++) {
arr[j][i] = sc.nextInt();
}
}
return arr;
}
private static long gcd(long a, long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
private static int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
private static int[] readIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
return arr;
}
private static double[] readDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextDouble();
}
return arr;
}
private static long[] readLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextLong();
}
return arr;
}
private static void debug(String[] arr) {
err.println(Arrays.toString(arr));
}
private static void debug(Boolean[][] arr) {
for (int i = 0; i < arr.length; i++) err.println(Arrays.toString(arr[i]));
}
private static void debug(int[] arr) {
err.println(Arrays.toString(arr));
}
private static void debug(long[] arr) {
err.println(Arrays.toString(arr));
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
BufferedWriter bw;
List<String> list = new ArrayList<>();
Set<String> set = new HashSet<>();
FastWriter() {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
<T> void print(T obj) throws IOException {
bw.write(obj.toString());
bw.flush();
}
void println() throws IOException {
print("\n");
}
<T> void println(T obj) throws IOException {
print(obj.toString() + "\n");
}
void print(int[] arr) throws IOException {
for (int x : arr) {
print(x + " ");
}
println();
}
void print(long[] arr) throws IOException {
for (long x : arr) {
print(x + " ");
}
println();
}
void print(double[] arr) throws IOException {
for (double x : arr) {
print(x + " ");
}
println();
}
void printCharN(char c, int n) throws IOException {
for (int i = 0; i < n; i++) {
print(c);
}
}
}
static class Dsu {
int[] parent, size;
Dsu(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
private int findParent(int u) {
if (parent[u] == u) return u;
return parent[u] = findParent(parent[u]);
}
private boolean union(int u, int v) {
// System.out.println("uf "+u+" "+v);
int pu = findParent(u);
// System.out.println("uf2 "+pu+" "+v);
int pv = findParent(v);
// System.out.println("uf3 " + u + " " + pv);
if (pu == pv) return false;
if (size[pu] <= size[pv]) {
parent[pu] = pv;
size[pv] += size[pu];
} else {
parent[pv] = pu;
size[pu] += size[pv];
}
return true;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 250756a93672a595bf416a14800bdab8 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class JavaApplication3 {
public static void main(String[]args){
Scanner in =new Scanner(System.in);
int t=in.nextInt();
for(int c=0;c<t;++c){
int n= in.nextInt();
int[]r =new int[n];
for(int i=0;i<n;++i)r[i]=in.nextInt();
int begin =-1;
int count =1;
int temp =0;
int top=0;
for(int i=1;i<n;++i){
if(r[i] != r[i-1]){
if(count>1){ if(begin == -1)begin=temp;top=temp+(count);}
count =1;
temp=i;
}
else ++count;
}
if(count>1){ if(begin == -1)begin=temp;top=temp+(count);}
int h = top -begin;
if(h<6)System.out.println((h-1)/2);
else if(h==6)System.out.println(3);
else System.out.println((h-3));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1d88b678cba2a9bac551052a720f29cd | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class B {
static FastScanner fs = new FastScanner();
static PrintWriter pw = new PrintWriter(System.out);
static StringBuilder sb = new StringBuilder("");
public static void main(String[] args) {
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = fs.nextInt();
int idx = -1, idx2 = -1;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
idx = i + 1;
break;
}
}
for (int i = n - 1; i >= 1; i--) {
if (a[i] == a[i - 1]) {
idx2 = i - 1;
break;
}
}
if (idx2 != -1 && idx != -1 && n == 3) idx2++;
if (idx2 != -1 && idx != -1 && idx == idx2) idx2++;
sb.append((idx2 - idx) != -1 ? idx2-idx +"\n" : 0 + "\n");
}
pw.print(sb.toString());
pw.close();
}
static void sort(int[] a) {
ArrayList<Integer> al = new ArrayList<>();
for (int i : a)
al.add(i);
Collections.sort(al);
for (int i = 0; i < a.length; i++)
a[i] = al.get(i);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {while (!st.hasMoreTokens()) try {st = new StringTokenizer(br.readLine());} catch (IOException e) {}return st.nextToken();}
int nextInt() {return Integer.parseInt(next());}
long nextLong() {return Long.parseLong(next());}
double nextDouble() {return Double.parseDouble(next());}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 7180a456b158158864516f530645dd00 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | // JAI SHREE RAM, HAR HAR MAHADEV, HARE KRISHNA
import java.util.*;
import java.util.Map.Entry;
import java.util.stream.*;
import java.lang.*;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.io.*;
public class CodeForces {
static private final String INPUT = "input.txt";
static private final String OUTPUT = "output.txt";
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st;
static PrintWriter out = new PrintWriter(System.out);
static DecimalFormat df = new DecimalFormat("0.00000");
final static int mod = (int) (1e9 + 7);
final static int MAX = Integer.MAX_VALUE;
final static int MIN = Integer.MIN_VALUE;
final static long INF = Long.MAX_VALUE;
final static long NEG_INF = Long.MIN_VALUE;
static Random rand = new Random();
// ======================= MAIN ==================================
public static void main(String[] args) throws IOException {
long time = System.currentTimeMillis();
boolean oj = System.getProperty("ONLINE_JUDGE") != null;
// ==== start ====
input();
preprocess();
int t = 1;
t = readInt();
while (t-- > 0) {
solve();
}
out.flush();
// ==== end ====
if (!oj)
System.out.println(Arrays.deepToString(new Object[] { System.currentTimeMillis() - time + " ms" }));
}
private static void solve() throws IOException {
int n = readInt();
int[] arr = readIntArray(n);
List<Integer> list = new ArrayList<>();
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
list.add(i);
}
if (list.size() <= 1)
out.println(0);
else if (list.size() == 2) {
if (list.get(0) == list.get(1) - 1)
out.println(1);
else
out.println(list.get(1) - list.get(0) - 1);
} else {
out.println(list.get(list.size() - 1) - list.get(0) - 1);
}
}
private static void preprocess() throws IOException {
}
// cd C:\Users\Eshan Bhatt\Visual Studio Code\Competitive Programming\CodeForces
// javac CodeForces.java
// java CodeForces
// javac CodeForces.java && java CodeForces
// ==================== CUSTOM CLASSES ================================
static class Pair {
int first, second;
Pair(int f, int s) {
first = f;
second = s;
}
public int compareTo(Pair o) {
if (this.first == o.first)
return this.second - o.second;
return this.first - o.first;
}
@Override
public boolean equals(Object obj) {
if (obj == this)
return true;
if (obj == null)
return false;
if (this.getClass() != obj.getClass())
return false;
Pair other = (Pair) (obj);
if (this.first != other.first)
return false;
if (this.second != other.second)
return false;
return true;
}
@Override
public int hashCode() {
return this.first ^ this.second;
}
@Override
public String toString() {
return this.first + " " + this.second;
}
}
static class DequeNode {
DequeNode prev, next;
int val;
DequeNode(int val) {
this.val = val;
}
DequeNode(int val, DequeNode prev, DequeNode next) {
this.val = val;
this.prev = prev;
this.next = next;
}
}
// ======================= FOR INPUT ==================================
private static void input() {
FileInputStream instream = null;
PrintStream outstream = null;
try {
instream = new FileInputStream(INPUT);
outstream = new PrintStream(new FileOutputStream(OUTPUT));
System.setIn(instream);
System.setOut(outstream);
} catch (Exception e) {
System.err.println("Error Occurred.");
}
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
static String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(readLine());
return st.nextToken();
}
static long readLong() throws IOException {
return Long.parseLong(next());
}
static int readInt() throws IOException {
return Integer.parseInt(next());
}
static double readDouble() throws IOException {
return Double.parseDouble(next());
}
static char readCharacter() throws IOException {
return next().charAt(0);
}
static String readString() throws IOException {
return next();
}
static String readLine() throws IOException {
return br.readLine().trim();
}
static int[] readIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = readInt();
return arr;
}
static int[][] read2DIntArray(int n, int m) throws IOException {
int[][] arr = new int[n][m];
for (int i = 0; i < n; i++)
arr[i] = readIntArray(m);
return arr;
}
static List<Integer> readIntList(int n) throws IOException {
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(readInt());
return list;
}
static long[] readLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = readLong();
return arr;
}
static long[][] read2DLongArray(int n, int m) throws IOException {
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++)
arr[i] = readLongArray(m);
return arr;
}
static List<Long> readLongList(int n) throws IOException {
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(readLong());
return list;
}
static char[] readCharArray(int n) throws IOException {
return readString().toCharArray();
}
static char[][] readMatrix(int n, int m) throws IOException {
char[][] mat = new char[n][m];
for (int i = 0; i < n; i++)
mat[i] = readCharArray(m);
return mat;
}
// ========================= FOR OUTPUT ==================================
private static void printIList(List<Integer> list) {
for (int i = 0; i < list.size(); i++)
out.print(list.get(i) + " ");
out.println(" ");
}
private static void printLList(List<Long> list) {
for (int i = 0; i < list.size(); i++)
out.print(list.get(i) + " ");
out.println(" ");
}
private static void printIArray(int[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println(" ");
}
private static void print2DIArray(int[][] arr) {
for (int i = 0; i < arr.length; i++)
printIArray(arr[i]);
}
private static void printLArray(long[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println(" ");
}
private static void print2DLArray(long[][] arr) {
for (int i = 0; i < arr.length; i++)
printLArray(arr[i]);
}
// ====================== TO CHECK IF STRING IS NUMBER ========================
private static boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
private static boolean isLong(String s) {
try {
Long.parseLong(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
// ==================== FASTER SORT ================================
private static void sort(int[] arr) {
int n = arr.length;
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void reverseSort(int[] arr) {
int n = arr.length;
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void sort(long[] arr) {
int n = arr.length;
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void reverseSort(long[] arr) {
int n = arr.length;
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
// ==================== MATHEMATICAL FUNCTIONS ===========================
private static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
private static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static int mod_pow(int a, int b, int mod) {
if (b == 0)
return 1;
int temp = mod_pow(a, b >> 1, mod);
temp %= mod;
temp = (int) ((1L * temp * temp) % mod);
if ((b & 1) == 1)
temp = (int) ((1L * temp * a) % mod);
return temp;
}
private static int multiply(int a, int b) {
return (int) ((((1L * a) % mod) * ((1L * b) % mod)) % mod);
}
private static int divide(int a, int b) {
return multiply(a, mod_pow(b, mod - 2, mod));
}
private static boolean isPrime(long n) {
for (long i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return true;
}
private static long nCr(long n, long r) {
if (n - r > r)
r = n - r;
long ans = 1L;
for (long i = r + 1; i <= n; i++)
ans *= i;
for (long i = 2; i <= n - r; i++)
ans /= i;
return ans;
}
private static List<Integer> factors(int n) {
List<Integer> list = new ArrayList<>();
for (int i = 1; 1L * i * i <= n; i++)
if (n % i == 0) {
list.add(i);
if (i != n / i)
list.add(n / i);
}
return list;
}
private static List<Long> factors(long n) {
List<Long> list = new ArrayList<>();
for (long i = 1; i * i <= n; i++)
if (n % i == 0) {
list.add(i);
if (i != n / i)
list.add(n / i);
}
return list;
}
// ==================== Primes using Seive =====================
private static List<Integer> getPrimes(int n) {
boolean[] prime = new boolean[n + 1];
Arrays.fill(prime, true);
for (int i = 2; 1L * i * i <= n; i++)
if (prime[i])
for (int j = i * i; j <= n; j += i)
prime[j] = false;
// return prime;
List<Integer> list = new ArrayList<>();
for (int i = 2; i <= n; i++)
if (prime[i])
list.add(i);
return list;
}
private static int[] SeivePrime(int n) {
int[] primes = new int[n];
for (int i = 0; i < n; i++)
primes[i] = i;
for (int i = 2; 1L * i * i < n; i++) {
if (primes[i] != i)
continue;
for (int j = i * i; j < n; j += i)
if (primes[j] == j)
primes[j] = i;
}
return primes;
}
// ==================== STRING FUNCTIONS ================================
private static boolean isPalindrome(String str) {
int i = 0, j = str.length() - 1;
while (i < j)
if (str.charAt(i++) != str.charAt(j--))
return false;
return true;
}
// check if a is subsequence of b
private static boolean isSubsequence(String a, String b) {
int idx = 0;
for (int i = 0; i < b.length() && idx < a.length(); i++)
if (a.charAt(idx) == b.charAt(i))
idx++;
return idx == a.length();
}
private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
private static String sortString(String str) {
int[] arr = new int[256];
for (char ch : str.toCharArray())
arr[ch]++;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < 256; i++)
while (arr[i]-- > 0)
sb.append((char) i);
return sb.toString();
}
// ==================== LIS & LNDS ================================
private static int LIS(int arr[], int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find1(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find1(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) >= val) {
ret = mid;
j = mid - 1;
} else {
i = mid + 1;
}
}
return ret;
}
private static int LNDS(int[] arr, int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find2(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find2(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) <= val) {
i = mid + 1;
} else {
ret = mid;
j = mid - 1;
}
}
return ret;
}
// =============== Lower Bound & Upper Bound ===========
// less than or equal
private static int lower_bound(List<Integer> list, int val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(List<Long> list, long val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(int[] arr, int val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(long[] arr, long val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
// greater than or equal
private static int upper_bound(List<Integer> list, int val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(List<Long> list, long val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(int[] arr, int val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(long[] arr, long val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
// ==================== UNION FIND =====================
private static int find(int x, int[] parent) {
if (parent[x] == x)
return x;
return parent[x] = find(parent[x], parent);
}
private static boolean union(int x, int y, int[] parent, int[] rank) {
int lx = find(x, parent), ly = find(y, parent);
if (lx == ly)
return false;
if (rank[lx] > rank[ly])
parent[ly] = lx;
else if (rank[lx] < rank[ly])
parent[lx] = ly;
else {
parent[lx] = ly;
rank[ly]++;
}
return true;
}
// ================== SEGMENT TREE (RANGE SUM & RANGE UPDATE) ==================
public static class SegmentTree {
int n;
long[] arr, tree, lazy;
SegmentTree(long arr[]) {
this.arr = arr;
this.n = arr.length;
this.tree = new long[(n << 2)];
this.lazy = new long[(n << 2)];
build(1, 0, n - 1);
}
void build(int id, int start, int end) {
if (start == end)
tree[id] = arr[start];
else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
build(left, start, mid);
build(right, mid + 1, end);
tree[id] = tree[left] + tree[right];
}
}
void update(int l, int r, long val) {
update(1, 0, n - 1, l, r, val);
}
void update(int id, int start, int end, int l, int r, long val) {
distribute(id, start, end);
if (end < l || r < start)
return;
if (start == end)
tree[id] += val;
else if (l <= start && end <= r) {
lazy[id] += val;
distribute(id, start, end);
} else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
update(left, start, mid, l, r, val);
update(right, mid + 1, end, l, r, val);
tree[id] = tree[left] + tree[right];
}
}
long query(int l, int r) {
return query(1, 0, n - 1, l, r);
}
long query(int id, int start, int end, int l, int r) {
if (end < l || r < start)
return 0L;
distribute(id, start, end);
if (start == end)
return tree[id];
else if (l <= start && end <= r)
return tree[id];
else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
return query(left, start, mid, l, r) + query(right, mid + 1, end, l, r);
}
}
void distribute(int id, int start, int end) {
if (start == end)
tree[id] += lazy[id];
else {
tree[id] += lazy[id] * (end - start + 1);
lazy[(id << 1)] += lazy[id];
lazy[(id << 1) + 1] += lazy[id];
}
lazy[id] = 0;
}
}
// ==================== TRIE ================================
static class Trie {
class Node {
Node[] children;
boolean isEnd;
Node() {
children = new Node[26];
}
}
Node root;
Trie() {
root = new Node();
}
void insert(String word) {
Node curr = root;
for (char ch : word.toCharArray()) {
if (curr.children[ch - 'a'] == null)
curr.children[ch - 'a'] = new Node();
curr = curr.children[ch - 'a'];
}
curr.isEnd = true;
}
boolean find(String word) {
Node curr = root;
for (char ch : word.toCharArray()) {
if (curr.children[ch - 'a'] == null)
return false;
curr = curr.children[ch - 'a'];
}
return curr.isEnd;
}
}
// ==================== FENWICK TREE ================================
static class FT {
long[] tree;
int n;
FT(int[] arr, int n) {
this.n = n;
this.tree = new long[n + 1];
for (int i = 1; i <= n; i++) {
update(i, arr[i - 1]);
}
}
void update(int idx, int val) {
while (idx <= n) {
tree[idx] += val;
idx += idx & -idx;
}
}
long query(int l, int r) {
return getSum(r) - getSum(l - 1);
}
long getSum(int idx) {
long ans = 0L;
while (idx > 0) {
ans += tree[idx];
idx -= idx & -idx;
}
return ans;
}
}
// ==================== BINARY INDEX TREE ================================
static class BIT {
long[][] tree;
int n, m;
BIT(int[][] mat, int n, int m) {
this.n = n;
this.m = m;
tree = new long[n + 1][m + 1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
update(i, j, mat[i - 1][j - 1]);
}
}
}
void update(int x, int y, int val) {
while (x <= n) {
int t = y;
while (t <= m) {
tree[x][t] += val;
t += t & -t;
}
x += x & -x;
}
}
long query(int x1, int y1, int x2, int y2) {
return getSum(x2, y2) - getSum(x1 - 1, y2) - getSum(x2, y1 - 1) + getSum(x1 - 1, y1 - 1);
}
long getSum(int x, int y) {
long ans = 0L;
while (x > 0) {
int t = y;
while (t > 0) {
ans += tree[x][t];
t -= t & -t;
}
x -= x & -x;
}
return ans;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | fef7c5d2c27ae1b5ba22e5ec9ad131e1 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class test {
static void solve(int n, long[] a) {
int cnt = 0;
for (int i = 2; i < n + 1; i++) {
if (a[i] == a[i - 1]) {
cnt++;
}
}
if (cnt <= 1) {
System.out.println(0);
return;
}
else {
ArrayList<Integer> s = new ArrayList<>();
for (int i = 2; i < n + 1; i++) {
if (a[i] == a[i - 1]) {
s.add(i);
}
}
cnt = s.get(s.size() - 1) - s.get(0) - 1;
System.out.println(Math.max(cnt, 1));
}
}
public static void main(String[] args) {
Scanner w = new Scanner(System.in);
int t = w.nextInt();
for (int i = 0; i < t; i++) {
int n = w.nextInt();
long[] a = new long[n + 1];
for (int j = 1; j < n + 1; j++) {
a[j] = w.nextLong();
}
solve(n, a);
}
}
//------------------//
static int gcd(int a, int b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
static int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
static void increment(ArrayList<Long> a) {
for (int i = 0; i < a.size(); i++) {
long val = a.get(i);
val++;
a.set(i, val);
}
}
static void changeStatus(String a) {
for (int i = 0; i < a.length(); i++) {
if (a.charAt(i) == '0') {
a.replace(a.charAt(i), '1');
}
else {
a.replace(a.charAt(i), '0');
}
}
}
static long factorial(long n) {
long res = 1;
for (long i = 2; i <= n; i++) {
res *= i;
}
return res;
}
static long cGeneral(long n, long r) {
return factorial(n)/(factorial(r) * factorial(n-r));
}
static String reverse(String s) {
StringBuilder input1 = new StringBuilder();
input1.append(s);
input1.reverse();
return input1.toString();
}
static void reverseArray(String[] a) {
for (int i = 0; i < a.length; i++) {
a[i] = reverse(a[i]);
}
}
static boolean equalsArray(int[] a, int[] b) {
for (int i = 0; i < a.length; i++) {
if (a[i] != b[i]) {
return false;
}
}
return true;
}
static long sum(ArrayList<Long> a) {
long sum = 0;
for (int i = 0; i < a.size(); i++) {
sum += a.get(i);
}
return sum;
}
static long sum(long[] a) {
long sum = 0;
for (int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static long max(ArrayList<Long> a) {
long max = Long.MIN_VALUE;
if (a.size() == 0) {
max = 0;
}
for (int i = 0; i < a.size(); i++) {
if (a.get(i) >= max) {
max = a.get(i);
}
}
return max;
}
static long min(ArrayList<Long> a) {
long min = Long.MAX_VALUE;
if (a.size() == 0) {
min = 0;
}
for (int i = 0; i < a.size(); i++) {
if (a.get(i) <= min) {
min = a.get(i);
}
}
return min;
}
static long max(long[] a) {
long max = Long.MIN_VALUE;
if (a.length == 0) {
max = 0;
}
for (int i = 0; i < a.length; i++) {
if (a[i] >= max) {
max = a[i];
}
}
return max;
}
static long min(long[] a) {
long min = Long.MAX_VALUE;
if (a.length == 0) {
min = 0;
}
for (int i = 0; i < a.length; i++) {
if (a[i] <= min) {
min = a[i];
}
}
return min;
}
static boolean isPrime(long num)
{
if(num<=1)
{
return false;
}
for(long i=2;i<=num/2;i++)
{
if((num%i)==0)
return false;
}
return true;
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 696d8287170224123c113fa5ac72df76 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class cf1672C {
// https://codeforces.com/problemset/problem/1672/C
public static void main(String[] args) {
Kattio io = new Kattio();
int t = io.nextInt(), n;
int[] arr;
for (int i = 0; i < t; i++) {
n = io.nextInt();
arr = new int[n];
int eq = 0;
int last = -1;
for (int k = 0; k < n; k++) {
arr[k] = io.nextInt();
if (k > 0) {
if (arr[k] == last) eq ++;
}
last = arr[k];
}
int moves = 0;
if (eq > 1) {
for (int k = 0; k < n - 2; k++) {
if (arr[k] == arr[k+1]) {
moves ++;
if (k + 4 <= n && (arr[k+2] == arr[k+3] || arr[k+1] == arr[k+2])) {
if (arr[k] == arr[k+2] && arr[k] == arr[k+3]) eq -=2;
else eq --;
if (eq <= 1) break;
}
arr[k+1] = 0;
arr[k+2] = 0;
}
}
}
io.println(moves);
}
io.close();
}
// Kattio
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() { this(System.in,System.out); }
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(problemName+".out");
r = new BufferedReader(new FileReader(problemName+".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) {}
return null;
}
public String nextLine() {
try {
st = null;
return r.readLine();
} catch (Exception e) {}
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 5e1b78923ba321d1f50f811ea7f87c7b | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
//import java.util.*;
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-->0) {
Integer n = Integer.parseInt(br.readLine());
Integer[] a = Arrays.stream(br.readLine().split("\\s+")).map(Integer::parseInt).toArray(Integer[]::new);
int firstfist = Integer.MAX_VALUE; int lastlast = -1;
int times = 0;
for(int i=1;i<n;i++){
if(a[i].equals(a[i - 1])){
times++;
firstfist = Math.min(firstfist, i-1);
lastlast = Math.max(lastlast, i);
}
}
if(times<=1) System.out.println(0);
else System.out.println(Math.max(1, lastlast-firstfist-2));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | f4c83919317d5a608ff550f6c154cecf | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
public class UnequalArray {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
int first = 0;
int second = 0;
boolean flag = false;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
first = i + 1;
flag = true;
break;
}
}
for (int i = n - 1; i >= 1; i--) {
if (a[i] == a[i - 1]) {
second = i - 1;
flag = true;
break;
}
}
if (flag) {
if (second < first) {
System.out.println(0);
} else if (first == second) {
System.out.println(1);
} else {
System.out.println(second - first);
}
} else {
System.out.println(0);
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | f3b5bde8eab48d8e97ca685dd2f75ec0 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.awt.*;
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.StringTokenizer;
public class C {
{
MULTI_TEST = true;
FILE_NAME = "";
NEED_FILE_IO = false;
INF = (long) 1e18;
} // fields
void solve() {
int n = ri();
int[] a = ria(n);
ArrayList<Point> list = new ArrayList<>();
int sum = 0;
int l = 0, r = 0;
while(r < n) {
while(r + 1 < n && a[r] == a[r + 1]) r++;
if(r - l >= 1) {
list.add(new Point(l, r));
sum += r - l + 1;
}
l = r + 1;
r = l;
}
if(sum > 2) {
if(list.size() != 1) {
int firstPos = list.get(0).x + 1;
int lastPos = list.get(list.size() - 1).y - 1;
int size = lastPos - firstPos + 1;
out.println(size - 1);
}
else {
int size = list.get(0).y - list.get(0).x + 1;
out.println(size == 3 ? 1 : size - 3);
}
} else {
out.println(0);
}
}
public static void main(String[] args) {
new C().run();
} //main
@SuppressWarnings("unused")
long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
@SuppressWarnings("unused")
long lcm(long a, long b) {
return a / gcd(a, b) * b;
}
@SuppressWarnings("unused")
int gcd(int a, int b) {
return (int) gcd((long) a, b);
}
@SuppressWarnings("unused")
int lcm(int a, int b) {
return (int) lcm(a, (long) b);
}
@SuppressWarnings("unused")
int sqrtInt(int x) {
return (int) sqrtLong(x);
}
@SuppressWarnings("unused")
long sqrtLong(long x) {
long root = (long) Math.sqrt(x);
while (root * root > x) --root;
while ((root + 1) * (root + 1) <= x) ++root;
return root;
}
@SuppressWarnings("unused")
int cbrtLong(long x) {
int cbrt = (int) Math.cbrt(x);
while ((long) cbrt * cbrt >= x) {
cbrt--;
}
while ((long) (cbrt + 1) * (cbrt + 1) <= x) cbrt++;
return cbrt;
}
@SuppressWarnings("unused")
long binpow(long a, long power) {
return binpow(a, power, Long.MAX_VALUE);
}
@SuppressWarnings("unused")
long binpow(long a, long power, long modulo) {
if (power == 0) return 1 % modulo;
if (power % 2 == 1) {
long b = binpow(a, power - 1, modulo) % modulo;
return ((a % modulo) * b) % modulo;
} else {
long b = binpow(a, power / 2, modulo) % modulo;
return (b * b) % modulo;
}
}
@SuppressWarnings("unused")
long fastMod(String s1, long n2) {
long num = 0;
for (int i = 0; i < s1.length() - 1; i++) {
num += Integer.parseInt(String.valueOf(s1.charAt(i)));
num *= 10;
if (num >= n2) {
num = num % n2;
}
}
return (num + Integer.parseInt(String.valueOf(s1.charAt(s1.length() - 1)))) % n2;
}
@SuppressWarnings("unused")
long factorialMod(long n, long p) {
long res = 1;
while (n > 1) {
res = (res * ((n / p) % 2 == 1 ? p - 1 : 1)) % p;
for (int i = 2; i <= n % p; ++i)
res = (res * i) % p;
n /= p;
}
return res % p;
}
@SuppressWarnings("unused")
boolean isPrime(int number) {
for (int i = 2; i * i <= number; i++) {
if (number % i == 0) {
return false;
}
}
return number > 1;
}
@SuppressWarnings("unused")
boolean[] primes(int border) {
boolean[] isPrimes = new boolean[border + 1];
Arrays.fill(isPrimes, true);
isPrimes[0] = false;
isPrimes[1] = false;
for (int i = 2; i < border + 1; i++) {
if (!isPrimes[i]) continue;
for (int k = i * 2; k < border; k += i) {
isPrimes[k] = false;
}
}
return isPrimes;
} //Number theory
@SuppressWarnings("unused")
void sort(int[] a) {
int n = a.length;
Integer[] arr = new Integer[n];
for (int i = 0; i < n; i++) {
arr[i] = a[i];
}
Arrays.sort(arr);
for (int i = 0; i < n; i++) {
a[i] = arr[i];
}
}
@SuppressWarnings("unused")
void sort(long[] a) {
int n = a.length;
Long[] arr = new Long[n];
for (int i = 0; i < n; i++) {
arr[i] = a[i];
}
Arrays.sort(arr);
for (int i = 0; i < n; i++) {
a[i] = arr[i];
}
}
@SuppressWarnings("unused")
int max(int[] a) {
int n = a.length;
int max = Integer.MIN_VALUE;
for (int j : a) {
if (j > max) max = j;
}
return max;
}
@SuppressWarnings("unused")
long max(long[] a) {
int n = a.length;
long max = Long.MIN_VALUE;
for (long l : a) {
if (l > max) max = l;
}
return max;
}
@SuppressWarnings("unused")
int maxIndex(int[] a) {
int n = a.length;
int max = Integer.MIN_VALUE;
int index = 0;
for (int i = 0; i < n; i++) {
if (a[i] > max) {
max = a[i];
index = i;
}
}
return index;
}
@SuppressWarnings("unused")
int maxIndex(long[] a) {
int n = a.length;
long max = Long.MIN_VALUE;
int index = 0;
for (int i = 0; i < n; i++) {
if (a[i] > max) {
max = a[i];
index = i;
}
}
return index;
}
@SuppressWarnings("unused")
int min(int[] a) {
int n = a.length;
int min = Integer.MAX_VALUE;
for (int j : a) {
if (j < min) min = j;
}
return min;
}
@SuppressWarnings("unused")
int minIndex(int[] a) {
int n = a.length;
int min = Integer.MAX_VALUE;
int index = 0;
for (int i = 0; i < n; i++) {
if (a[i] < min) {
min = a[i];
index = i;
}
}
return index;
}
@SuppressWarnings("unused")
int minIndex(long[] a) {
int n = a.length;
long min = Long.MAX_VALUE;
int index = 0;
for (int i = 0; i < n; i++) {
if (a[i] < min) {
min = a[i];
index = i;
}
}
return index;
}
@SuppressWarnings("unused")
long min(long[] a) {
int n = a.length;
long min = Long.MAX_VALUE;
for (long l : a) {
if (l < min) min = l;
}
return min;
}
@SuppressWarnings("unused")
long sum(int[] a) {
int n = a.length;
long sum = 0;
for (int j : a) {
sum += j;
}
return sum;
}
@SuppressWarnings("unused")
long sum(long[] a) {
int n = a.length;
long sum = 0;
for (long l : a) {
sum += l;
}
return sum;
} //Arrays Operations
@SuppressWarnings("unused")
String readLine() {
try {
return in.readLine();
} catch (Exception e) {
throw new InputMismatchException();
}
}
@SuppressWarnings("unused")
String rs() {
while (!tok.hasMoreTokens()) {
tok = new StringTokenizer(readLine());
}
return tok.nextToken();
}
@SuppressWarnings("unused")
int ri() {
return Integer.parseInt(rs());
}
@SuppressWarnings("unused")
long rl() {
return Long.parseLong(rs());
}
@SuppressWarnings("unused")
double rd() {
return Double.parseDouble(rs());
}
@SuppressWarnings("unused")
int[] ria(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = ri();
}
return a;
}
@SuppressWarnings("unused")
int[] riawd(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = ri() - 1;
}
return a;
}
@SuppressWarnings("unused")
long[] rla(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = rl();
}
return a;
}
@SuppressWarnings("unused")
private boolean yesNo(boolean yes, String yesString, String noString) {
out.println(yes ? yesString : noString);
return yes;
} //fastIO
void run() {
try {
long start = System.currentTimeMillis();
initIO();
if (MULTI_TEST) {
long t = rl();
while (t-- > 0) {
solve();
}
} else {
solve();
}
out.close();
System.err.println("Time(ms) - " + (System.currentTimeMillis() - start));
} catch (Exception e) {
e.printStackTrace();
System.exit(-1);
}
}
void initConsoleIO() {
out = new PrintWriter(System.out);
in = new BufferedReader(new InputStreamReader(System.in));
}
void initFileIO(String inName, String outName) throws FileNotFoundException {
out = new PrintWriter(outName);
in = new BufferedReader(new FileReader(inName));
}
void initIO() throws FileNotFoundException {
if (!FILE_NAME.isEmpty()) {
initFileIO(FILE_NAME + ".in", FILE_NAME + ".out");
} else {
if (NEED_FILE_IO && new File("input.txt").exists()) {
initFileIO("input.txt", "output.txt");
} else {
initConsoleIO();
}
}
tok = new StringTokenizer("");
} //initIO
private final String FILE_NAME;
private final boolean MULTI_TEST;
private final boolean NEED_FILE_IO;
private final long INF;
BufferedReader in;
PrintWriter out;
StringTokenizer tok; //fields
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 27289055f428adfb3c32036dfeedc966 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
public class Codeforces {
static int max = (int) 1e5 + 5;
public static void main(String[] args) {
FastReader fastReader = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int tt = fastReader.nextInt();
while (tt-- > 0) {
int n = fastReader.nextInt();
int a[] = fastReader.ria(n);
int min = IMAX, max = 0;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
min = min(min, i);
max = max(max, i + 1);
}
}
if (min == IMAX) {
out.println(0);
continue;
} else {
int ans = max - min + 1;
if (ans == 3) {
out.println(1);
} else {
out.println(max(0, ans - 3));
}
}
}
out.close();
}
// constants
static final int IBIG = 1000000007;
static final int IMAX = 2147483647;
static final long LMAX = 9223372036854775807L;
static Random __r = new Random();
// math util
static int minof(int a, int b, int c) {
return min(a, min(b, c));
}
static int minof(int... x) {
if (x.length == 1)
return x[0];
if (x.length == 2)
return min(x[0], x[1]);
if (x.length == 3)
return min(x[0], min(x[1], x[2]));
int min = x[0];
for (int i = 1; i < x.length; ++i)
if (x[i] < min)
min = x[i];
return min;
}
static long minof(long a, long b, long c) {
return min(a, min(b, c));
}
static long minof(long... x) {
if (x.length == 1)
return x[0];
if (x.length == 2)
return min(x[0], x[1]);
if (x.length == 3)
return min(x[0], min(x[1], x[2]));
long min = x[0];
for (int i = 1; i < x.length; ++i)
if (x[i] < min)
min = x[i];
return min;
}
static int maxof(int a, int b, int c) {
return max(a, max(b, c));
}
static int maxof(int... x) {
if (x.length == 1)
return x[0];
if (x.length == 2)
return max(x[0], x[1]);
if (x.length == 3)
return max(x[0], max(x[1], x[2]));
int max = x[0];
for (int i = 1; i < x.length; ++i)
if (x[i] > max)
max = x[i];
return max;
}
static long maxof(long a, long b, long c) {
return max(a, max(b, c));
}
static long maxof(long... x) {
if (x.length == 1)
return x[0];
if (x.length == 2)
return max(x[0], x[1]);
if (x.length == 3)
return max(x[0], max(x[1], x[2]));
long max = x[0];
for (int i = 1; i < x.length; ++i)
if (x[i] > max)
max = x[i];
return max;
}
static int powi(int a, int b) {
if (a == 0)
return 0;
int ans = 1;
while (b > 0) {
if ((b & 1) > 0)
ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static long powl(long a, int b) {
if (a == 0)
return 0;
long ans = 1;
while (b > 0) {
if ((b & 1) > 0)
ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static int fli(double d) {
return (int) d;
}
static int cei(double d) {
return (int) ceil(d);
}
static long fll(double d) {
return (long) d;
}
static long cel(double d) {
return (long) ceil(d);
}
static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
static int[] exgcd(int a, int b) {
if (b == 0)
return new int[] { 1, 0 };
int[] y = exgcd(b, a % b);
return new int[] { y[1], y[0] - y[1] * (a / b) };
}
static long[] exgcd(long a, long b) {
if (b == 0)
return new long[] { 1, 0 };
long[] y = exgcd(b, a % b);
return new long[] { y[1], y[0] - y[1] * (a / b) };
}
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static long mix(long x) {
x += 0x9e3779b97f4a7c15L;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebL;
return x ^ (x >> 31);
}
public static boolean[] findPrimes(int limit) {
assert limit >= 2;
final boolean[] nonPrimes = new boolean[limit];
nonPrimes[0] = true;
nonPrimes[1] = true;
int sqrt = (int) Math.sqrt(limit);
for (int i = 2; i <= sqrt; i++) {
if (nonPrimes[i])
continue;
for (int j = i; j < limit; j += i) {
if (!nonPrimes[j] && i != j)
nonPrimes[j] = true;
}
}
return nonPrimes;
}
// array util
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i)
ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i)
ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i)
ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i)
ans[i] = a[i];
return ans;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] ria(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = Integer.parseInt(next());
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] rla(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = Long.parseLong(next());
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output |
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