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 | c8b7e28feb0621e163d6f444c51d0649 | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class A {
public static void main(String[] args) {
FastReader scan = new FastReader();
PrintWriter out = new PrintWriter(System.out);
Task solver = new Task();
int t = scan.nextInt();
for(int tt = 1; tt <= t; tt++) solver.solve(tt, scan, out);
out.close();
}
static class Task {
public void solve(int testNumber, FastReader scan, PrintWriter out) {
long a = scan.nextLong(), b = scan.nextLong();
if(b < a) a ^= b ^ (b = a);
if(a == 0 && b == 0) out.println(0);
else if(a == b) out.println(1);
else if((b - a) % 2 == 0) out.println(2);
else out.println(-1);
}
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws FileNotFoundException {
br = new BufferedReader(new FileReader(new File(s)));
}
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;
}
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | ea1fcf0ab092d227ab9add8a7eaa621a | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class A {
public static void main(String[]args){
MyScanner sc = new MyScanner();
int n = sc.nextInt();
for(int i=0;i<n;i++){
int c = sc.nextInt();
int d = sc.nextInt();
if(c==0&&d==0){
System.out.println(0);
}
else if(Math.abs(c)==Math.abs(d)){
System.out.println(1);
}
else if(Math.abs(c-d)%2==0){
System.out.println(2);
}
else{
System.out.println(-1);
}
}
}
}
class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
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;
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 813745a098597a63cb9d7c8275382acf | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.IOException;
import java.util.Scanner;
public class Main {
public static void main(String[] args) throws IOException {
new Main().FunctionCall();
}
void FunctionCall(){
Scanner scanner=new Scanner(System.in);
int t=scanner.nextInt();
while(t-->0){
int a=scanner.nextInt();
int b=scanner.nextInt();
solve(a, b);
}
}
void solve(int a, int b){
if(b>a){
solve(b , a);
return;
}
if(a==b && a!=0){
System.out.println(1);
return;
}
if(a==b && a==0){
System.out.println(0);
return;
}
int diff=a-b;
if(diff%2!=0){
System.out.println(-1);
return;
}
System.out.println(2);
}
}
| Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 70f610274f4851ff06dc49c3499f25fd | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for (int i=0; i<t; i++) {
String[] ab = br.readLine().split(" ");
int a = Integer.parseInt(ab[0]);
int b = Integer.parseInt(ab[1]);
if (Math.abs(a-b) % 2 == 1) {
System.out.println(-1);
continue;
}
if (a == 0 && b == 0) {
System.out.println(0);
continue;
}
if (a==b) {
System.out.println(1);
continue;
}
System.out.println(2);
}
}
}
| Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 5328b89c9a5f0f6512f477287a7b8779 | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.time.LocalTime;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
public class A {
public static void main(String[] args) {
FastScanner scan = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int T = 1;
T = scan.nextInt();
for(int tt=0; tt<T; tt++) {
int c = scan.nextInt(), d = scan.nextInt();
int diff = Math.abs(c - d);
if(c == 0 && diff == 0) System.out.println(0);
else if(diff % 2 != 0) System.out.println(-1);
else {
System.out.println(diff == 0 ? 1 : 2);
}
}
}
/*
static class IntComparator implements Comparator<Integer> {
public int compare(Integer o1, Integer o2) {
// return Integer.compare(a[o1], a[o2]);
return 0;
}
}
*/
public static void printIntArray(int [] a) {
for(int i: a) System.out.print(i + " ");
System.out.println();
}
public static void swapInt(int a, int b) {
int temp = a;
a = b;
b = temp;
}
public static void sort(int [] a) {
ArrayList<Integer> b = new ArrayList<>();
for(int i: a) b.add(i);
Collections.sort(b);
for(int i=0; i<a.length; i++) a[i]= b.get(i);
}
public static void reverse(int [] a) {
ArrayList<Integer> b = new ArrayList<>();
for(int i: a) b.add(i);
Collections.reverse(b);
for(int i=0; i<a.length; i++) a[i]= b.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\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 4c89c8159c87951e5609f441fcbabfaf | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | //package lovejava;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class LoveJava {
static int sum1 = 0;
static int sum2 = 0;
public static void main(String[] args) {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
String s;
String arr[];
String arr2[];
try {
int t=Integer.parseInt(br.readLine());
while(t-->0) {
// int n = Integer.parseInt(br.readLine());
arr = (br.readLine().replaceAll("//s+$", " ").split(" "));
/*
int myScore[] = new int[arr.length];
arr2 = (br.readLine().replaceAll("//s+$", " ").split(" "));
int[] IllScore = new int[arr2.length];
*/
int c=Integer.parseInt(arr[0]);
int d=Integer.parseInt(arr[1]);
if(((int)Math.abs(d-c)&1)==1)//odd
pw.println(-1);
else {
if(c==d&c==0)
pw.println(0);
else if(c==d)
pw.println(1);
else pw.println(2);
}
}
pw.close();
} catch (IOException ex) {
System.out.println("input mismatch");
}
}
public static String balancedSums(List<Integer> arr) {
// Write your code here
// int pow=(int)Math.pow(10,5)+1;
if(arr.size()==1)
return "YES";
long[]prefixSum=new long[arr.size()];
long[]suffixSum=new long[arr.size()];
long sum=0l;
for(int i=0;i<arr.size();i++){
sum+=arr.get(i);
prefixSum[i]=sum;
}
sum=0l;
for(int i=arr.size()-1;i>=0;i--){
sum+=arr.get(i);
suffixSum[i]=sum;
}
for(int i=0;i<arr.size();i++){
if(i-1>=0&&i+1<arr.size()&&prefixSum[i-1]==suffixSum[i+1]){
return "YES";
}
else if((i>=0&&i<arr.size())&&((i==0&&suffixSum[i+1]==0)||(i==arr.size()-1&&prefixSum[i-1]==0)))
return "YES";
}
return "NO";
}
public static void solution(String[]arr){
int n = Integer.parseInt(arr[0]);
int k = Integer.parseInt(arr[1]);
StringBuilder stb = new StringBuilder(arr[0]);
int[] array = new int[10];
Arrays.fill(array, -1);
int count = 0;
for (int i = 0; i < stb.length(); i++) {
array[Integer.parseInt(String.valueOf(stb.charAt(i)))] = Integer.parseInt(String.valueOf(stb.charAt(i)));
}
for (int i = 0; i < array.length; i++) {
if (array[i] != -1)
count++;
}
if (count <= k)
System.out.println(n);
else {
Arrays.sort(array);
int i;
boolean flag;
for (i = 0; i < array.length; i++)
if (array[i] == -1)
;//pointer at digits
if (k == 1) {
flag = true;
while (flag) {
for (int j = 0; j < stb.length(); j++)
stb.setCharAt(j, (char) array[i]);
if (Integer.parseInt(stb.toString()) > n) {
System.out.println((Integer.parseInt(stb.toString())));
flag=false;
}
else i++;
}
}
else {
char newchar = '\u0000';
int l = 0;
int pOfDiff;
char ch = stb.charAt(0);
for (int j = 1; j < stb.length(); j++) {
if (stb.charAt(j) == ch) ;
else {
newchar = stb.charAt(j);
l = j;
while (j < stb.length()) {
stb.setCharAt(j, newchar);
}
}
}
if (Integer.parseInt(stb.toString()) < n) {
flag=true;
while (flag) {
for (; l < stb.length(); l++)
stb.setCharAt(l, (char) array[i]);
if (Integer.parseInt(stb.toString()) > n) {
System.out.println((Integer.parseInt(stb.toString())));
flag = false;
} else i++;
}
}
}
}
}
public static int fun(int[]myScore,int[]illyScore){
Arrays.sort(myScore);
Arrays.sort(illyScore);
int k=myScore.length-(myScore.length/4);
int mySum=0;
int illSum=0;
for(int i=1;i<=k;i++){
mySum+=myScore[myScore.length-i];
}
for(int i=1;i<=k;i++){
illSum+=illyScore[illyScore.length-i];
}
if(mySum>=illSum)
return 0;
else {
int l=(int) Math.ceil((illSum-mySum)/(double)100);
int h=(int)Math.pow(10,5)*5;
while (h-l>1){
int mid=l+(h-l)/2;
if(binarysearch(mid,myScore,illyScore)==true)
h=mid;
else l=mid+1;
}
if(binarysearch(l,myScore,illyScore)==true)
return l-myScore.length;
else return h-myScore.length;
}
}
public static boolean binarysearch(int chosenNum,int[]myScore,int[]ilyScore){
int x=chosenNum-myScore.length;
int t=chosenNum-chosenNum/4;
int mySum=0;
int illysum=0;
if(x>=t)
mySum=t*100;
else {
mySum = x * 100;
for(int i=1;i<=t-x;i++){
mySum+= myScore[myScore.length-i];
}
}
if(t>=ilyScore.length){
for(int i=0;i<ilyScore.length;i++){
illysum+= ilyScore[i];
}
}
else {
for(int i=1;i<=t;i++){
illysum+= ilyScore[ilyScore.length-i];
}
}
if(mySum>=illysum)
return true;
else return false;
}
public static void func2(int r,int c) {
int[][] board = new int[r][c];
int[] arrR = {-1, -1, -1, 0, 1, 1, 1, 0};
int[] arrC = {-1, 0, 1, 1, 1, 0, -1, -1};
for (int[] e : board) {
for (int i : e) {
i = 0;
}
}
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (i == 0 || i == r - 1 || j == 0 || j == c - 1) {
if (boolFun(board, arrR, arrC, i, j, r, c) == false)
board[i][j] = 1;
}
}
}
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
System.out.print(board[i][j]);
}
System.out.println();
}
System.out.println();
}
public static boolean boolFun(int[][] board,int[]r,int[]c,int i,int j,int rmax,int cmax){
for(int k=0;k<r.length;k++){
if(i+r[k]<rmax&&i+r[k]>=0&&j+c[k]<cmax&&j+c[k]>=0)//row
if(board[i+r[k]][j+c[k]]==1)
return true;
}
return false;
}
public static int func(int n){
if(n==0)
return 0;
StringBuilder stb=new StringBuilder(String.valueOf(n));
for(int i=0;i<stb.length();i++){
if(stb.charAt(i)=='0'||stb.charAt(i)=='1')
;
else {
while (i<stb.length()){
stb.setCharAt(i,'1');
i++;
}
}
}
return 1+func(n-Integer.parseInt(stb.toString()));
}
}
/*
class Solution {
public int maxSubArray(int[] nums) {
}
}
*/
class Rectangle2{
int len=1;
int bre=1;
Rectangle2(int len,int bre){
this.len=len;
this.bre=bre;
}
}
class Solution {
public long nthStair(int n) {
// Code here
long[] dp = new long[n + 1];
Arrays.fill(dp, 0l);
dp[0] = 1l;
for (int i = 1; i <= n; i++) {
dp[i] += dp[i - 1];
}
for (int i = 2; i <= n; i++) {
dp[i] += dp[i - 2];
}
return dp[n];
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | fa45e631fa0a85b37cd4c72b929b4b9a | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.util.Arrays;
public class codeforces {
static long[]fac=new long[200100];
static long[] two= new long[200100] ;
static long mod=((int)1e9)+9;
static String[]pow=new String[63];
static int[][]perm;
static int[]a,pe;
static int x=0;
static int n,r;
public static void main(String[] args) throws IOException{
int t=sc.nextInt();
while(t-->0) {
int c=sc.nextInt();
int d=sc.nextInt();
int diff=Math.abs(c-d);
if(diff%2!=0)pw.println(-1);
else {
if(diff==0&&c==0)pw.println(0);
else if(diff==0)pw.println(1);
else {
pw.println(2);
}
}
}
pw.close();
}
public static boolean reachable(double x1,double y1,double x2,double y2) {
double m=(y2-y1)/(x2-x1);
double c=y1-x1*m;
double a=m*m+1;
double b=2*m*c;
c=c*c-r*r;
double sol1=(-b+Math.sqrt(b*b-4*a*c))/(2*a);
double sol2=(-b-Math.sqrt(b*b-4*a*c))/(2*a);
return Math.abs(x2-x1)-Math.min(Math.abs(sol1-x1), Math.abs(sol2-x1))<0.00000001;
}
public static boolean isplus1(int n) {
return (n-1)%6==0;
}
public static void permutation(int idx,int v) {
if(v==(1<<n)-1) {
perm[x++]=pe.clone();
return ;
}
for (int i = 0; i < n; i++) {
if((v&1<<i)==0) {
pe[idx]=a[i];
permutation(idx+1, v|1<<i);
}
}
return ;
}
public static void pre2() {
for (int i = 0; i < pow.length; i++) {
long x=1l<<i;
pow[i]=x+"";
}
}
public static void sort(int[]a) {
mergesort(a, 0, a.length-1);
}
public static void sortIdx(long[]a,long[]idx) {
mergesortidx(a, idx, 0, a.length-1);
}
public static long C(int a,int b) {
long x=fac[a];
long y=fac[a-b]*fac[b];
return x*pow2(y,mod-2)%mod;
}
public static long pow2(long a,long b) {
long ans=1;a%=mod;
for(long i=b;i>0;i/=2) {
if((i&1)!=0)
ans=ans*a%mod;
a=a*a%mod;
}
return ans;
}
public static void pre(){
fac[0]=1;
for (int i = 1; i < fac.length; i++) {
fac[i]=fac[i-1]*i%mod;
}
two[0]=1;
for (int i = 1; i < fac.length; i++) {
two[i]=two[i-1]*2%mod;
}
}
static class Node{
int x;
Node next;
Node previous;
public Node(int x) {
this.x=x;
next=null;
previous=null;
}
@Override
public String toString() {
return x+"";
}
}
static class LinkedL{
Node head;
Node last;
public LinkedL() {
head=last=null;
}
public void insertLast(int k) {
Node n=new Node(k);
n.previous=last;
last.next=n;
last=n;
}
public void insertFirst(int k) {
if(head==null) {
head=last=new Node(k);
}else {
Node n=new Node(k);
n.next=head;
head.previous=n;
head=n;
}
}
@Override
public String toString() {
String s="";
Node c=head;
while(c!=null) {
s+=c+" ";
c=c.next;
}
return s;
}
}
public static long p(int n,int k) {
long res=1;
if(k<=n/2) {
while(k-->0) {
res*=n;
n--;
}
}else {
res=fac(n)/fac(n-k);
}
return res;
}
public static long eval(String s) {
long p=1;
long res=0;
for (int i = 0; i < s.length(); i++) {
res+=p*(s.charAt(s.length()-1-i)=='1'?1:0);
p*=2;
}
return res;
}
public static String binary(long x) {
String s="";
while(x!=0) {
s=(x%2)+s;
x/=2;
}
return s;
}
public static boolean allSame(String s) {
char x=s.charAt(0);
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i)!=x)return false;
}
return true;
}
public static boolean isPalindrom(String s) {
int l=0;
int r=s.length()-1;
while(l<r) {
if(s.charAt(r--)!=s.charAt(l++))return false;
}
return true;
}
public static String putAtFront(String s,char c) {
if(s.length()==0)return s;
if(s.charAt(s.length()-1)==c) {
return s.charAt(s.length()-1)+putAtFront(s.substring(0,s.length()-1), c);
}else {
return putAtFront(s.substring(0,s.length()-1), c)+s.charAt(s.length()-1);
}
}
public static boolean isSubString(String s,String t) {
int ls=s.length();
int lt=t.length();
boolean res=false;
for (int i = 0; i <=lt-ls; i++) {
if(t.substring(i, i+ls).equals(s)) {
res=true;
break;
}
}
return res;
}
public static boolean isSorted(int[]a) {
for (int i = 0; i < a.length-1; i++) {
if(a[i]>a[i+1])return false;
}
return true;
}
public 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;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
public static long power(long x,long k) {
long res=1;
long mod=((int)1e9)+7;
for (int i = 0; i < k; i++) {
res*=x;
res=res%mod;
}
return res;
}
public static int whichPower(int x) {
int res=0;
for (int j = 0; j < 31; j++) {
if((1<<j&x)!=0) {
res=j;
break;
}
}
return res;
}
public static long evaln(String x,int n) {
long res=0;
for (int i = 0; i < x.length(); i++) {
res+=Long.parseLong(x.charAt(x.length()-1-i)+"")*Math.pow(n, i);
}
return res;
}
static void merge(int[] arr,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
int[] l=new int[len1];
int[] r=new int[len2];
for(int i=0;i<len1;i++)l[i]=arr[b+i];
for(int i=0;i<len2;i++)r[i]=arr[m+1+i];
int i=0,j=0,k=b;
while(i<len1 && j<len2) {
if(l[i]<r[j])arr[k++]=l[i++];
else arr[k++]=r[j++];
}
while(i<len1)arr[k++]=l[i++];
while(j<len2)arr[k++]=r[j++];
return;
}
static void mergesortidx(long[] arr,long[]idx,int b,int e) {
if(b<e) {
int m=b+(e-b)/2;
mergesortidx(arr,idx,b,m);
mergesortidx(arr,idx,m+1,e);
mergeidx(arr,idx,b,m,e);
}
return;
}
static void mergeidx(long[] arr,long[]idx,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
long[] l=new long[len1];
long[] lidx=new long[len1];
long[] r=new long[len2];
long[] ridx=new long[len2];
for(int i=0;i<len1;i++) {
l[i]=arr[b+i];
lidx[i]=idx[b+i];
}
for(int i=0;i<len2;i++) {
r[i]=arr[m+1+i];
ridx[i]=idx[m+1+i];
}
int i=0,j=0,k=b;
while(i<len1 && j<len2) {
if(l[i]<r[j]) {
arr[k++]=l[i++];
idx[k-1]=lidx[i-1];
}
else {
arr[k++]=r[j++];
idx[k-1]=ridx[j-1];
}
}
while(i<len1) {
idx[k]=lidx[i];
arr[k++]=l[i++];
}
while(j<len2) {
idx[k]=ridx[j];
arr[k++]=r[j++];
}
return;
}
static void mergesort(int[] arr,int b,int e) {
if(b<e) {
int m=b+(e-b)/2;
mergesort(arr,b,m);
mergesort(arr,m+1,e);
merge(arr,b,m,e);
}
return;
}
static long mergen(int[] arr,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
int[] l=new int[len1];
int[] r=new int[len2];
for(int i=0;i<len1;i++)l[i]=arr[b+i];
for(int i=0;i<len2;i++)r[i]=arr[m+1+i];
int i=0,j=0,k=b;
long c=0;
while(i<len1 && j<len2) {
if(l[i]<r[j])arr[k++]=l[i++];
else {
arr[k++]=r[j++];
c=c+(long)(len1-i);
}
}
while(i<len1)arr[k++]=l[i++];
while(j<len2)arr[k++]=r[j++];
return c;
}
static long mergesortn(int[] arr,int b,int e) {
long c=0;
if(b<e) {
int m=b+(e-b)/2;
c=c+(long)mergesortn(arr,b,m);
c=c+(long)mergesortn(arr,m+1,e);
c=c+(long)mergen(arr,b,m,e);
}
return c;
}
public static long fac(int n) {
if(n==0)return 1;
return n*fac(n-1);
}
public static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static long summ(long x) {
long sum=0;
while(x!=0) {
sum+=x%10;
x=x/10;
}
return sum;
}
public static ArrayList<Integer> findDivisors(int n){
ArrayList<Integer>res=new ArrayList<Integer>();
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
// If divisors are equal, print only one
if (n/i == i)
res.add(i);
else {
res.add(i);
res.add(n/i);
}
}
}
return res;
}
public static void sort2darray(Integer[][]a){
Arrays.sort(a,Comparator.<Integer[]>comparingInt(x -> x[0]).thenComparingInt(x -> x[1]));
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(String file) throws FileNotFoundException {
br = new BufferedReader(new FileReader(file));
}
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 {
return Double.parseDouble(next());
}
public boolean ready() throws IOException {
return br.ready();
}
public int[] nextArrint(int size) throws IOException {
int[] a=new int[size];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextInt();
}
return a;
}
public long[] nextArrlong(int size) throws IOException {
long[] a=new long[size];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextLong();
}
return a;
}
public int[][] next2dArrint(int rows,int columns) throws IOException{
int[][]a=new int[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j]=sc.nextInt();
}
}
return a;
}
public long[][] next2dArrlong(int rows,int columns) throws IOException{
long[][]a=new long[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j]=sc.nextLong();
}
}
return a;
}
}
static class Side{
Point a;
Point b;
public Side(Point a,Point b) {
this.a=a;
this.b=b;
}
@Override
public boolean equals(Object obj) {
Side s=(Side)obj;
return (s.a.equals(a)&&s.b.equals(b))||(s.b.equals(a)&&s.a.equals(b));
}
@Override
public String toString() {
return "("+a.toString()+","+b.toString()+")";
}
}
static class Point{
int x;
int y;
int z;
public Point(int x,int y,int z) {
this.x=x;
this.y=y;
this.z=z;
}
@Override
public boolean equals(Object obj) {
Point p=(Point)obj;
return x==p.x&&y==p.y&&z==p.z;
}
@Override
public String toString() {
return "("+x+","+y+","+z+")";
}
}
static class Pair{
long x;
long y;
public Pair(long x,long y) {
this.x=x;
this.y=y;
}
}
static Scanner sc=new Scanner(System.in);
static PrintWriter pw=new PrintWriter(System.out);
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | e5601810dcbbb1eb22eed96d3ceac8ce | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
public class Practice {
public static long mod = (long) Math.pow(10, 9) + 7;
// public static long mod2 = 998244353;
// public static int tt = 1;
public static ArrayList<Long> one;
public static ArrayList<Long> two;
// public static long[] fac = new long[200005];
// public static long[] invfac = new long[200005];
public static void main(String[] args) throws Exception {
PrintWriter pw = new PrintWriter(System.out);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
int p = 1;
while (t-- > 0) {
String[] s1 = br.readLine().split(" ");
int c = Integer.valueOf(s1[0]);
int d = Integer.valueOf(s1[1]);
int x = Math.abs(c - d);
if (c == 0 && d == 0) {
pw.println(0);
} else {
if (x == 0) {
pw.println(1);
} else if (x % 2 == 0) {
pw.println(2);
} else {
pw.println(-1);
}
}
}
pw.close();
}
//
// private static long power(long a, long p) {
// // TODO Auto-generated method stub
// long res = 1;
// while (p > 0) {
// if (p % 2 == 1) {
// res = (res * a) % mod;
// }
// p = p / 2;
// a = (a * a) % mod;
// }
// return res;
// }
// private static int kmp(String str) {
// // TODO Auto-generated method stub
// // System.out.println(str);
// int[] pi = new int[str.length()];
// pi[0] = 0;
// for (int i = 1; i < str.length(); i++) {
// int j = pi[i - 1];
// while (j > 0 && str.charAt(i) != str.charAt(j)) {
// j = pi[j - 1];
// }
// if (str.charAt(j) == str.charAt(i)) {
// j++;
// }
// pi[i] = j;
// System.out.print(pi[i]);
// }
// System.out.println();
// return pi[str.length() - 1];
// }
}
// private static void getFac(long n, PrintWriter pw) {
// // TODO Auto-generated method stub
// int a = 0;
// while (n % 2 == 0) {
// a++;
// n = n / 2;
// }
// if (n == 1) {
// a--;
// }
// for (int i = 3; i <= Math.sqrt(n); i += 2) {
// while (n % i == 0) {
// n = n / i;
// a++;
// }
// }
// if (n > 1) {
// a++;
// }
// if (a % 2 == 0) {
// pw.println("Bob");
// } else {
// pw.println("Alice");
// }
// //System.out.println(a);
// return;
// }
// private static long power(long a, long p) {
// // TODO Auto-generated method stub
// long res = 1;
// while (p > 0) {
// if (p % 2 == 1) {
// res = (res * a) % mod;
// }
// p = p / 2;
// a = (a * a) % mod;
// }
// return res;
// }
//
// private static void fac() {
// fac[0] = 1;
// // TODO Auto-generated method stub
// for (int i = 1; i < fac.length; i++) {
// if (i == 1) {
// fac[i] = 1;
// } else {
// fac[i] = i * fac[i - 1];
// }
// if (fac[i] > mod) {
// fac[i] = fac[i] % mod;
// }
// }
// }
//
// private static int getLower(Long long1, Long[] st) {
// // TODO Auto-generated method stub
// int left = 0, right = st.length - 1;
// int ans = -1;
// while (left <= right) {
// int mid = (left + right) / 2;
// if (st[mid] <= long1) {
// ans = mid;
// left = mid + 1;
// } else {
// right = mid - 1;
// }
// }
// return ans;
// }
//private static long getGCD(long l, long m) {
//
// long t1 = Math.min(l, m);
// long t2 = Math.max(l, m);
// if (t1 == 0) {
// return t2;
// }
// while (true) {
// long temp = t2 % t1;
// if (temp == 0) {
// return t1;
// }
// t2 = t1;
// t1 = temp;
// }
//} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 9a4b83aa7a4f4e01145109c89310510c | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | //package codeforces;
import java.util.*;
import java.lang.*;
import java.io.*;
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 ni() {
return Integer.parseInt(next());
}
long nl() {
return Long.parseLong(next());
}
double nd() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
// MAIN FUNCTION
public static void main(String[] args) throws java.lang.Exception {
FastReader fr = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = fr.ni();
while(t-->0) {
int a = fr.ni();
int b = fr.ni();
int ans = -1;
if((Math.abs(a-b)&1) == 0) {
if(a == 0 && b == 0)ans = 0;
else if(a == b)ans = 1;
else ans = 2;
}
out.println(ans);
}
out.close();
}
}
| Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | d1925c81972aad109f790d2745650cd7 | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.util.*;
public class Leetcode {
public static void main(String[] args) {
int a;
int b;
int c;
Scanner scanner=new Scanner(System.in);
c=scanner.nextInt();
while(c-->0){
a= scanner.nextInt();
b=scanner.nextInt();
if(a==b&&a==0)
System.out.println(0);
else if(a==b)
System.out.println(1);
else if(Math.abs(a-b)%2==0){
System.out.println(2);
}else{
System.out.println(-1);
}
}
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 619e18a86784766ab080bb1d3f966556 | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes |
import java.util.*;
public class Main {
public static void main(String args[] ) throws Exception {
Scanner sc = new Scanner(System.in);
try {
int t = sc.nextInt();
while((t--)!=0){
int a=sc.nextInt();
int b=sc.nextInt();
if(a==b && a!=0 && b!=0) System.out.println("1");
else if(a==b && b==0) System.out.println("0");
else if((a%2==1 && b%2==0) || (a%2==0 && b%2==1)) System.out.println("-1");
else System.out.println("2");
}
} catch(Exception e) {
}
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 6b92a8b931df57b241eba8e406223c3b | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Sol {
final static int LINE_SIZE = 128;
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out), false);
void solve() throws Exception {
long c = in.nextInt();
long d = in.nextInt();
long dif = Math.abs(Math.abs(c) - Math.abs(d));
if (dif % 2 == 1) {
out.println(-1);
return;
}
if (c == 0 && d == 0) {
out.println(0);
return;
}
if (Math.abs(c) == Math.abs(d)) {
out.println(1);
return;
}
if (c == 0 || d == 0) {
out.println(2);
return;
}
out.println(2);
}
void runCases() throws Exception {
int tt = 1;
tt = in.nextInt();
while (tt-- > 0) {
solve();
}
in.close();
out.close();
}
public static void main(String[] args) throws Exception {
Sol solver = new Sol();
solver.runCases();
}
static class Pii implements Comparable<Pii> {
public int fi, se;
public Pii(int fi, int se) {
this.fi = fi;
this.se = se;
}
public int compareTo(Pii other) {
if (fi == other.fi) return se - other.se;
return fi - other.fi;
}
}
static class Pll implements Comparable<Pll> {
public long fi, se;
public Pll(long fi, long se) {
this.fi = fi;
this.se = se;
}
public int compareTo(Pll other) {
if (fi == other.fi) return se < other.se ? -1 : 1;
return fi < other.fi ? -1 : 1;
}
}
static class Pair<T extends Comparable<T>, U extends Comparable<U>> implements Comparable<Pair<T, U>> {
T fi;
U se;
public Pair(T fi, U se) {
this.fi = fi;
this.se = se;
}
public int compareTo(Pair<T, U> other) {
if (fi == other.fi) return se.compareTo(other.se);
return fi.compareTo(other.fi);
}
}
static class FastReader {
private final DataInputStream din = new DataInputStream(System.in);
private final byte[] buffer = new byte[65536];
private int bufferPointer = 0, bytesRead = 0;
public String readLine() throws Exception {
byte[] buf = new byte[LINE_SIZE];
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 Exception {
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');
return neg ? -ret : ret;
}
public long nextLong() throws Exception {
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');
return neg ? -ret : ret;
}
public double nextDouble() throws Exception {
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);
return neg ? -ret : ret;
}
private void fillBuffer() throws Exception {
bytesRead = din.read(buffer, bufferPointer = 0, 65536);
if (bytesRead == -1) buffer[0] = -1;
}
private byte read() throws Exception {
if (bufferPointer == bytesRead) fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws Exception {
din.close();
}
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 50083139414a674d5ae4f5723b567657 | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class _practise {
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());
}
int[] ia(int n)
{
int a[]=new int[n];
for(int i=0;i<n;i++)a[i]=nextInt();
return a;
}
int[][] ia(int n , int m)
{
int a[][]=new int[n][m];
for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextInt();
return a;
}
long[][] la(int n , int m)
{
long a[][]=new long[n][m];
for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextLong();
return a;
}
char[][] ca(int n , int m)
{
char a[][]=new char[n][m];
for(int i=0;i<n;i++)
{
String x =next();
for(int j=0;j<m ;j++) a[i][j]=x.charAt(j);
}
return a;
}
long[] la(int n)
{
long a[]=new long[n];
for(int i=0;i<n;i++)a[i]=nextLong();
return a;
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static void sort(long[] a)
{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {long oi=r.nextInt(n), temp=a[i];a[i]=a[(int)oi];a[(int)oi]=temp;}Arrays.sort(a);}
static void sort(int[] a)
{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {int oi=r.nextInt(n), temp=a[i];a[i]=a[oi];a[oi]=temp;}Arrays.sort(a);}
public static long sum(long a[])
{long sum=0; for(long i : a) sum+=i; return(sum);}
public static long count(long a[] , long x)
{long c=0; for(long i : a) if(i==x) c++; return(c);}
public static int sum(int a[])
{ int sum=0; for(int i : a) sum+=i; return(sum);}
public static int count(int a[] ,int x)
{int c=0; for(int i : a) if(i==x) c++; return(c);}
public static int count(String s ,char ch)
{int c=0; char x[] = s.toCharArray(); for(char i : x) if(ch==i) c++; return(c);}
public static boolean prime(int n)
{for(int i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
public static boolean prime(long n)
{for(long i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
public static int gcd(int n1, int n2)
{ if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
public static long gcd(long n1, long n2)
{ if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
public static int[] freq(int a[], int n)
{ int f[]=new int[n+1]; for(int i:a) f[i]++; return f;}
public static int[] pos(int a[], int n)
{ int f[]=new int[n+1]; for(int i=0; i<n ;i++) f[a[i]]=i; return f;}
public static long mindig(long n)
{
long ans=9; while(n>0) { if(n%10<ans) ans=n%10; n/=10; }
return ans;
}
public static long maxdig(long n)
{
long ans=0; while(n>0) { if(n%10>ans) ans=n%10; n/=10; }
return ans;
}
public static boolean palin(String s)
{
StringBuilder sb = new StringBuilder();
sb.append(s);
String str=String.valueOf(sb.reverse());
if(s.equals(str))
return true;
else return false;
}
public static void main(String args[])
{
FastReader in=new FastReader();
PrintWriter so = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
_practise ob = new _practise();
int T = in.nextInt();
//int T = 1;
tc : while(T-->0)
{
long c = in.nextLong();
long d = in.nextLong();
if((c+d)%2!=0) so.println(-1);
else
{
if(c==d)
{
if(c==0) so.println(0);
else so.println(1);
}
else so.println(2);
}
}
so.flush();
/*String s = in.next();
* Arrays.stream(f).min().getAsInt()
* BigInteger f = new BigInteger("1"); 1 ke jagah koi bhi value ho skta jo aap
* initial value banan chahte
int a[] = new int[n];
Stack<Integer> stack = new Stack<Integer>();
Deque<Integer> q = new LinkedList<>(); or Deque<Integer> q = new ArrayDeque<Integer>();
PriorityQueue<Long> pq = new PriorityQueue<Long>();
ArrayList<Integer> al = new ArrayList<Integer>();
StringBuilder sb = new StringBuilder();
HashSet<Integer> st = new LinkedHashSet<Integer>();
Set<Integer> s = new HashSet<Integer>();
Map<Long,Integer> hm = new HashMap<Long, Integer>(); //<key,value>
for(Map.Entry<Integer, Integer> i :hm.entrySet())
HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
so.println("HELLO");
Arrays.sort(a,Comparator.comparingDouble(o->o[0]))
Arrays.sort(a, (aa, bb) -> Integer.compare(aa[1], bb[1]));
Set<String> ts = new TreeSet<>();*/
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | c30b5f694b7bb2cefda64eab5808fc83 | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.util.*;
public class HelloWorld{
public static void main(String []args){
Scanner sc = new Scanner(System.in);
int test =sc.nextInt();
int a=0;
int b=0;
for(int i=0;i<test;i++)
{
int c = sc.nextInt();
int d = sc.nextInt();
a=c+d;
if(c==0 && d==0)
{
System.out.println(0);
}
else if(Math.abs(c-d)%2==1)
{
System.out.println(-1);
}
else if(Math.abs(a)/2==c)
{
System.out.println(1);
}
else
{
System.out.println(2);
}
}
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 41053fdb914896fcfff27076f12d44fd | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
final static FastScanner scan = new FastScanner();
static void test() {
int a = scan.nextInt();
int b = scan.nextInt();
if(Math.abs(a - b) % 2 == 1) {
System.out.println(-1);
}
else {
if(a == b) {
System.out.println(a == 0 ? 0 : 1);
}
else System.out.println(2);
}
}
public static void main(String args[]) {
int T = 1;
T = scan.nextInt();
for(int tc = 0; tc < T; ++ tc) {
test();
}
}
}
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());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | b4f328af15110ed2a25a42999659db4b | train_107.jsonl | 1630247700 | William has two numbers $$$a$$$ and $$$b$$$ initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer $$$k$$$ is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer $$$k$$$) add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or add number $$$k$$$ to $$$b$$$ and subtract $$$k$$$ from $$$a$$$. Note that after performing operations, numbers $$$a$$$ and $$$b$$$ may become negative as well.William wants to find out the minimal number of operations he would have to perform to make $$$a$$$ equal to his favorite number $$$c$$$ and $$$b$$$ equal to his second favorite number $$$d$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) throws IOException
{
FastScanner f= new FastScanner();
int ttt=1;
ttt=f.nextInt();
PrintWriter out=new PrintWriter(System.out);
outer: for(int tt=0;tt<ttt;tt++) {
int a=f.nextInt();
int b=f.nextInt();
if(a==0 && b==0) System.out.println(0);
else if (a==b) System.out.println(1);
else if(Math.abs(a-b)%2==0) System.out.println(2);
else System.out.println(-1);
}
out.close();
}
static void sort(int[] p) {
ArrayList<Integer> q = new ArrayList<>();
for (int i: p) q.add( i);
Collections.sort(q);
for (int i = 0; i < p.length; i++) p[i] = q.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());
}
double nextDouble() {
return Double.parseDouble(next());
}
long[] readLongArray(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
} | Java | ["6\n1 2\n3 5\n5 3\n6 6\n8 0\n0 0"] | 1 second | ["-1\n2\n2\n1\n2\n0"] | NoteLet us demonstrate one of the suboptimal ways of getting a pair $$$(3, 5)$$$: Using an operation of the first type with $$$k=1$$$, the current pair would be equal to $$$(1, 1)$$$. Using an operation of the third type with $$$k=8$$$, the current pair would be equal to $$$(-7, 9)$$$. Using an operation of the second type with $$$k=7$$$, the current pair would be equal to $$$(0, 2)$$$. Using an operation of the first type with $$$k=3$$$, the current pair would be equal to $$$(3, 5)$$$. | Java 8 | standard input | [
"math"
] | 8e7c0b703155dd9b90cda706d22525c9 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into. | 800 | For each test case output a single number, which is the minimal number of operations which William would have to perform to make $$$a$$$ equal to $$$c$$$ and $$$b$$$ equal to $$$d$$$, or $$$-1$$$ if it is impossible to achieve this using the described operations. | standard output | |
PASSED | 862b2dc4f8fe28ca3ea84b07cadd7942 | train_107.jsonl | 1630247700 | William really likes puzzle kits. For one of his birthdays, his friends gifted him a complete undirected edge-weighted graph consisting of $$$n$$$ vertices.He wants to build a spanning tree of this graph, such that for the first $$$k$$$ vertices the following condition is satisfied: the degree of a vertex with index $$$i$$$ does not exceed $$$d_i$$$. Vertices from $$$k + 1$$$ to $$$n$$$ may have any degree.William wants you to find the minimum weight of a spanning tree that satisfies all the conditions.A spanning tree is a subset of edges of a graph that forms a tree on all $$$n$$$ vertices of the graph. The weight of a spanning tree is defined as the sum of weights of all the edges included in a spanning tree. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.ArrayList;
import java.io.OutputStream;
import java.io.PrintStream;
import java.io.IOException;
import java.lang.reflect.Field;
import java.nio.charset.StandardCharsets;
import java.io.UncheckedIOException;
import java.util.function.Consumer;
import java.util.List;
import java.io.Closeable;
import java.util.Comparator;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) throws Exception {
Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 29);
thread.start();
thread.join();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastInput in = new FastInput(inputStream);
FastOutput out = new FastOutput(outputStream);
HDIYTree solver = new HDIYTree();
solver.solve(1, in, out);
out.close();
}
}
static class HDIYTree {
Debug debug = new Debug(true);
public void solve(int testNumber, FastInput in, FastOutput out) {
int n = in.ri();
int k = in.ri();
int[] d = in.ri(k);
int n2 = n * (n - 1) / 2;
int ck = k * (k - 1) / 2;
IntegerArrayList usList = new IntegerArrayList(n2);
IntegerArrayList vsList = new IntegerArrayList(n2);
LongArrayList wsList = new LongArrayList(n2);
IntegerArrayList edgeList = new IntegerArrayList(ck);
List<Edge> outside = new ArrayList<>(n2);
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (i >= k && j >= k) {
Edge e = new Edge(i, j, in.ri());
outside.add(e);
} else {
usList.add(i);
vsList.add(j);
wsList.add(in.ri());
}
if (i < k && j < k) {
edgeList.add(usList.size() - 1);
}
}
}
DSU dsu = new DSU(n);
dsu.init();
outside.sort(Comparator.comparingInt(x -> x.w));
debug.debug("size", usList.size());
for (Edge e : outside) {
if (dsu.find(e.a) == dsu.find(e.b)) {
continue;
}
dsu.merge(e.a, e.b);
usList.add(e.a);
vsList.add(e.b);
wsList.add(e.w);
}
int[] us = usList.toArray();
int[] vs = vsList.toArray();
long[] ws = wsList.toArray();
int[] edges = edgeList.toArray();
int m = edges.length;
long inf = (long) 1e9;
long[] wsSnapshot = ws.clone();
int[] type = new int[us.length];
for (int i = 0; i < us.length; i++) {
if (us[i] < k && vs[i] >= k) {
type[i] = us[i];
} else if (vs[i] < k && us[i] >= k) {
type[i] = vs[i];
} else {
type[i] = k;
}
}
debug.debug("size", us.length);
long best = inf;
for (int i = 0; i < 1 << m; i++) {
boolean ok = true;
dsu.init();
System.arraycopy(wsSnapshot, 0, ws, 0, ws.length);
int[] cap = new int[k + 1];
System.arraycopy(d, 0, cap, 0, k);
cap[k] = n;
for (int j = 0; j < m; j++) {
int e = edges[j];
if (Bits.get(i, j) == 0) {
ws[e] += inf;
} else {
int u = us[e];
int v = vs[e];
ws[e] -= inf;
cap[u]--;
cap[v]--;
if (dsu.find(u) == dsu.find(v)) {
ok = false;
}
dsu.merge(u, v);
}
}
for (int j = 0; j < ws.length; j++) {
ws[j] = -ws[j];
}
for (int j = 0; j <= k; j++) {
if (cap[j] < 0) {
ok = false;
}
}
if (!ok) {
continue;
}
MatroidIndependentSet container = MatroidIndependentSet.ofColorContainers(type, cap);
MatroidIndependentSet tree = MatroidIndependentSet.ofSpanningTree(n, new int[][]{us, vs});
MaximumWeightMatroidIntersect mi = new MaximumWeightMatroidIntersect(us.length, ws);
boolean[] sol = mi.intersect(container, tree);
long sum = 0;
int occur = 0;
for (int j = 0; j < sol.length; j++) {
if (sol[j]) {
occur++;
sum += ws[j];
}
}
sum = -sum;
sum += Integer.bitCount(i) * inf;
if (sum >= inf || occur < n - 1) {
continue;
}
best = Math.min(best, sum);
}
out.println(best);
}
}
static class LongArrayList implements Cloneable {
private int size;
private int cap;
private long[] data;
private static final long[] EMPTY = new long[0];
public LongArrayList(int cap) {
this.cap = cap;
if (cap == 0) {
data = EMPTY;
} else {
data = new long[cap];
}
}
public LongArrayList(long[] data) {
this(0);
addAll(data);
}
public LongArrayList(LongArrayList list) {
this.size = list.size;
this.cap = list.cap;
this.data = Arrays.copyOf(list.data, size);
}
public LongArrayList() {
this(0);
}
public void ensureSpace(int req) {
if (req > cap) {
while (cap < req) {
cap = Math.max(cap + 10, 2 * cap);
}
data = Arrays.copyOf(data, cap);
}
}
public void add(long x) {
ensureSpace(size + 1);
data[size++] = x;
}
public void addAll(long[] x) {
addAll(x, 0, x.length);
}
public void addAll(long[] x, int offset, int len) {
ensureSpace(size + len);
System.arraycopy(x, offset, data, size, len);
size += len;
}
public void addAll(LongArrayList list) {
addAll(list.data, 0, list.size);
}
public long[] toArray() {
return Arrays.copyOf(data, size);
}
public String toString() {
return Arrays.toString(toArray());
}
public boolean equals(Object obj) {
if (!(obj instanceof LongArrayList)) {
return false;
}
LongArrayList other = (LongArrayList) obj;
return SequenceUtils.equal(data, 0, size - 1, other.data, 0, other.size - 1);
}
public int hashCode() {
int h = 1;
for (int i = 0; i < size; i++) {
h = h * 31 + Long.hashCode(data[i]);
}
return h;
}
public LongArrayList clone() {
LongArrayList ans = new LongArrayList();
ans.addAll(this);
return ans;
}
}
static interface IntegerDeque extends IntegerStack {
}
static interface IntegerStack {
}
static class IntegerMultiWayStack {
private int[] values;
private int[] next;
private int[] heads;
private int alloc;
private int stackNum;
public IntegerIterator iterator(final int queue) {
return new IntegerIterator() {
int ele = heads[queue];
public boolean hasNext() {
return ele != 0;
}
public int next() {
int ans = values[ele];
ele = next[ele];
return ans;
}
};
}
private void doubleCapacity() {
int newSize = Math.max(next.length + 10, next.length * 2);
next = Arrays.copyOf(next, newSize);
values = Arrays.copyOf(values, newSize);
}
public void alloc() {
alloc++;
if (alloc >= next.length) {
doubleCapacity();
}
next[alloc] = 0;
}
public void clear() {
alloc = 0;
Arrays.fill(heads, 0, stackNum, 0);
}
public boolean isEmpty(int qId) {
return heads[qId] == 0;
}
public IntegerMultiWayStack(int qNum, int totalCapacity) {
values = new int[totalCapacity + 1];
next = new int[totalCapacity + 1];
heads = new int[qNum];
stackNum = qNum;
}
public void addLast(int qId, int x) {
alloc();
values[alloc] = x;
next[alloc] = heads[qId];
heads[qId] = alloc;
}
public String toString() {
StringBuilder builder = new StringBuilder();
for (int i = 0; i < stackNum; i++) {
if (isEmpty(i)) {
continue;
}
builder.append(i).append(": ");
for (IntegerIterator iterator = iterator(i); iterator.hasNext(); ) {
builder.append(iterator.next()).append(",");
}
if (builder.charAt(builder.length() - 1) == ',') {
builder.setLength(builder.length() - 1);
}
builder.append('\n');
}
return builder.toString();
}
}
static class MatroidIntersect {
protected IntegerDequeImpl dq;
protected int[] dists;
protected boolean[] added;
protected boolean[][] adj1;
protected boolean[][] adj2;
protected int n;
protected boolean[] x1;
protected boolean[] x2;
protected static int distInf = (int) 1e9;
protected int[] pre;
protected Consumer<boolean[]> callback;
protected static Consumer<boolean[]> nilCallback = x -> {
};
public void setCallback(Consumer<boolean[]> callback) {
if (callback == null) {
callback = nilCallback;
}
this.callback = callback;
}
public MatroidIntersect(int n) {
this.n = n;
dq = new IntegerDequeImpl(n);
dists = new int[n];
added = new boolean[n];
adj1 = new boolean[n][];
adj2 = new boolean[n][];
x1 = new boolean[n];
x2 = new boolean[n];
pre = new int[n];
setCallback(nilCallback);
}
protected boolean adj(int i, int j) {
if (added[i]) {
return adj1[i][j];
} else {
return adj2[j][i];
}
}
protected boolean expand(MatroidIndependentSet a, MatroidIndependentSet b, int round) {
Arrays.fill(x1, false);
Arrays.fill(x2, false);
a.prepare(added);
b.prepare(added);
a.extend(added, x1);
b.extend(added, x2);
for (int i = 0; i < n; i++) {
if (x1[i] && x2[i]) {
pre[i] = -1;
xorPath(i);
return true;
}
}
for (int i = 0; i < n; i++) {
if (added[i]) {
Arrays.fill(adj1[i], false);
Arrays.fill(adj2[i], false);
}
}
a.computeAdj(added, adj1);
b.computeAdj(added, adj2);
Arrays.fill(dists, distInf);
Arrays.fill(pre, -1);
dq.clear();
for (int i = 0; i < n; i++) {
if (added[i]) {
continue;
}
//right
if (x1[i]) {
dists[i] = 0;
dq.addLast(i);
}
}
int tail = -1;
while (!dq.isEmpty()) {
int head = dq.removeFirst();
if (x2[head]) {
tail = head;
break;
}
for (int j = 0; j < n; j++) {
if (added[head] != added[j] && adj(head, j) && dists[j] > dists[head] + 1) {
dists[j] = dists[head] + 1;
dq.addLast(j);
pre[j] = head;
}
}
}
if (tail == -1) {
return false;
}
xorPath(tail);
return true;
}
protected void xorPath(int tail) {
boolean[] last1 = new boolean[n];
boolean[] last2 = new boolean[n];
for (boolean add = true; tail != -1; tail = pre[tail], add = !add) {
assert added[tail] != add;
added[tail] = add;
if (add) {
adj1[tail] = last1;
adj2[tail] = last2;
} else {
last1 = adj1[tail];
last2 = adj2[tail];
adj1[tail] = null;
adj2[tail] = null;
}
}
}
public boolean[] intersect(MatroidIndependentSet a, MatroidIndependentSet b) {
Arrays.fill(added, false);
int round = 0;
callback.accept(added);
while (expand(a, b, round)) {
round++;
callback.accept(added);
}
return added;
}
}
static class Edge {
int a;
int b;
int w;
public Edge(int a, int b, int w) {
this.a = a;
this.b = b;
this.w = w;
}
}
static class Bits {
private Bits() {
}
public static int get(int x, int i) {
return (x >>> i) & 1;
}
}
static interface IntegerIterator {
boolean hasNext();
int next();
}
static class DSU {
protected int[] p;
public int[] size;
protected int n;
public DSU(int n) {
p = new int[n];
size = new int[n];
}
public void init() {
init(p.length);
}
public void init(int n) {
this.n = n;
for (int i = 0; i < n; i++) {
p[i] = i;
size[i] = 1;
}
}
public final int find(int a) {
if (p[a] == p[p[a]]) {
return p[a];
}
find(p[a]);
preAccess(a);
return p[a] = p[p[a]];
}
protected void preAccess(int a) {
}
protected void preMerge(int a, int b) {
size[a] += size[b];
}
public final void merge(int a, int b) {
a = find(a);
b = find(b);
if (a == b) {
return;
}
if (size[a] < size[b]) {
int tmp = a;
a = b;
b = tmp;
}
preMerge(a, b);
p[b] = a;
}
public String toString() {
return Arrays.toString(Arrays.copyOf(p, n));
}
}
static class FastOutput implements AutoCloseable, Closeable, Appendable {
private static final int THRESHOLD = 32 << 10;
private OutputStream writer;
private StringBuilder cache = new StringBuilder(THRESHOLD * 2);
private static Field stringBuilderValueField;
private char[] charBuf = new char[THRESHOLD * 2];
private byte[] byteBuf = new byte[THRESHOLD * 2];
static {
try {
stringBuilderValueField = StringBuilder.class.getSuperclass().getDeclaredField("value");
stringBuilderValueField.setAccessible(true);
} catch (Exception e) {
stringBuilderValueField = null;
}
stringBuilderValueField = null;
}
public FastOutput append(CharSequence csq) {
cache.append(csq);
return this;
}
public FastOutput append(CharSequence csq, int start, int end) {
cache.append(csq, start, end);
return this;
}
private void afterWrite() {
if (cache.length() < THRESHOLD) {
return;
}
flush();
}
public FastOutput(OutputStream writer) {
this.writer = writer;
}
public FastOutput append(char c) {
cache.append(c);
afterWrite();
return this;
}
public FastOutput append(long c) {
cache.append(c);
afterWrite();
return this;
}
public FastOutput println(long c) {
return append(c).println();
}
public FastOutput println() {
return append('\n');
}
public FastOutput flush() {
try {
if (stringBuilderValueField != null) {
try {
byte[] value = (byte[]) stringBuilderValueField.get(cache);
writer.write(value, 0, cache.length());
} catch (Exception e) {
stringBuilderValueField = null;
}
}
if (stringBuilderValueField == null) {
int n = cache.length();
if (n > byteBuf.length) {
//slow
writer.write(cache.toString().getBytes(StandardCharsets.UTF_8));
// writer.append(cache);
} else {
cache.getChars(0, n, charBuf, 0);
for (int i = 0; i < n; i++) {
byteBuf[i] = (byte) charBuf[i];
}
writer.write(byteBuf, 0, n);
}
}
writer.flush();
cache.setLength(0);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
return this;
}
public void close() {
flush();
try {
writer.close();
} catch (IOException e) {
throw new UncheckedIOException(e);
}
}
public String toString() {
return cache.toString();
}
}
static class IntegerArrayList implements Cloneable {
private int size;
private int cap;
private int[] data;
private static final int[] EMPTY = new int[0];
public int[] getData() {
return data;
}
public IntegerArrayList(int cap) {
this.cap = cap;
if (cap == 0) {
data = EMPTY;
} else {
data = new int[cap];
}
}
public IntegerArrayList(int[] data) {
this(0);
addAll(data);
}
public IntegerArrayList(IntegerArrayList list) {
this.size = list.size;
this.cap = list.cap;
this.data = Arrays.copyOf(list.data, size);
}
public IntegerArrayList() {
this(0);
}
public void ensureSpace(int req) {
if (req > cap) {
while (cap < req) {
cap = Math.max(cap + 10, 2 * cap);
}
data = Arrays.copyOf(data, cap);
}
}
public void add(int x) {
ensureSpace(size + 1);
data[size++] = x;
}
public void addAll(int[] x) {
addAll(x, 0, x.length);
}
public void addAll(int[] x, int offset, int len) {
ensureSpace(size + len);
System.arraycopy(x, offset, data, size, len);
size += len;
}
public void addAll(IntegerArrayList list) {
addAll(list.data, 0, list.size);
}
public int size() {
return size;
}
public int[] toArray() {
return Arrays.copyOf(data, size);
}
public void clear() {
size = 0;
}
public String toString() {
return Arrays.toString(toArray());
}
public boolean equals(Object obj) {
if (!(obj instanceof IntegerArrayList)) {
return false;
}
IntegerArrayList other = (IntegerArrayList) obj;
return SequenceUtils.equal(data, 0, size - 1, other.data, 0, other.size - 1);
}
public int hashCode() {
int h = 1;
for (int i = 0; i < size; i++) {
h = h * 31 + Integer.hashCode(data[i]);
}
return h;
}
public IntegerArrayList clone() {
IntegerArrayList ans = new IntegerArrayList();
ans.addAll(this);
return ans;
}
}
static class SequenceUtils {
public static boolean equal(int[] a, int al, int ar, int[] b, int bl, int br) {
if ((ar - al) != (br - bl)) {
return false;
}
for (int i = al, j = bl; i <= ar; i++, j++) {
if (a[i] != b[j]) {
return false;
}
}
return true;
}
public static boolean equal(long[] a, int al, int ar, long[] b, int bl, int br) {
if ((ar - al) != (br - bl)) {
return false;
}
for (int i = al, j = bl; i <= ar; i++, j++) {
if (a[i] != b[j]) {
return false;
}
}
return true;
}
}
static class MaximumWeightMatroidIntersect extends MatroidIntersect {
protected long[] weight;
protected long[] pathWeight;
protected static final long weightInf = (long) 1e18;
protected boolean[] inq;
protected long[] fixWeight;
public MaximumWeightMatroidIntersect(int n, long[] weight) {
super(n);
this.weight = weight;
pathWeight = new long[n];
inq = new boolean[n];
fixWeight = new long[n];
}
protected boolean expand(MatroidIndependentSet a, MatroidIndependentSet b, int round) {
a.prepare(added);
b.prepare(added);
Arrays.fill(x1, false);
Arrays.fill(x2, false);
a.extend(added, x1);
b.extend(added, x2);
for (int i = 0; i < n; i++) {
if (added[i]) {
Arrays.fill(adj1[i], false);
Arrays.fill(adj2[i], false);
fixWeight[i] = weight[i];
} else {
fixWeight[i] = -weight[i];
}
}
a.computeAdj(added, adj1);
b.computeAdj(added, adj2);
Arrays.fill(dists, MatroidIntersect.distInf);
Arrays.fill(pathWeight, weightInf);
Arrays.fill(pre, -1);
dq.clear();
for (int i = 0; i < n; i++) {
if (added[i]) {
continue;
}
//right
if (x1[i]) {
dists[i] = 0;
pathWeight[i] = fixWeight[i];
dq.addLast(i);
inq[i] = true;
}
}
while (!dq.isEmpty()) {
int head = dq.removeFirst();
inq[head] = false;
for (int j = 0; j < n; j++) {
if (added[head] != added[j] && adj(head, j)) {
int comp = Long.compare(pathWeight[j], pathWeight[head] + fixWeight[j]);
if (comp == 0) {
comp = Integer.compare(dists[j], dists[head] + 1);
}
if (comp <= 0) {
continue;
}
dists[j] = dists[head] + 1;
pathWeight[j] = pathWeight[head] + fixWeight[j];
pre[j] = head;
if (!inq[j]) {
inq[j] = true;
dq.addLast(j);
}
}
}
}
int tail = -1;
for (int i = 0; i < n; i++) {
if (!x2[i] || !x1[i] && pre[i] == -1) {
continue;
}
if (tail == -1 || pathWeight[i] < pathWeight[tail] || pathWeight[i] == pathWeight[tail] &&
dists[i] < dists[tail]) {
tail = i;
}
}
if (tail == -1) {
return false;
}
xorPath(tail);
return true;
}
}
static class IntegerDequeImpl implements IntegerDeque {
private int[] data;
private int bpos;
private int epos;
private static final int[] EMPTY = new int[0];
private int n;
public IntegerDequeImpl(int cap) {
if (cap == 0) {
data = EMPTY;
} else {
data = new int[cap];
}
bpos = 0;
epos = 0;
n = cap;
}
private void expandSpace(int len) {
while (n < len) {
n = Math.max(n + 10, n * 2);
}
int[] newData = new int[n];
if (bpos <= epos) {
if (bpos < epos) {
System.arraycopy(data, bpos, newData, 0, epos - bpos);
}
} else {
System.arraycopy(data, bpos, newData, 0, data.length - bpos);
System.arraycopy(data, 0, newData, data.length - bpos, epos);
}
epos = size();
bpos = 0;
data = newData;
}
public IntegerIterator iterator() {
return new IntegerIterator() {
int index = bpos;
public boolean hasNext() {
return index != epos;
}
public int next() {
int ans = data[index];
index = IntegerDequeImpl.this.next(index);
return ans;
}
};
}
public int removeFirst() {
int ans = data[bpos];
bpos = next(bpos);
return ans;
}
public void addLast(int x) {
ensureMore();
data[epos] = x;
epos = next(epos);
}
public void clear() {
bpos = epos = 0;
}
private int next(int x) {
return x + 1 >= n ? 0 : x + 1;
}
private void ensureMore() {
if (next(epos) == bpos) {
expandSpace(n + 1);
}
}
public int size() {
int ans = epos - bpos;
if (ans < 0) {
ans += data.length;
}
return ans;
}
public boolean isEmpty() {
return bpos == epos;
}
public String toString() {
StringBuilder builder = new StringBuilder();
for (IntegerIterator iterator = iterator(); iterator.hasNext(); ) {
builder.append(iterator.next()).append(' ');
}
return builder.toString();
}
}
static class FastInput {
private final InputStream is;
private byte[] buf = new byte[1 << 13];
private int bufLen;
private int bufOffset;
private int next;
public FastInput(InputStream is) {
this.is = is;
}
public void populate(int[] data) {
for (int i = 0; i < data.length; i++) {
data[i] = readInt();
}
}
private int read() {
while (bufLen == bufOffset) {
bufOffset = 0;
try {
bufLen = is.read(buf);
} catch (IOException e) {
bufLen = -1;
}
if (bufLen == -1) {
return -1;
}
}
return buf[bufOffset++];
}
public void skipBlank() {
while (next >= 0 && next <= 32) {
next = read();
}
}
public int ri() {
return readInt();
}
public int[] ri(int n) {
int[] ans = new int[n];
populate(ans);
return ans;
}
public int readInt() {
boolean rev = false;
skipBlank();
if (next == '+' || next == '-') {
rev = next == '-';
next = read();
}
int val = 0;
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
return rev ? val : -val;
}
}
static class Debug {
private boolean offline;
private PrintStream out = System.err;
public Debug(boolean enable) {
offline = enable && System.getSecurityManager() == null;
}
public Debug debug(String name, int x) {
if (offline) {
debug(name, "" + x);
}
return this;
}
public Debug debug(String name, String x) {
if (offline) {
out.printf("%s=%s", name, x);
out.println();
}
return this;
}
}
static interface MatroidIndependentSet {
void computeAdj(boolean[] added, boolean[][] adj);
void extend(boolean[] added, boolean[] extendable);
void prepare(boolean[] added);
static MatroidIndependentSet ofColorContainers(int[] type, int[] cap) {
int[] size = new int[cap.length];
return new MatroidIndependentSet() {
/**
* O(rn)
* @param added
* @param adj
*/
public void computeAdj(boolean[] added, boolean[][] adj) {
for (int i = 0; i < added.length; i++) {
if (!added[i]) {
continue;
}
for (int j = 0; j < added.length; j++) {
if (added[j]) {
continue;
}
if (size[type[j]] < cap[type[j]] ||
type[i] == type[j]) {
adj[i][j] = true;
}
}
}
}
/**
* O(n)
* @param added
* @param extendable
*/
public void extend(boolean[] added, boolean[] extendable) {
for (int i = 0; i < added.length; i++) {
if (!added[i] && size[type[i]] < cap[type[i]]) {
extendable[i] = true;
}
}
}
public void prepare(boolean[] added) {
Arrays.fill(size, 0);
for (int i = 0; i < added.length; i++) {
if (added[i]) {
size[type[i]]++;
}
}
}
};
}
static MatroidIndependentSet ofSpanningTree(int n, int[][] edges) {
return new MatroidIndependentSet() {
DSU dsu = new DSU(n);
IntegerMultiWayStack g = new IntegerMultiWayStack(n, edges[0].length);
IntegerArrayList inset = new IntegerArrayList(edges[0].length);
int[] p = new int[n];
int[] depth = new int[n];
public void dfs(int root, int fa, int d) {
p[root] = fa;
depth[root] = d;
for (IntegerIterator iterator = g.iterator(root); iterator.hasNext(); ) {
int e = iterator.next();
if (e == fa) {
continue;
}
dfs(opponent(e, root), e, d + 1);
}
}
public int opponent(int i, int root) {
return edges[0][i] == root ? edges[1][i] : edges[0][i];
}
/**
* O(rn)
*/
public void computeAdj(boolean[] added, boolean[][] adj) {
int[] insetData = inset.getData();
int m = inset.size();
for (int i = 0; i < added.length; i++) {
if (added[i]) {
continue;
}
if (dsu.find(edges[0][i]) != dsu.find(edges[1][i])) {
for (int j = 0; j < m; j++) {
adj[insetData[j]][i] = true;
}
} else {
int a = edges[0][i];
int b = edges[1][i];
while (a != b) {
if (depth[a] < depth[b]) {
int tmp = a;
a = b;
b = tmp;
}
adj[p[a]][i] = true;
a = opponent(p[a], a);
}
}
}
}
/**
* O(n)
*/
public void extend(boolean[] added, boolean[] extendable) {
for (int i = 0; i < added.length; i++) {
if (!added[i]) {
extendable[i] = dsu.find(edges[0][i]) != dsu.find(edges[1][i]);
}
}
}
/**
* O(r+n)
*/
public void prepare(boolean[] added) {
g.clear();
dsu.init();
inset.clear();
for (int i = 0; i < added.length; i++) {
if (added[i]) {
dsu.merge(edges[0][i], edges[1][i]);
g.addLast(edges[0][i], i);
g.addLast(edges[1][i], i);
inset.add(i);
}
}
Arrays.fill(p, -1);
for (int i = 0; i < n; i++) {
if (p[i] == -1) {
dfs(i, -1, 0);
}
}
}
};
}
}
}
| Java | ["10 5\n5 3 4 2 1\n29 49 33 12 55 15 32 62 37\n61 26 15 58 15 22 8 58\n37 16 9 39 20 14 58\n10 15 40 3 19 55\n53 13 37 44 52\n23 59 58 4\n69 80 29\n89 28\n48"] | 6 seconds | ["95"] | null | Java 11 | standard input | [
"graphs",
"greedy",
"math",
"probabilities"
] | d870600853fa87cce3e0b4a985bcc5c8 | The first line of input contains two integers $$$n$$$, $$$k$$$ ($$$2 \leq n \leq 50$$$, $$$1 \leq k \leq min(n - 1, 5)$$$). The second line contains $$$k$$$ integers $$$d_1, d_2, \ldots, d_k$$$ ($$$1 \leq d_i \leq n$$$). The $$$i$$$-th of the next $$$n - 1$$$ lines contains $$$n - i$$$ integers $$$w_{i,i+1}, w_{i,i+2}, \ldots, w_{i,n}$$$ ($$$1 \leq w_{i,j} \leq 100$$$): weights of edges $$$(i,i+1),(i,i+2),\ldots,(i,n)$$$. | 3,300 | Print one integer: the minimum weight of a spanning tree under given degree constraints for the first $$$k$$$ vertices. | standard output | |
PASSED | 7740066b3ed4d6dc5e2d0f9474b0bae1 | train_107.jsonl | 1630247700 | As mentioned previously William really likes playing video games. In one of his favorite games, the player character is in a universe where every planet is designated by a binary number from $$$0$$$ to $$$2^n - 1$$$. On each planet, there are gates that allow the player to move from planet $$$i$$$ to planet $$$j$$$ if the binary representations of $$$i$$$ and $$$j$$$ differ in exactly one bit.William wants to test you and see how you can handle processing the following queries in this game universe: Destroy planets with numbers from $$$l$$$ to $$$r$$$ inclusively. These planets cannot be moved to anymore. Figure out if it is possible to reach planet $$$b$$$ from planet $$$a$$$ using some number of planetary gates. It is guaranteed that the planets $$$a$$$ and $$$b$$$ are not destroyed. | 1024 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.ArrayList;
import java.io.OutputStream;
import java.io.PrintStream;
import java.io.IOException;
import java.lang.reflect.Field;
import java.nio.charset.StandardCharsets;
import java.io.UncheckedIOException;
import java.util.List;
import java.io.Closeable;
import java.util.Comparator;
import java.util.Collections;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) throws Exception {
new TaskAdapter().run();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastInput in = new FastInput(inputStream);
FastOutput out = new FastOutput(outputStream);
GGatesToAnotherWorld solver = new GGatesToAnotherWorld();
solver.solve(1, in, out);
out.close();
}
}
static class GGatesToAnotherWorld {
Debug debug = new Debug(false);
DSU dsu;
List<Node> nodeList;
public void solve(int testNumber, FastInput in, FastOutput out) {
int n = in.ri();
int m = in.ri();
List<Query> qs = new ArrayList<>(m);
List<Query> updates = new ArrayList<>(m);
for (int i = 0; i < m; i++) {
int t = in.rs().equals("block") ? 0 : 1;
Query q = new Query(t, in.rl(), in.rl());
qs.add(q);
if (q.type == 0) {
updates.add(q);
}
}
debug.debug("updates", updates);
List<Query> revQs = new ArrayList<>(qs);
Collections.reverse(revQs);
debug.debug("revQs", revQs);
nodeList = new ArrayList<>(m * 2);
updates.sort(Comparator.comparingLong(x -> x.l));
long last = 0;
for (Query q : updates) {
long l = last;
long r = q.l - 1;
addNode(l, r, false);
addNode(q.l, q.r, true);
last = q.r + 1;
}
addNode(last, (1L << n) - 1, false);
debug.elapse("createNode");
Node[] nodes = nodeList.toArray(new Node[0]);
debug.debug("nodes", nodes);
for (int i = 0; i < nodes.length; i++) {
nodes[i].id = i;
}
assert SortUtils.notStrictAscending(nodes, 0, nodes.length - 1, Comparator.comparingLong(x -> x.l));
dac(nodes, 0, nodes.length - 1, 0, (1L << n) - 1, n - 1);
dsu = new DSU(nodes.length);
dsu.init();
for (Node root : nodes) {
recalc(root);
}
debug.elapse("build");
// search(nodes, nodes[0], nodes[12]);
// for(Node cur = nodes[12]; cur != null; cur = cur.prev){
// debug.debug("cur", cur);
// }
for (Query q : revQs) {
Node a = search(nodes, q.l);
Node b = search(nodes, q.r);
if (q.type == 0) {
assert a.del;
assert b.del;
a.del = false;
b.del = false;
recalc(a);
recalc(b);
} else {
assert !a.del;
assert !b.del;
q.ans = dsu.find(a.id) == dsu.find(b.id);
}
}
debug.elapse("answer");
for (Query q : qs) {
if (q.type == 1) {
out.println(q.ans ? 1 : 0);
}
}
}
public boolean intersect(long l, long r, long L, long R) {
return l <= R && r >= L;
}
public boolean checkInRange(Node[] nodes, int l, int r, long L, long R) {
for (int i = l; i <= r; i++) {
if (nodes[i].l < L || nodes[i].r > R) {
return false;
}
}
return true;
}
public void dac(Node[] nodes, int l, int r, long L, long R, int D) {
assert checkInRange(nodes, l, r, L, R);
if (l >= r) {
return;
}
if (D < 0) {
assert l == r;
}
long M = (L + R) / 2;
int ml = l - 1;
while (ml + 1 <= r && nodes[ml + 1].r <= M) {
ml++;
}
int mr = r + 1;
while (mr - 1 >= l && nodes[mr - 1].l > M) {
mr--;
}
dac(nodes, l, ml, L, M, D - 1);
dac(nodes, mr, r, M + 1, R, D - 1);
if (ml + 1 < mr) {
for (int j = ml + 1; j < mr; j++) {
for (int i = l; i <= ml; i++) {
addEdge(nodes[i], nodes[j]);
}
for (int i = mr; i <= r; i++) {
addEdge(nodes[i], nodes[j]);
}
}
} else {
int lIter = l;
int rIter = mr;
long mask = (1L << D) - 1;
while (lIter <= ml && rIter <= r) {
int comp = Long.compare(nodes[lIter].r & mask, nodes[rIter].r & mask);
if (intersect(nodes[lIter].l & mask, nodes[lIter].r & mask,
nodes[rIter].l & mask, nodes[rIter].r & mask)) {
addEdge(nodes[lIter], nodes[rIter]);
}
if (comp <= 0) {
lIter++;
}
if (comp >= 0) {
rIter++;
}
}
}
}
public void recalc(Node root) {
if (root.del) {
return;
}
for (Node node : root.adj) {
if (node.del) {
continue;
}
dsu.merge(root.id, node.id);
}
}
public boolean check(Node a, Node b) {
for (int i = 0; i < 50; i++) {
for (long j = a.l; j <= a.r; j++) {
long to = j ^ (1L << i);
if (b.l <= to && b.r >= to) {
return true;
}
}
}
return false;
}
public void addEdge(Node a, Node b) {
assert check(a, b);
a.adj.add(b);
b.adj.add(a);
}
public Node search(Node[] nodes, long x) {
int l = 0;
int r = nodes.length - 1;
while (l < r) {
int m = (l + r + 1) >> 1;
if (nodes[m].l <= x) {
l = m;
} else {
r = m - 1;
}
}
return nodes[l];
}
void addNode(long l, long r, boolean del) {
if (r < l) {
return;
}
if (l == r) {
addNode0(l, r, del);
return;
}
int index = Bits.theFirstDifferentIndex(l, r);
long prefix = (l >> (index + 1)) << (index + 1);
long mid = prefix | (1L << index);
addNode0(l, mid - 1, del);
addNode0(mid, r, del);
}
void addNode0(long l, long r, boolean del) {
if (r < l) {
return;
}
nodeList.add(new Node(l, r, del));
}
}
static class Bits {
private Bits() {
}
public static int theFirstDifferentIndex(long x, long y) {
return Log2.floorLog(x ^ y);
}
}
static class FastOutput implements AutoCloseable, Closeable, Appendable {
private static final int THRESHOLD = 32 << 10;
private OutputStream writer;
private StringBuilder cache = new StringBuilder(THRESHOLD * 2);
private static Field stringBuilderValueField;
private char[] charBuf = new char[THRESHOLD * 2];
private byte[] byteBuf = new byte[THRESHOLD * 2];
static {
try {
stringBuilderValueField = StringBuilder.class.getSuperclass().getDeclaredField("value");
stringBuilderValueField.setAccessible(true);
} catch (Exception e) {
stringBuilderValueField = null;
}
stringBuilderValueField = null;
}
public FastOutput append(CharSequence csq) {
cache.append(csq);
return this;
}
public FastOutput append(CharSequence csq, int start, int end) {
cache.append(csq, start, end);
return this;
}
private void afterWrite() {
if (cache.length() < THRESHOLD) {
return;
}
flush();
}
public FastOutput(OutputStream writer) {
this.writer = writer;
}
public FastOutput append(char c) {
cache.append(c);
afterWrite();
return this;
}
public FastOutput append(int c) {
cache.append(c);
afterWrite();
return this;
}
public FastOutput println(int c) {
return append(c).println();
}
public FastOutput println() {
return append('\n');
}
public FastOutput flush() {
try {
if (stringBuilderValueField != null) {
try {
byte[] value = (byte[]) stringBuilderValueField.get(cache);
writer.write(value, 0, cache.length());
} catch (Exception e) {
stringBuilderValueField = null;
}
}
if (stringBuilderValueField == null) {
int n = cache.length();
if (n > byteBuf.length) {
//slow
writer.write(cache.toString().getBytes(StandardCharsets.UTF_8));
// writer.append(cache);
} else {
cache.getChars(0, n, charBuf, 0);
for (int i = 0; i < n; i++) {
byteBuf[i] = (byte) charBuf[i];
}
writer.write(byteBuf, 0, n);
}
}
writer.flush();
cache.setLength(0);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
return this;
}
public void close() {
flush();
try {
writer.close();
} catch (IOException e) {
throw new UncheckedIOException(e);
}
}
public String toString() {
return cache.toString();
}
}
static class Debug {
private boolean offline;
private PrintStream out = System.err;
private long time = System.currentTimeMillis();
static int[] empty = new int[0];
public Debug(boolean enable) {
offline = enable && System.getSecurityManager() == null;
}
public Debug elapse(String name) {
if (offline) {
debug(name, System.currentTimeMillis() - time);
time = System.currentTimeMillis();
}
return this;
}
public Debug debug(String name, long x) {
if (offline) {
debug(name, "" + x);
}
return this;
}
public Debug debug(String name, String x) {
if (offline) {
out.printf("%s=%s", name, x);
out.println();
}
return this;
}
public Debug debug(String name, Object x) {
return debug(name, x, empty);
}
public Debug debug(String name, Object x, int... indexes) {
if (offline) {
if (x == null || !x.getClass().isArray()) {
out.append(name);
for (int i : indexes) {
out.printf("[%d]", i);
}
out.append("=").append("" + x);
out.println();
} else {
indexes = Arrays.copyOf(indexes, indexes.length + 1);
if (x instanceof byte[]) {
byte[] arr = (byte[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof short[]) {
short[] arr = (short[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof boolean[]) {
boolean[] arr = (boolean[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof char[]) {
char[] arr = (char[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof int[]) {
int[] arr = (int[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof float[]) {
float[] arr = (float[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof double[]) {
double[] arr = (double[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else if (x instanceof long[]) {
long[] arr = (long[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
} else {
Object[] arr = (Object[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug(name, arr[i], indexes);
}
}
}
}
return this;
}
}
static class DSU {
protected int[] p;
public int[] size;
protected int n;
public DSU(int n) {
p = new int[n];
size = new int[n];
}
public void init() {
init(p.length);
}
public void init(int n) {
this.n = n;
for (int i = 0; i < n; i++) {
p[i] = i;
size[i] = 1;
}
}
public final int find(int a) {
if (p[a] == p[p[a]]) {
return p[a];
}
find(p[a]);
preAccess(a);
return p[a] = p[p[a]];
}
protected void preAccess(int a) {
}
protected void preMerge(int a, int b) {
size[a] += size[b];
}
public final void merge(int a, int b) {
a = find(a);
b = find(b);
if (a == b) {
return;
}
if (size[a] < size[b]) {
int tmp = a;
a = b;
b = tmp;
}
preMerge(a, b);
p[b] = a;
}
public String toString() {
return Arrays.toString(Arrays.copyOf(p, n));
}
}
static class Log2 {
public static int floorLog(long x) {
if (x <= 0) {
throw new IllegalArgumentException();
}
return 63 - Long.numberOfLeadingZeros(x);
}
}
static class Query {
int type;
long l;
long r;
boolean ans;
public Query(int type, long l, long r) {
this.type = type;
this.l = l;
this.r = r;
}
public String toString() {
return "Query{" +
"type=" + type +
", l=" + l +
", r=" + r +
'}';
}
}
static class SortUtils {
private SortUtils() {
}
public static <T> boolean notStrictAscending(T[] data, int l, int r, Comparator<T> comp) {
for (int i = l + 1; i <= r; i++) {
if (comp.compare(data[i], data[i - 1]) < 0) {
return false;
}
}
return true;
}
}
static class Node {
long l;
long r;
boolean del;
int id;
List<Node> adj = new ArrayList<>();
public Node(long l, long r, boolean del) {
this.l = l;
this.r = r;
this.del = del;
}
public String toString() {
return "Node{" +
"l=" + l +
", r=" + r +
", del=" + del +
", id=" + id +
'}';
}
}
static class FastInput {
private final InputStream is;
private StringBuilder defaultStringBuf = new StringBuilder(1 << 13);
private byte[] buf = new byte[1 << 13];
private int bufLen;
private int bufOffset;
private int next;
public FastInput(InputStream is) {
this.is = is;
}
private int read() {
while (bufLen == bufOffset) {
bufOffset = 0;
try {
bufLen = is.read(buf);
} catch (IOException e) {
bufLen = -1;
}
if (bufLen == -1) {
return -1;
}
}
return buf[bufOffset++];
}
public void skipBlank() {
while (next >= 0 && next <= 32) {
next = read();
}
}
public int ri() {
return readInt();
}
public int readInt() {
boolean rev = false;
skipBlank();
if (next == '+' || next == '-') {
rev = next == '-';
next = read();
}
int val = 0;
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
return rev ? val : -val;
}
public long rl() {
return readLong();
}
public long readLong() {
boolean rev = false;
skipBlank();
if (next == '+' || next == '-') {
rev = next == '-';
next = read();
}
long val = 0;
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
return rev ? val : -val;
}
public String rs() {
return readString();
}
public String readString(StringBuilder builder) {
skipBlank();
while (next > 32) {
builder.append((char) next);
next = read();
}
return builder.toString();
}
public String readString() {
defaultStringBuf.setLength(0);
return readString(defaultStringBuf);
}
}
}
| Java | ["3 3\nask 0 7\nblock 3 6\nask 0 7", "6 10\nblock 12 26\nask 44 63\nblock 32 46\nask 1 54\nblock 27 30\nask 10 31\nask 11 31\nask 49 31\nblock 31 31\nask 2 51"] | 4 seconds | ["1\n0", "1\n1\n0\n0\n1\n0"] | NoteThe first example test can be visualized in the following way: Response to a query ask 0 7 is positive.Next after query block 3 6 the graph will look the following way (destroyed vertices are highlighted): Response to a query ask 0 7 is negative, since any path from vertex $$$0$$$ to vertex $$$7$$$ must go through one of the destroyed vertices. | Java 11 | standard input | [
"bitmasks",
"data structures",
"dsu",
"two pointers"
] | 327718077f69c0ed1c48acaf09b8e576 | The first line contains two integers $$$n$$$, $$$m$$$ ($$$1 \leq n \leq 50$$$, $$$1 \leq m \leq 5 \cdot 10^4$$$), which are the number of bits in binary representation of each planets' designation and the number of queries, respectively. Each of the next $$$m$$$ lines contains a query of two types: block l r — query for destruction of planets with numbers from $$$l$$$ to $$$r$$$ inclusively ($$$0 \le l \le r < 2^n$$$). It's guaranteed that no planet will be destroyed twice. ask a b — query for reachability between planets $$$a$$$ and $$$b$$$ ($$$0 \le a, b < 2^n$$$). It's guaranteed that planets $$$a$$$ and $$$b$$$ hasn't been destroyed yet. | 3,300 | For each query of type ask you must output "1" in a new line, if it is possible to reach planet $$$b$$$ from planet $$$a$$$ and "0" otherwise (without quotation marks). | standard output | |
PASSED | f47ce88eb6d38583ef4ae2a9f905daf3 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.util.*;
public class InterestingSum
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
String[] s=new String[650];
//System.out.println("Enter the number of test cases:");
int n1=sc.nextInt();
String[] s1=new String[n1];
//System.out.println("Enter the test cases");
for(int i=0;i<n1;i++)
{
s1[i]=sc.next();
}
int i=0;
for(int j='a';j<='z';j++)
{
for(int k='a';k<='z';k++)
{
if(j!=k)
{
s[i]=Character. toString(j)+Character. toString(k);
i++;
}
}
}
for(int j=0;j<n1;j++)
{
for(int k=0;k<650;k++)
{
if(s1[j].equals(s[k]))
{
System.out.println(k+1);
}
}
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 6ca231797fba82a36e5c9ea2668b7d65 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
String s = sc.next();
int res = (26 * (s.charAt(0) - 'a')) + (s.charAt(1) - 'a');
if (s.charAt(0) > 'b')
res -= (s.charAt(0) - 'a') - 1;
if (s.charAt(0) != 'a' && s.charAt(1) > s.charAt(0))
res -= 1;
System.out.println(res);
}
/*
* ul
zy
ge
fd
nu
sd
yn
*/
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | bf474b8116264a774c2ced9cb532b860 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class B_Dictionary {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- != 0) {
String s = sc.next();
int a = s.charAt(0) - 'a' + 1, b = s.charAt(1) - 'a' + 1;
int tf = 25 * (a - 1) + b;
if (b > a)
tf--;
System.out.println(tf);
}
sc.close();
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | be80197e0d82937a005a421913e2d205 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class B_Dictionary {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- != 0) {
String s = sc.next();
int a = s.charAt(0) - 'a' + 1, b = s.charAt(1) - 'a' + 1;
int tf = 25 * (a - 1);
if (b > a) {
tf += b - 1;
} else {
tf += b;
}
System.out.println(tf);
}
sc.close();
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | bc83599005aa75ca2a5713215fc0da84 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class Main {
private static class Task {
public void solve(int testNumber, InputReader in, PrintWriter out) {
char[] word = in.next().toCharArray();
int p = word[0] - 'a', q = word[1] - 'a';
if (p == 0) {
out.println(q);
} else {
out.println(p * 25 + (q > p ? q : q + 1));
}
}
}
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task solver = new Task();
// solver.solve(1, in, out);
int testNumber = in.nextInt();
for (int i = 1; i <= testNumber; i++) {
solver.solve(i, in, out);
}
out.close();
}
private static class InputReader {
private final BufferedReader reader;
private StringTokenizer tokenizer;
private InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
private String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
private String nextLine() {
String str;
try {
str = reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
return str;
}
private boolean hasNext() {
while (tokenizer != null && !tokenizer.hasMoreTokens()) {
String nextLine = null;
try {
nextLine = reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
private int nextInt() {
return Integer.parseInt(next());
}
private long nextLong() {
return Long.parseLong(next());
}
private double nextDouble() {
return Double.parseDouble(next());
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 89153a6846d5ce4d2b180f824ef1f16e | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
import java.util.HashMap;
public class Solution {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int index = 1;
HashMap<String, Integer> dictionary = new HashMap<>();
for (char i = 65; i <= 90; i++) {
for (char j = 65; j <= 90; j++) {
if (i == j) {
continue;
}
StringBuilder builder = new StringBuilder();
builder.append(i);
builder.append(j);
dictionary.put(builder.toString(), index++);
}
}
int t = scn.nextInt();
for(int i = 0; i < t; i++) {
String str = scn.next();
System.out.println(dictionary.get(str.toUpperCase()));
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 839a6d9a8ef4497266a365196804b9ae | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | //package com.company;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
public class Demo {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-- > 0){
String s = in.next();
int sum = (s.charAt(0)-97)*25;
if(s.charAt(1) < s.charAt(0)){
sum += (s.charAt(1)-96);
}else {
sum += (s.charAt(1)-97);
}
System.out.println(sum);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 3c767f2e986e93aab1439bed3d2fab11 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.util.*;
public class _1674B_Dictionary {
public static void main(String[] args) {
ArrayList<String> dic = new ArrayList<>();
for (char i = 'a'; i <= 'z'; i++) {
for (char j = 'a'; j <= 'z'; j++) {
if (i == j) {
continue;
}
dic.add("" + i + j);
}
}
Scanner input = new Scanner(System.in);
int test = input.nextInt();
while (test-- > 0) {
String num = input.next();
System.out.println(dic.indexOf(num) + 1);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | b16de7d10d49ec508b395f471d66fdf3 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args){
MyScanner scanner = new MyScanner();
int testCount = scanner.nextInt();
for(int testIdx = 1; testIdx <= testCount; testIdx++){
String s = scanner.nextLine();
int res = (s.charAt(0) - 'a') * 25 + (s.charAt(1) - 'a') + 1;
if(s.charAt(1) - s.charAt(0) > 0){
res--;
}
out.println(res);
}
out.close();
}
public static void printResult(int idx, long res){
out.println("Case #" + idx + ": " + res);
}
static void print1DArray(int[] arr){
for(int i = 0; i < arr.length; i++){
out.print(arr[i]);
if(i < arr.length - 1){
out.print(' ');
}
}
out.print('\n');
}
static void print1DArrayList(List<Integer> arrayList){
for(int i = 0; i < arrayList.size(); i++){
out.print(arrayList.get(i));
if(i < arrayList.size() - 1){
out.print(' ');
}
}
out.print('\n');
}
public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
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;
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 66a387562591e16065472363198f8859 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
public class Dictionary {
public static void main(String[]args) {
Scanner sn = new Scanner(System.in);
int t = sn.nextInt();
sn.nextLine();
for(int i =0; i<t; i++) {
String word = sn.nextLine();
char ch1 = word.charAt(0);
char ch2 = word.charAt(1);
int sum = 0;
sum += (ch1-'a')*25;
sum += ch2-'a';
if(ch1>ch2) {
sum++;
}
System.out.println(sum);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | e31d2d627bf5d7975fb244ad16511cae | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.Stream;
public class Main {
public static void main(String[] args) throws IOException {
ArrayList<String> list = new ArrayList<>();
for(int j = 97; j <= 122; j++){
for(int k = 97; k <=122; k++){
if(j == k)
continue;
StringBuilder sb = new StringBuilder();
sb.append((char)j);
sb.append((char)k);
list.add(sb.toString());
}
}
Scanner sc = new Scanner(System.in);
String[] t = new String[sc.nextInt()];
sc.nextLine();
for( int i = 0; i < t.length; i ++){
t[i] = sc.next();
}
for( int m = 0; m < t.length; m++){
System.out.println((1+list.indexOf(t[m])));
}
}
}
// ascii 97 - 122 = a-z
// x -> 26-(122 - charat(i))
// y -> 25-(122 - charat(i))
// xy -> abs pos
/*
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
long k = sc.nextLong();
if(n%2==0)
{
if(k==n/2)
System.out.prlongln(n-1);
if(k>n/2)
System.out.prlongln(2*(k-n/2));
if(k<n/2)
System.out.prlongln((2*k)-1);
}
else
{
if(k==n/2+1)
System.out.prlongln(n);
if(k>n/2+1)
System.out.prlongln(2*(k-(n/2+1)));
if(k<n/2+1)
System.out.prlongln((2*k)-1);
}
}
*/
/* interesting Math problem for function
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
if (n % 2 == 0){
System.out.prlongln(n/2);
}
else {
System.out.prlongln((n-1)/2-n);
}
*/
/*
//The key point here is to look at y, not at x -> y-(x mod y) is the formula to calculate the distance of x to x with remainder 0
public static void main(String[] args) throws IOException{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long[] res = new long[n];
sc.nextLine();
for(int i = 0; i < n; i ++){
long p = 0;
long a = sc.nextLong();
long b = sc.nextLong();
if(a < b) {
res[i] = b - a;
continue;
}
if(a%b == 0){
res[i] = 0;
continue;
}
res[i] = b-(a%b);
}
for(long j : res){
System.out.println(j);
}
}
}
*/
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 0ed8913eb527a6a6b800c20f3e54aeba | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class MyClass {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int tc= sc.nextInt();
while(tc > 0){
tc--;
String s = sc.next();
int c1 = s.charAt(0)-'a';
int c2 = s.charAt(1)-'a';
int m = c1*25;
int ans =0;
if(c2 > c1){
ans += (m+c2);
}else{
ans += (m + c2+1);
}
System.out.println(ans);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 9c4eb7bc1ba09fa393f2d0d7d6d3ed0a | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Solution{
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
int n=s.nextInt();
s.nextLine();
while(n!=0)
{
--n;
String r=s.nextLine();
int y=(r.charAt(0)-97)*25+(r.charAt(1)-97);
if(r.charAt(1)<r.charAt(0))
{
y+=1;
}
System.out.println(y);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 132707b40944b15631ce73aaa03e4ba7 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.nio.charset.StandardCharsets;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class CF1674B {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in, StandardCharsets.UTF_8);
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
String s = scanner.next();
System.out.println(solve(s));
}
}
private static Map<String, Integer> idxMap;
private static String solve(String s) {
if (idxMap == null) {
idxMap = new HashMap<>();
int idx = 1;
for (char i = 'a'; i <= 'z'; i++) {
for (char j = 'a'; j <= 'z'; j++) {
if (i != j) {
idxMap.put("" + i + j, idx++);
}
}
}
}
return String.valueOf(idxMap.get(s));
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | c7358020f1b967cddb0487ea780a16cc | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Testing {
public static void main(String[] args) {
Scanner s = new Scanner (System.in);
int t=s.nextInt();
for(int i=0;i<t;i++){
String input=s.next();
int x=(int)(input.charAt(1)-97)-(input.charAt(0)-97);
if(input.charAt(1)<input.charAt(0)) x++;
System.out.println(((int)input.charAt(0)%97)*26+x);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | fd5db711bbdc6f5245adf6aea2a1d067 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Solution {
private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private static PrintWriter pw = new PrintWriter(System.out);
static void getRes() throws Exception{
String inp = br.readLine();
char first = inp.charAt(0), second = inp.charAt(1);
pw.println((first - 'a') * 25 + (second < first ? second - 'a' + 1 : second - 'a'));
}
public static void main(String[] args) throws Exception {
int tests = Integer.parseInt(br.readLine());
while (tests-- > 0) {
getRes();
}
pw.flush();
pw.close();
}
static int[] inputIntArray(int n) throws Exception{
int[] arr = new int[n];
String[] inp = br.readLine().split(" ");
for (int x = 0; x < n; x++)
arr[x] = Integer.parseInt(inp[x]);
return arr;
}
static long[] inputLongArray(int n) throws Exception{
long[] arr = new long[n];
String[] inp = br.readLine().split(" ");
for (int x = 0; x < n; x++)
arr[x] = Integer.parseInt(inp[x]);
return arr;
}
static void printArray(int[] arr) {
for (int x = 0; x < arr.length; x++)
pw.print(arr[x] + " ");
pw.println();
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 17 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 8b4d050348424507b0bdae3aee68f337 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
public class Diccionary {
public static void main(String args[]){
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
while(n-->0){
int c, d;
String a = scanner.next();
if(a.charAt(0)=='a')
c = ((a.charAt(0)-'a')*25);
else
c = ((a.charAt(0)-'a')*25)+1;
if(a.charAt(1) >= a.charAt(0)+1 && a.charAt(0)!='a')
d = ((a.charAt(1)-'a')-1);
else
d = (a.charAt(1)-'a');
System.out.println(c+d);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 0c3d069f0b8c30ac94b8f55bcef2772b | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.util.HashMap;
import java.util.Scanner;
public class codeforcesdiv3 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
scan.nextLine();
for(int i = 0; i < n; i++){
String s = scan.nextLine();
System.out.println(allSum(s));
}
}
public static int allSum(String s){
HashMap<Character , Integer> hm = new HashMap<>();
HashMap<Character , Integer> h = new HashMap<>();
String ss = "abcdefghijklmnopqrstuvwxyz";
int pl = 0;
int kl = 0;
int sum = 0;
if(((int)(s.charAt(0))-97) > ((int)(s.charAt(1))-97)){
sum += 1;
}
for(int i = 0; i < ss.length(); i++){
hm.put(ss.charAt(i) , pl);
pl += 25;
}
for(int i = 0; i < ss.length(); i++){
h.put(ss.charAt(i) , kl++);
}
for(int j = 0; j < s.length(); j++){
if(j % 2 == 0) {
sum += hm.get(s.charAt(j));
}else{
sum += h.get(s.charAt(j));
}
}
return sum;
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 800a4607719983d1e3cd56f40732e800 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 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
{
public static void solution(String str)
{
HashMap<Character,Integer> hm=new HashMap<>();
char ch='a';
for(int i=1;i<=26;i++)
{
hm.put(ch,i);
ch+=1;
}
char a=str.charAt(0);
char b=str.charAt(1);
int sol=0;
if(hm.get(a)<hm.get(b))
sol=((hm.get(a)-1)*25) + (hm.get(b)-1);
else
sol= ((hm.get(a)-1)*25) + (hm.get(b));
System.out.println(sol);
}
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int test=sc.nextInt();
while(test-->0)
{
String str=sc.next();
solution(str);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | b69ef37dfc23689ed6fc562815fa9f60 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Practice {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
while(t--!=0){
String val = sc.nextLine();
int ans = (val.charAt(0)-'a')*25;
ans+=val.charAt(1)-'a'+1;
if(val.charAt(0)<val.charAt(1))
ans--;
System.out.println(ans);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 5eb1b15ad028f42e5abe6fc6aa0ac3e6 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.PrintWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
public class Test {
public static void main(String[] args) throws Exception {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));
int tc = Integer.parseInt(reader.readLine());
char[][] chars = new char[tc][];
for (int i = 0; i < tc; i++) {
chars[i] = reader.readLine().toCharArray();
}
for (int i = 0; i < tc; i++) {
writer.println(25 * (chars[i][0] - 'a') + (chars[i][1] - 'a') + (chars[i][0] < chars[i][1] ? 0 : 1));
}
writer.flush();
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | d0c24065d88ca167513499d806a8a73c | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class DictionarySolution2 {
public static void main(String[] args) throws Exception {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int tc = Integer.parseInt(reader.readLine());
char[][] chars = new char[tc][];
for (int i = 0; i < tc; i++) {
chars[i] = reader.readLine().toCharArray();
}
for (int i = 0; i < tc; i++) {
System.out.println(25 * (chars[i][0] - 'a') + (chars[i][1] - 'a') + (chars[i][0] < chars[i][1] ? 0 : 1));
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | edba3367d037dd1d80a10bf681ecff07 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class DictionarySolution1 {
public static void main(String[] args) throws Exception {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int tc = Integer.parseInt(reader.readLine());
while (tc-- > 0) {
char[] chars = reader.readLine().toCharArray();
System.out.println(25 * (chars[0] - 'a') + (chars[1] - 'a') + (chars[0] < chars[1] ? 0 : 1));
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 43e9cabf3779ca2f9f969f74c689fc65 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class dict
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
sc.nextLine();
while(t-->0)
{
String s=sc.nextLine();
if(s.charAt(0)>s.charAt(1))
System.out.println(25*(int)(s.charAt(0)-'a')+(int)s.charAt(1)-'a'+1);
else
System.out.println(25*(int)(s.charAt(0)-'a')+(int)s.charAt(1)-'a');
}
sc.close();
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 20678b2ba32e21af90cbf207501e3d30 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import javax.lang.model.util.ElementScanner6;
import javax.sound.midi.SysexMessage;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
public class Solution {
static HashMap<Integer, Integer> createHashMap(int arr[]) {
HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
for (int i = 0; i < arr.length; i++) {
Integer c = hmap.get(arr[i]);
if (hmap.get(arr[i]) == null) {
hmap.put(arr[i], 1);
}
else {
hmap.put(arr[i], ++c);
}
}
return hmap;
}
static int parseInt(String str) {
return Integer.parseInt(str);
}
static String noZeros(char[] num) {
String str = "";
for (int i = 0; i < num.length; i++) {
if (num[i] == '0')
continue;
else
str += num[i];
}
return str;
}
public static void Sort2DArrayBasedOnColumnNumber(int[][] array, final int columnNumber) {
Arrays.sort(array, new Comparator<int[]>() {
@Override
public int compare(int[] first, int[] second) {
if (first[columnNumber - 1] > second[columnNumber - 1])
return 1;
else
return -1;
}
});
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tk;
int t = parseInt(br.readLine());
String str;
int preindex;
int index;
for (int i = 1; i <= t; i++) {
str = br.readLine();
preindex = (str.charAt(0) - 96) * 25;
index = 25 - ((str.charAt(1) - 96));
index = preindex - index;
if (str.charAt(1) - str.charAt(0) > 0)
System.out.println(index - 1);
else
System.out.println(index);
}
}
}
/*
* 96
* 8
* 1 8 2 7 3 6 4 5
* -1 0 2 7 9 12 16 20
* count=16
*
* BRWBRWBRW
*
*
* x=7
* 2 1 2
* count=3
* 3 2 3
* count=5
* 4 3 4
* count=7
* 5 4 5
* count=8
* 6 5 6
* 7 6 7
* 8 7 8
* 9 8 9
* count=11
* ans+=(x-sum)/(i+1)+1
* -1 2 0
* -1 0 2
* sum=0
* sum=-1
* 1 2 4 2 1
* 4 2 2 1 1
* 4 2 2 1 1
*
* 2 4 5 5 2
* 4 5 5 2 2
* 5 5 2 2 4
*
* 2
* 2 2 4 5 5
*
*
*
* 4
* 2 10 1 7
* 3
* sum=-1
* sum
* 7 9
* 8
* 6
* 3
* 7
* 12
* 6
* -1
* 7
* 16
*
*
*
* 10 10
* 9
* 11
* 14
* 10
* 5
* 11
* 18
* 10
* 1
* 11
* +10 -5=5
* 5 7
* 6
* 4
* 1
* 5
* 10
* 6
* -1
* 3
*
* 2
*
*
*
* sum=15
* sum<=x
* count+=n
* count=5
* sum+=n
* sun=20
* count=10
* sum=25
* sum=18
*
*
*
*
*
*
* indeces={0,1,6,7}
* 5-1=4
*/ | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | bb22e79e37948944c8c772f467d0aa74 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.lang.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;
public class Last {
static Scanner sc = new Scanner(System.in);
static int I() {
return sc.nextInt();
}
static long L() {
return sc.nextLong();
}
static String S() {
return sc.next();
}
static long mod = (long) (1e9 + 7);
public static void main(String[] args) {
int t = I();
HashMap<String,Integer> hm=new HashMap<>();
int idx=1;
for(int i=0;i<26;i++)
for(int j=0;j<26;j++)
if(i!=j){
String s=(char)(i+'a')+""+(char)(j+'a');
hm.put(s, idx++);
}
while (t-- > 0) {
String s=S();
System.out.println(hm.get(s));
}
}
}
/*
(b^a)*x=y
*/
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | bdeb3c82dbba7f0c4e1a6f5acd98f354 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | /*Author: Jatin Yadav*/
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public 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();
}
}
public static void main(String args[])throws IOException{
Reader s = new Reader();
long t = s.nextLong();
//int b[] = new int[100005];
while(t-->0){
int sum=0;
String str = s.readLine();
int as1 = ((int)(str.charAt(0))-97);
int as2 = ((int)(str.charAt(1))-97);
int ans = as1*25+as2;
if(as2<as1){
ans+=1;
}
System.out.println(ans);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 2b791fd48e855847d5d8115cc6c81f64 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class Solution{
public static void main(String[] args) throws IOException
{
FastScanner f= new FastScanner();
int ttt=1;
ttt=f.nextInt();
PrintWriter out=new PrintWriter(System.out);
outer: for(int tt=0;tt<ttt;tt++) {
char[] l=f.next().toCharArray();
int ans=(l[0]-'a')*26+l[1]-'a';
ans-=Math.max(0, l[0]-'a'-1);
if(l[0]<l[1] && l[0]!='a') ans--;
System.out.println(ans);
}
out.close();
}
static void sort(int[] p) {
ArrayList<Integer> q = new ArrayList<>();
for (int i: p) q.add( i);
Collections.sort(q);
for (int i = 0; i < p.length; i++) p[i] = q.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());
}
double nextDouble() {
return Double.parseDouble(next());
}
long[] readLongArray(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 1d812ecba6098b778b37b5694a41e66e | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
public class Dictionary {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));
String Eng [] = {"a" , "b" , "c" , "d" , "e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
int x = Integer.parseInt(reader.readLine());
int indexF = 0;
int indexS = 0;
for (int i = 0;i<x;i++){
String line = reader.readLine();
String split[] = line.split("");
for (int j = 0;j<Eng.length;j++){
if (split[0].equals(Eng[j])){
indexF = j;
}
if (split[1].equals(Eng[j]))
indexS = j;
}
if (indexF == 0 || indexF < indexS)
writer.println((25 * indexF) + indexS);
else
writer.println((25 * indexF) + indexS+1);
}
writer.flush();
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 6c4a73e10ca20036c9ae29dffedf662b | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
/*
* Author: Atuer
*/
public class Main
{
// ==== Solve Code ====//
static int INF = 2000000010;
public static void csh()
{
}
public static void main(String[] args) throws IOException
{
// csh();
int t = in.nextInt();
while (t-- > 0)
{
solve();
out.flush();
}
out.close();
}
public static void solve()
{
char[] s = in.next().toCharArray();
out.println((s[0] - 'a') * 25 + s[1] - 'a' + (s[1] < s[0] ? 1 : 0));
}
public static class Node
{
int x, y, k;
public Node(int x, int y, int k)
{
this.x = x;
this.y = y;
this.k = k;
}
}
// ==== Solve Code ====//
// ==== Template ==== //
public static long cnm(int a, int b)
{
long sum = 1;
int i = a, j = 1;
while (j <= b)
{
sum = sum * i / j;
i--;
j++;
}
return sum;
}
public static int gcd(int a, int b)
{
return b == 0 ? a : gcd(b, a % b);
}
public static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
public static void gbSort(int[] a, int l, int r)
{
if (l < r)
{
int m = (l + r) >> 1;
gbSort(a, l, m);
gbSort(a, m + 1, r);
int[] t = new int[r - l + 1];
int idx = 0, i = l, j = m + 1;
while (i <= m && j <= r)
if (a[i] <= a[j])
t[idx++] = a[i++];
else
t[idx++] = a[j++];
while (i <= m)
t[idx++] = a[i++];
while (j <= r)
t[idx++] = a[j++];
for (int z = 0; z < t.length; z++)
a[l + z] = t[z];
}
}
// ==== Template ==== //
// ==== IO ==== //
static InputStream inputStream = System.in;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(System.out);
static class InputReader
{
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream)
{
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next()
{
while (tokenizer == null || !tokenizer.hasMoreTokens())
{
try
{
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e)
{
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
boolean hasNext()
{
while (tokenizer == null || !tokenizer.hasMoreTokens())
{
try
{
tokenizer = new StringTokenizer(reader.readLine());
} catch (Exception e)
{
return false;
// TODO: handle exception
}
}
return true;
}
public String nextLine()
{
String str = null;
try
{
str = reader.readLine();
} catch (IOException e)
{
e.printStackTrace();
}
return str;
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
public Double nextDouble()
{
return Double.parseDouble(next());
}
public BigInteger nextBigInteger()
{
return new BigInteger(next());
}
public BigDecimal nextBigDecimal()
{
return new BigDecimal(next());
}
}
// ==== IO ==== //
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 60dc012e51f8aeef72f8285071ea668e | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
/*
* Author: Atuer
*/
public class Main
{
// ==== Solve Code ====//
static int INF = 2000000010;
public static void csh()
{
}
public static void main(String[] args) throws IOException
{
// csh();
int t = in.nextInt();
while (t-- > 0)
{
solve();
out.flush();
}
out.close();
}
public static void solve()
{
char[] a = in.next().toCharArray();
int b = a[0] - 'a';
int c = a[1] - 'a';
int d = b*25+c;
out.println(b < c ? d : d + 1);
}
public static class Node
{
int x, y, k;
public Node(int x, int y, int k)
{
this.x = x;
this.y = y;
this.k = k;
}
}
// ==== Solve Code ====//
// ==== Template ==== //
public static long cnm(int a, int b)
{
long sum = 1;
int i = a, j = 1;
while (j <= b)
{
sum = sum * i / j;
i--;
j++;
}
return sum;
}
public static int gcd(int a, int b)
{
return b == 0 ? a : gcd(b, a % b);
}
public static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
public static void gbSort(int[] a, int l, int r)
{
if (l < r)
{
int m = (l + r) >> 1;
gbSort(a, l, m);
gbSort(a, m + 1, r);
int[] t = new int[r - l + 1];
int idx = 0, i = l, j = m + 1;
while (i <= m && j <= r)
if (a[i] <= a[j])
t[idx++] = a[i++];
else
t[idx++] = a[j++];
while (i <= m)
t[idx++] = a[i++];
while (j <= r)
t[idx++] = a[j++];
for (int z = 0; z < t.length; z++)
a[l + z] = t[z];
}
}
// ==== Template ==== //
// ==== IO ==== //
static InputStream inputStream = System.in;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(System.out);
static class InputReader
{
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream)
{
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next()
{
while (tokenizer == null || !tokenizer.hasMoreTokens())
{
try
{
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e)
{
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
boolean hasNext()
{
while (tokenizer == null || !tokenizer.hasMoreTokens())
{
try
{
tokenizer = new StringTokenizer(reader.readLine());
} catch (Exception e)
{
return false;
// TODO: handle exception
}
}
return true;
}
public String nextLine()
{
String str = null;
try
{
str = reader.readLine();
} catch (IOException e)
{
e.printStackTrace();
}
return str;
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
public Double nextDouble()
{
return Double.parseDouble(next());
}
public BigInteger nextBigInteger()
{
return new BigInteger(next());
}
public BigDecimal nextBigDecimal()
{
return new BigDecimal(next());
}
}
// ==== IO ==== //
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 8af4eaee534b4174665dd6c5d1a3e614 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import javax.sound.midi.Soundbank;
public class dictionar
{
public static void main(String[] args)
{
String[] str= new String[650];
for(int i=0;i<650;i++)
{
str[i]="";
}
int c=0;
for(char ch='a';ch<='z';ch++)
{
for(char ch1='a';ch1<='z';ch1++)
{
if(ch!=ch1)
{
str[c]=str[c]+(char)ch+(char)ch1;
c++;
}
}
}
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
String s=sc.next();
for(int i=0;i<650;i++)
{
if(s.compareTo(str[i])==0)
{
System.out.println(i+1);
}
}
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 7547ac5a71b2ee8d189c931d4b9cc6d2 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class Dispatcher {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
scan.nextLine();
for(int i=0;i<t;i++) {
String a = scan.nextLine();
int c = a.valueOf(a).charAt(0) - 97;
int d = a.valueOf(a).charAt(1) - 97;
if(c>d) d++;
System.out.println(c*25 + d);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 414e45f7158ad03789302a9781b2d35a | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import java.io.*;
public class lab_1 {
public static void main(String[] args) throws IOException{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
int start = 0;
int diff = 0;
int dec = 0;
int second = 0;
for(int i = 0;i<t;i++)
{dec = 0;
String s = sc.nextLine();
start = s.charAt(0) - 97;
second = s.charAt(1) - 97;
if((start)<( second) && s.charAt(0)!='a')
{
dec = -1;
}
if(s.charAt(0) - 97 != 0)
{
second = s.charAt(1) - 96;
}
else
{
second = s.charAt(1) - 97;
}
diff = start*25 + second + dec ;
System.out.println(diff);
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
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 boolean ready() throws IOException {
return br.ready();
}
public int[] readArr(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(next());
}
return arr;
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | f8ea5e915290bb035f74b5499e555f50 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.util.Scanner;
public class Dictionary {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while (t-->0){
String s=sc.next();
char first = s.charAt(0);
char second = s.charAt(1);
int index = (first - 'a') * 25 + 1;
index += second - 'a';
if (second > first) {
index--;
}
System.out.println(index);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | f3de5f8434b41b3dc7355e52913c23e5 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
int t= sc.nextInt();
while(t-- >0){
String str= sc.next();
int ans = 25*(str.charAt(0)-97)+(str.charAt(1)-97);
if(str.charAt(0) > str.charAt(1)){
ans++;
}
System.out.println(ans);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 409195beb7125f7cc801f4ea62787e90 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.lang.reflect.Array;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.StringTokenizer;
import java.util.Collections;
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 {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
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 close() throws IOException {
bw.close();
}
}
static long power(long x,long n)
{
long value=(long)(1e9+7);
long res=1;
while(n>0)
{
if(n%2!=0)
res=(res*x)%value;
x=(x*x)%value;
n=n/2;
}
return res;
}
static boolean isPrime(long 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;
}
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 sortDescending(long[] a)
{
ArrayList<Long> l=new ArrayList<>();
for (long i:a)
l.add(i);
l.sort(Collections.reverseOrder());
for (int i=0; i<a.length; i++)
a[i]=l.get(i);
}
public static boolean Equals(ArrayList<Long> a, ArrayList<Long> b, long n)
{
for(int i=0;i<n;i++)
{
if(!b.contains(a.get(i)))
return false;
}
for(int i=0;i<n;i++)
{
if(!a.contains(b.get(i)))
return false;
}
return true;
}
static int[] SetBit(long n,int []arr)
{
int res=0;
int k=0;
while(n>0)
{
if(n%2!=0)
arr[k]++;
k++;
n=n/2;
}
return arr;
}
static int josephus(int n, int k)
{
if (n == 1)
return 0;
else
return ((josephus((n - 1), k) + k ) % n );
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// method to return LCM of two numbers
static long lcm(long a, long b)
{
return (a / gcd(a, b)) * b;
}
static long Sum(long n)
{
long sum=0;
while(n>0)
{
sum+=n%10;
n/=10;
}
return sum;
}
static boolean isSubsequence(String s, String a)
{
int j=0;
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)==a.charAt(j))
j++;
if(j==a.length())
return true;
}
if(j!=a.length())
return false;
return true;
}
static boolean chechPalind(long n)
{
String s=String.valueOf(n);
int low=0,high=s.length()-1;
while(low<high)
{
if(s.charAt(low)!=s.charAt(high))
return false;
low++;
high--;
}
return true;
}
public static void main (String[] args) throws Exception {
FastReader sc = new FastReader();
FastWriter out = new FastWriter();
// Scanner sc=new Scanner(System.in);
long T=sc.nextLong();
int tc=1;
while(T-- >0)
{
String s=sc.next();
int ans=((s.charAt(0)-97)*25);
if(s.charAt(1)<s.charAt(0))
ans+=s.charAt(1)-96;
else
ans+=s.charAt(1)-96-1;
out.println(ans);
}
out.close();
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 17fa54781dcce8a0d5f6ee60e42148b2 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
public class CODE_FORCES {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int size = scanner.nextInt();
String[] words = new String[size];
for (int i = 0; i < size; i++) {
words[i] = scanner.next();
}
for (int i = 0; i < size; i++) {
if (words[i].charAt(0) > words[i].charAt(1)){
int aa = words[i].charAt(0) - 'a';
int great = words[i].charAt(1) - words[i].charAt(0);
System.out.println(aa*26+great+1);
}else if(words[i].charAt(0) < words[i].charAt(1)){
int great = words[i].charAt(1)-words[i].charAt(0);
if(words[i].charAt(0) == 'a' ) {
System.out.println(great);
}else{
int aa = words[i].charAt(0) - 'a';
int tt = words[i].charAt(1) - words[i].charAt(0);
System.out.println(aa*26+tt);
}
}else{
}
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 9218901c1f3b8a2f448d536e37252b6d | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class B_1674 {
public static void main(String args[]) {
Scanner sc=new Scanner(System.in);
String[] arr=new String[650];
int n=0;
for(int i=97;i<123;i++) {
for(int j=97;j<123;j++) {
if(i==j) continue;
arr[n]=String.valueOf((char)i)+String.valueOf((char)j);
n++;
}
}
int t=sc.nextInt();
sc.nextLine();
while(t-->0) {
String s=sc.nextLine();
int low=0;
int high=650;
int mid=0;
while(low<=high) {
mid=(low+high)/2;
if(arr[mid].equals(s)) break;
if(arr[mid].compareTo(s)<0) low=mid;
else high=mid;
}
System.out.println(mid+1);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | f5daf03686e77629d42016b270228254 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class B_1674 {
public static void main(String args[]) {
Scanner sc=new Scanner(System.in);
String[] arr=new String[650];
int n=0;
for(int i=97;i<123;i++) {
for(int j=97;j<123;j++) {
if(i==j) continue;
arr[n]=String.valueOf((char)i)+String.valueOf((char)j);
n++;
}
}
int t=sc.nextInt();
sc.nextLine();
while(t-->0) {
String s=sc.nextLine();
for(int i=0;i<650;i++) {
if(arr[i].equals(s)) {
System.out.println(i+1);
break;
}
}
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | a0ad7f3f98af5e4fad6bb2410247fee9 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class CF_1674B {
static void shinchan() {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
while(T>0) {
String s = sc.next();
int res = (s.charAt(0)-'a')*26+(s.charAt(1)-'a');
if(s.charAt(0)<=s.charAt(1)) {
res-=Math.max(0, (s.charAt(0)-'a'));
}else {
res-=Math.max(0, (s.charAt(0)-'a')-1);
}
System.out.println(res);
T--;
}
}
public static void main(String[] args) {
shinchan();
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 72488877d016cfe091e95cb52056fea0 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class class428 {
public static void main(String arg[])
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
String s=sc.next();
int val=s.charAt(0)-'a';
int f=25*val;
for(int i=0;i<26;i++)
{
char c=(char)('a'+i);
if(c==s.charAt(0))
continue;
f++;
if(s.charAt(1)==c)
break;
}
System.out.println(f);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 3ac29ed2d98b2aa3286f19e60c002a82 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.HashMap;
import java.util.StringTokenizer;
public class B {
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
static HashMap<String, Integer> map = new HashMap<>();
static void genWords() {
int i = 1;
for (char c1 = 'a'; c1 <= 'z'; c1++)
for (char c2 = 'a'; c2 <= 'z'; c2++)
if (c1 != c2)
map.put(String.valueOf(c1) + c2, i++);
}
public static void main(String[] args) throws IOException {
genWords();
int t = sc.nextInt();
while (t-- > 0) {
String s = sc.nextLine();
pw.println(map.get(s));
}
pw.flush();
pw.close();
}
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 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 | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 43539dbb3e4404851d0df62df53d1e23 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
/**
* User: Jameel Sawafta
* Date: 5/2/2022
* Time: 8:02 PM
* ProblemLink:https://codeforces.com/contest/1674/problem/B
*/
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Random;
import java.util.Arrays;
import java.util.Collections;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
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 Dictionary {
static void solve() throws Exception {
int t = scanInt();
while (t-- > 0) {
String s = scanString();
char[] ch = s.toCharArray();
if ((int)ch[0]<(int) ch[1]) {
out.println((ch[0]-97)*25+(ch[1]-97));
}
else out.println(((ch[0] - 97) * 25 + (ch[1] - 97))+1);
}
}
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);
}
}
// for array
public static void swap(int index, int index2, int[] a) {
int t = a[index2];
a[index2] = a[index];
a[index] = t;
}
//for Arraylist
public static <T> void swap(int index, int index2, ArrayList<T> a) {
T t = a.get(index2);
a.set(index2, a.get(index));
a.set(index, t);
}
private static void addRandomNumbers(ArrayList<Integer> a, int noOfElements, int bound) {
Random rand = new Random();
for (int i = 0; i < noOfElements; i++) {
a.add(rand.nextInt(bound));
}
}
public static void printArray(ArrayList<Integer> a) {
for (int i = 0; i < a.size(); i++) {
Integer y = a.get(i);
System.out.print(y + "\t");
}
System.out.println();
}
private static boolean isSorted(ArrayList<Integer> array) {
for (int i = 0; i < array.size() - 1; i++) {
if (array.get(i) > array.get(i + 1))
return false;
}
return true;
}
private static void copyArrays(int[] a, int start, int end, int[] temp) {
for (int i = start; i <= end; i++) {
temp[i - start] = a[i];
}
}
private static boolean isEven(int num) {
return ((num & 1) == 0 ? true : false);
}
/* Method to check if x is power of 2*/
private static boolean isPowerOfTwo(int x) {
return x != 0 && ((x & (x - 1)) == 0);
}
//Calculating the number of digits
private static int numOfDigits(int num) {
return (int) Math.floor(Math.log10(num)) + 1;
}
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 int gcd(int a, int b) {
BigInteger b1 = BigInteger.valueOf(a);
BigInteger b2 = BigInteger.valueOf(b);
BigInteger gcd = b1.gcd(b2);
return gcd.intValue();
}
private static void countSort(int[] arr) {
int max = Arrays.stream(arr).max().getAsInt();
int min = Arrays.stream(arr).min().getAsInt();
int range = max - min + 1;
int count[] = new int[range];
int output[] = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
count[arr[i] - min]++;
}
for (int i = 1; i < count.length; i++) {
count[i] += count[i - 1];
}
for (int i = arr.length - 1; i >= 0; i--) {
output[count[arr[i] - min] - 1] = arr[i];
count[arr[i] - min]--;
}
for (int i = 0; i < arr.length; i++) {
arr[i] = output[i];
}
}
private static void countSort(ArrayList<Integer> arr) {
int max = Collections.max(arr);
int min = Collections.min(arr);
int range = max - min + 1;
int count[] = new int[range];
int output[] = new int[arr.size()];
for (int i = 0; i < arr.size(); i++) {
count[arr.get(i) - min]++;
}
for (int i = 1; i < count.length; i++) {
count[i] += count[i - 1];
}
for (int i = arr.size() - 1; i >= 0; i--) {
output[count[arr.get(i) - min] - 1] = arr.get(i);
count[arr.get(i) - min]--;
}
for (int i = 0; i < arr.size(); i++) {
arr.set(i, output[i]);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | d0850ef43be9929309ee5ccd00d1e2bc | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Dictionary_1674B {
private static Map<String,Integer> map = new HashMap<>();
private static void dict(){
int cnt = 1;
String s = "abcdefghijklmnopqrstuvwxyz";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < 26; j++) {
if(i!=j){
map.put(s.charAt(i)+""+s.charAt(j), cnt++);
}
}
}
// System.out.println(map);
}
public static void main(String[] args) {
dict();
Scanner in = new Scanner(System.in);
int t = in.nextInt();
in.nextLine();
while(t-->0){
String s = in.nextLine();
System.out.println(map.get(s));
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | a9f28a60ee61cf1bde519890d167f096 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Dictionary_1674B {
private static Map<String,Integer> map = new HashMap<>();
private static void dict(){
int cnt = 1;
String s = "abcdefghijklmnopqrstuvwxyz";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < 26; j++) {
if(i!=j){
map.put(s.charAt(i)+""+s.charAt(j), cnt++);
}
}
}
// System.out.println(map);
}
public static void main(String[] args) {
dict();
Scanner in = new Scanner(System.in);
int t = in.nextInt();
in.nextLine();
while(t-->0){
String s = in.nextLine();
System.out.println(map.get(s));
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | dd0e6e0b4fafc4263605221b2b4bdf7e | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
String z = sc.nextLine();
while(t >0){
t--;
String s = sc.nextLine();
char first = s.charAt(0);
char second = s.charAt(1);
int answer = 0;
answer = (first - 'a') * 25;
if(second < first){
answer += second - 'a' + 1;
}else{
answer += second - 'a';
}
System.out.println(answer);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | e5a07523fb15c103ca20a8f9de714ebf | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
public class Dictionary {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int test = input.nextInt();
for (int i = 0; i < test; i++) {
String number = input.next();
String[] nums = number.split("");
String first = nums[0];
String second = nums[1];
if ("a".equals(first) == true) {
System.out.println((int) number.charAt(1) - 97);
}
if ("b".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 0);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 1);
}
}
if ("c".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 1);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 2);
}
}
if ("d".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 2);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 3);
}
}
if ("e".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 3);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 4);
}
}
if ("f".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 4);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 5);
}
}
if ("g".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 5);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 6);
}
}
if ("h".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 6);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 7);
}
}
if ("i".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 7);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 8);
}
}
if ("j".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 8);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 9);
}
}
if ("k".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 9);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 10);
}
}
if ("l".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 10);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 11);
}
}
if ("m".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 11);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 12);
}
}
if ("n".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 12);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 13);
}
}
if ("o".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 13);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 14);
}
}
if ("p".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 14);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 15);
}
}
if ("q".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 15);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 16);
}
}
if ("r".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 16);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 17);
}
}
if ("s".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 17);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 18);
}
}
if ("t".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 18);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 19);
}
}
if ("u".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 19);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 20);
}
}
if ("v".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 20);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 21);
}
}
if ("w".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 21);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 22);
}
}
if ("x".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 22);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 23);
}
}
if ("y".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 23);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 24);
}
}
if ("z".equals(first) == true) {
if ((int) number.charAt(0) > (int) number.charAt(1)) {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 24);
} else if ((int) number.charAt(0) == (int) number.charAt(1)) {
System.out.println(0);
} else {
System.out.println((int) ((((int) number.charAt(0)) - 97) * 26) + (number.charAt(1) - 97) - 25);
}
}
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 705cc2eb23a80b2fb791edeed5337380 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.*;
import java.lang.Math;
import java.util.StringTokenizer;
public class B {
public static void main(String[] args) {
int b, t;
char a[];
String str;
FastReader in = new FastReader();
t = in.nextInt();
while(t-- > 0) {
a = in.next().toCharArray();
int first = a[0];
int second = a[1];
for(int i = 0; i < 26;i++) {
if(first == 97 + i) {
first = i;
}
}
int ans = 0;
for(int i = 0; i < 26;i++) {
if(second == 97 + i) {
second = i;
}
}
if(first == 0) {
System.out.println(second);
continue;
}
if(second > first) {
second -= 1;
}
if(first == 1) {
System.out.println(26 + second);
continue;
}
for(int i = 0; i < first;i++) {
ans += 25;
}
System.out.println(ans + second + 1);
}
}
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());
}
int[] readArray(int n){
int[] a = new int[n];
for(int i = 0; i < n;i++){
a[i] = nextInt();
}
return a;
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | fda342e7ecf207be28df5216d8c73d2b | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.ArrayList;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
ArrayList<String> arrayList = new ArrayList<>();
scanner.nextLine();
for (int i = 0; i<n; i++) {
arrayList.add(scanner.nextLine());
}
ArrayList<String> vocabulary = new ArrayList<>();
int start = (int) 'a';
int finish = (int) 'z' + 1;
for (int i = start; i<finish; i++) {
String res = ((char)i)+"";
for (int j = start; j<finish; j++) {
if ((char)j != (char)i) {
res+=((char)j);
vocabulary.add(res);
}
res = ((char)i)+"";
}
}
for (int i = 0; i<arrayList.size(); i++) {
System.out.println(vocabulary.indexOf(arrayList.get(i))+1);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 980ea8880da1758051108297935a1b11 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Dictionary {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int tests = scn.nextInt();
scn.nextLine();
for(int t=0;t<tests;t++){
String str = scn.nextLine();
int a = str.charAt(0)-96;
int b = str.charAt(1) - 96;
int index;
if(b<=a){
index = 26*(a-1) + (b+1) - a;
}else{
index = 26*(a-1) + (b-a);
}
System.out.println(index);
}
scn.close();
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | ba42ab3545436aa8e6426a61696de855 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
import javax.swing.plaf.synth.SynthScrollPaneUI;
public class B_Dictionary{
static Scanner sc = new Scanner(System.in);
// @Harshit Maurya
public static void reverse(int[] arr, int l, int r) {
int d = (r - l + 1) / 2;
for (int i = 0; i < d; i++) {
int t = arr[l + i];
arr[l + i] = arr[r - i];
arr[r - i] = t;
}
}
// QUICK MATHS
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;
}
public static void main(String[] args) {
solve();
}
// BIT MAGIC
public int getMSB(int b) {
return (int)(Math.log(b) / Math.log(2));
}
//HELPER FUNCTIONS
private static int[] nextIntArray(int n){
int arr[]=new int[n];
for(int i=0; i<n; i++){
arr[i]=sc.nextInt();
}
return arr;
}
private static int[][] nextInt2DArray(int m,int n){
int arr[][]=new int[m][n];
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
arr[i][j]=sc.nextInt();
}
}
return arr;
}
private static long[] nextLongArray(int n){
long arr[]=new long[n];
for(int i=0; i<n; i++){
arr[i]=sc.nextLong();
}
return arr;
}
private static double[] nextDoubleArray(int n){
double arr[]=new double[n];
for(int i=0; i<n; i++){
arr[i]=sc.nextDouble();
}
return arr;
}
static int[] copy(int A[]) {
int B[]=new int[A.length];
for(int i=0;i<A.length;i++) {
B[i]=A[i];
}
return B;
}
static long[] copy(long A[]) {
long B[]=new long[A.length];
for(int i=0;i<A.length;i++) {
B[i]=A[i];
}
return B;
}
static long sum(int A[]) {
long sum=0;
for(int i : A) {
sum+=i;
}
return sum;
}
static long sum(long A[]) {
long sum=0;
for(long i : A) {
sum+=i;
}
return sum;
}
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 solve() {
int t = sc.nextInt();
while (t-- > 0) {
String str=sc.next();
char ch1=str.charAt(0),ch2=str.charAt(1);
int pow=ch1-'a';
int plus=ch2-'a';
long num=(25*pow)+(ch2<ch1?plus+1:plus);
System.out.println(num);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | a2f442d5db47d144c0c1505b0f275bde | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | //package Codeforces;
import java.io.*;
import java.util.*;
public class B {
public static void main (String[] Z) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder op = new StringBuilder();
StringTokenizer stz;
int T = Integer.parseInt(br.readLine());
HashMap<String, Integer> map = new HashMap<>();
char[] s = {'a', 'b'};
char first = 'a';
char second = 'b';
int count = 0;
for (int i = 0 ; i < 26; i++) {
for (int j = 0; j < 26; j++) {
if(i == j) continue;
++count;
s[0] = (char) ('a' + i);
s[1] = (char) ('a' + j);
map.put(String.valueOf(s), count);
}
}
// System.out.println(map);
while(T-- > 0) {
String sb = br.readLine();
int ans = map.get(sb);
op.append(ans);
op.append("\n");
}
System.out.println(op);
// END OF CODE
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | caecb22bcd06ce8e1440db214df0e8cc | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.lang.Math.*;
public class Main {
public static void main(String[] args) throws IOException {
int t = nextInt();
for (int i = 0; i < t; i++) {
String s = next();
if(s.charAt(0) < s.charAt(1)) {
out.println((s.charAt(0) - 'a') * 25 + s.charAt(1) - 'a');
}else{
out.println((s.charAt(0) - 'a') * 25 + s.charAt(1) - 'a' + 1);
}
}
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 String next() throws IOException {
while (in == null || !in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public static int nextInt() throws IOException {
return Integer.parseInt(next());
}
public static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public static long nextLong() throws IOException {
return Long.parseLong(next());
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 0097c4870a655b446dc0c2fdaca8ef2c | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.security.KeyPair;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;
/* Name of the class has to be "Main" only if the class is public. */
public class temp
{
abstract class sort implements Comparator<ArrayList<Integer>>{
@Override
public int compare(ArrayList<Integer> a,ArrayList<Integer> b) {
return a.get(0)-b.get(0);
}
}
private static Comparator sort;
public static void main (String[] args) throws Exception
{
FastScanner sc= new FastScanner();
int tt = sc.nextInt();
while(tt-- >0){
char arr[]=sc.next().toCharArray();
int last=arr[0]-'a';
int ans=last*25;
int a=arr[0]-'a';
int b=arr[1]-'a';
if(b<a) {
b++;
}
System.out.println(ans+b);
}
}
public static int fac(int n) {
if(n==0) return 1;
return n*fac(n-1);
}
public static boolean palin(String res) {
StringBuilder sb=new StringBuilder(res);
String temp=sb.reverse().toString();
return(temp.equals(res));
}
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 int gcd(int a, int b)
{
if(a > b)
a = (a+b)-(b=a);
if(a == 0L)
return b;
return gcd(b%a, a);
}
public static ArrayList<Integer> findDiv(int N)
{
//gens all divisors of N
ArrayList<Integer> ls1 = new ArrayList<Integer>();
ArrayList<Integer> ls2 = new ArrayList<Integer>();
for(int i=1; i <= (int)(Math.sqrt(N)+0.00000001); i++)
if(N%i == 0)
{
ls1.add(i);
ls2.add(N/i);
}
Collections.reverse(ls2);
for(int b: ls2)
if(b != ls1.get(ls1.size()-1))
ls1.add(b);
return ls1;
}
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 long power(long x, long y, long p)
{
//0^0 = 1
long res = 1L;
x = x%p;
while(y > 0)
{
if((y&1)==1)
res = (res*x)%p;
y >>= 1;
x = (x*x)%p;
}
return res;
}
static class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 6296261f3f858a09f4d4dfca41bb9477 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | //package com.company;
import java.io.*;
import java.util.*;
public class Cf_0 {
private static class FastReader {
InputStream is;
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
public FastReader(InputStream is) {
this.is = is;
}
public 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++];
}
public boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b)) ;
return b;
}
public String next() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public String nextLine() {
int c = skip();
StringBuilder sb = new StringBuilder();
while (!isEndOfLine(c)) {
sb.appendCodePoint(c);
c = readByte();
}
return sb.toString();
}
public int nextInt() {
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 << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public long nextLong() {
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 << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public double nextDouble() {
return Double.parseDouble(next());
}
public char[] next(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);
}
public char readChar() {
return (char) skip();
}
public long[] readArrayL(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = nextLong();
return arr;
}
public int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = nextInt();
return arr;
}
}
private static PrintWriter pw= new PrintWriter(System.out);
private static FastReader fr= new FastReader(System.in);
public static void main(String[] args) throws IOException {
//System.out.println((int)'a'-96);
int t= fr.nextInt();
for(int i=0;i<t;i++){
String str= fr.next();
pwl(solve800(str)+"");
}
pw.close();
}
public static int solve800(String str){
int res=0;
int f=(int)str.charAt(0)-97;
res=25*f;
int l=(int)str.charAt(1)-96;
res+=l;
if(str.charAt(1)>str.charAt(0))
res--;
return res;
}
public static long[]readLong(int n){
long []a= new long[n];
for(int it=0;it<n;it++){
a[it]=fr.nextLong();
}
return a;
}
public static int[] readArray(int n){
int []a= new int[n];
for(int it=0;it<n;it++){
a[it]=fr.nextInt();
}
return a;
}
public static int min(int []a){
int min=a[0];
for(int i=1;i<a.length;i++){
min=Math.min(min,a[i]);
}
return min;
}
public static int max(int []a){
int max=a[0];
for(int i=1;i<a.length;i++){
max=Math.max(max,a[i]);
}
return max;
}
public static int getDown(int n){
if(n==0)
return 9;
else return --n;
}
public static int getUp(int n){
if(n==9)
return 0;
else return ++n;
}
public static long findGCD(long a, long b) {
while(b != 0)
{
if(a > b)
{
a = a - b;
}
else
{
b = b - a;
}
}
return a;
}
public static long lcm(long a, long b) {
return (a / findGCD(a, b)) * b;
}
public static void print(int []a){
StringBuilder sb= new StringBuilder();
for(int t:a){
sb.append(t+" ");
}
pwl(sb.toString());
}
public static void pwl(String s){
pw.println(s);
}
public static void pw(String s){
pw.print(s);
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 9d2b9d31a505bc56e031d3d280b86c1f | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import java.io.*;
import static java.lang.Math.*;
public class Main {
public static void main(String[] args) throws IOException {
OutputStreamWriter osr = new OutputStreamWriter(System.out);
PrintWriter o = new PrintWriter(osr);
FastReader fr = new FastReader();
int t = fr.nextInt();
while (t--!=0)
{
String s = fr.next();
int x = (s.charAt(0) - 'a');
int y = (s.charAt(1) - 'a');
if(y < x)
y++;
o.println((x*25) + y);
}
o.close();
}
}
class FastReader {
// Attributes :
BufferedReader br;
StringTokenizer st;
// Constructor :
public FastReader() {
InputStreamReader isr = new InputStreamReader(System.in);
br = new BufferedReader(isr);
}
// Operations :
// #01 :
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
// #02 :
public String nextLine() throws IOException {
return br.readLine();
}
// #03 :
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
// #04 :
public long nextLong() throws IOException {
return Long.parseLong(next());
}
// #05 :
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
// #06 :
public int [] intArray (int size) throws IOException{
int [] arr = new int[size];
for (int i = 0 ; i < size; i++)
arr[i] = nextInt();
return arr;
}
// #07 :
public char [] charArray() throws IOException {
return nextLine().toCharArray();
}
}
class Pair {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
static class Compare implements Comparator<Pair> {
@Override
public int compare(Pair o1, Pair o2) {
return (o1.y - o2.y);
}
}
}
/*
if(arr[i] > arr[i+1] && !lid[i] && lid[i+1])
{
}
* */ | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 7ce38f6a193f97a4f40a06fa810beb44 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import java.io.*;
public class Dictionary {
public static void main(String[] args) {
try {
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int testcases = in.nextInt();
while (testcases-- > 0) {
String s = in.next();
char[] a = s.toCharArray();
int ans = (a[0] - 'a') * 26 + a[1] - 'a' + 1;
//dbg(ans);
ans = a[0] < a[1] ? ans - (a[0] - 'a' + 1) : ans - (a[0] - 'a');
out.println(ans);
}
out.close();
} catch (Exception e) {
// TODO: handle exception
return;
}
}
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());
}
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;
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static boolean LOCAL = System.getProperty("ONLINE_JUDGE") == null;
static void dbg(Object... o) {
if (LOCAL)
System.err.println(Arrays.deepToString(o));
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 6191087a77793c2ef67f4ce93960821b | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import java.io.*;
public class B {
public static void main(String[] args) throws IOException {
FastReader in = new FastReader(System.in);
PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
int t = in.nextInt();
for (int tt = 0; tt < t; tt++) {
String s = "abcdefghijklmnopqrstuvwxyz";
char[] ch = s.toCharArray();
// debug(ch);
ArrayList<String> lst = new ArrayList<>();
for (int i = 0; i < 26; i++) {
for (int j = 0; j < 26; j++) {
if (i == j) continue;
String a = String.valueOf(ch[i]);
a += ch[j];
lst.add(a);
}
}
Collections.sort(lst);
String str = in.next();
for (int i = 0; i < lst.size(); i++) {
if (Objects.equals(lst.get(i), str)) {
pw.println(i + 1);
break;
}
}
}
pw.close();
}
public static int gcd(int a, int b) {
if (b == 0) return a;
else return gcd(b, a % b);
}
static boolean checkPalindrome(String str) {
int len = str.length();
for (int i = 0; i < len / 2; i++) {
if (str.charAt(i) !=
str.charAt(len - i - 1))
return false;
}
return true;
}
public static void Sort(int[] a) {
ArrayList<Integer> lst = new ArrayList<>();
for (int i : a) lst.add(i);
Collections.sort(lst);
for (int i = 0; i < lst.size(); i++) a[i] = lst.get(i);
}
static void debug(Object... obj) {
System.err.println(Arrays.deepToString(obj));
}
static class FastReader {
InputStream is;
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
public FastReader(InputStream is) {
this.is = is;
}
public 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++];
}
public boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b)) ;
return b;
}
public String next() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public String nextLine() {
int c = skip();
StringBuilder sb = new StringBuilder();
while (!isEndOfLine(c)) {
sb.appendCodePoint(c);
c = readByte();
}
return sb.toString();
}
public int nextInt() {
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 << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public long nextLong() {
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 << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public double nextDouble() {
return Double.parseDouble(next());
}
public char[] next(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);
}
public char readChar() {
return (char) skip();
}
public long[] readArrayL(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = nextLong();
return arr;
}
public int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = nextInt();
return arr;
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 39b27fc8268353135590952df49de4be | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
Integer t = scanner.nextInt();
while (t-- > 0)
{
Integer sum =0;
String text = scanner.next();
int firstChar = text.charAt(0);
int secondChar = text.charAt(1);
sum+= 25 *(firstChar - 97);
if (firstChar < secondChar)
sum+=(secondChar - 98);
else
sum+=(secondChar - 97);
System.out.println(sum+1);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 8cf39da56f61fab8e1a3f2fe3fe00332 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args){
MyScanner scanner = new MyScanner();
int testCount = scanner.nextInt();
for(int testIdx = 1; testIdx <= testCount; testIdx++){
String s = scanner.nextLine();
int res = (s.charAt(0) - 'a') * 25 + (s.charAt(1) - 'a') + 1;
if(s.charAt(1) - s.charAt(0) > 0){
res--;
}
out.println(res);
}
out.close();
}
public static void printResult(int idx, long res){
out.println("Case #" + idx + ": " + res);
}
static void print1DArray(int[] arr){
for(int i = 0; i < arr.length; i++){
out.print(arr[i]);
if(i < arr.length - 1){
out.print(' ');
}
}
out.print('\n');
}
static void print1DArrayList(List<Integer> arrayList){
for(int i = 0; i < arrayList.size(); i++){
out.print(arrayList.get(i));
if(i < arrayList.size() - 1){
out.print(' ');
}
}
out.print('\n');
}
public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
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;
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | c87cb28463aac04f85c38c8a109dc544 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | //package combinatorics;
import java.util.Scanner;
public class CF1674B {
public static Scanner scanner;
public static void main(String[] args) {
scanner = new Scanner(System.in);
int n = scanner.nextInt();
for(int i = 0; i < n; i++) {
String str = scanner.next();
int res = (str.charAt(0) - 'a') * 26 + (str.charAt(1) - 'a');
int minus = str.charAt(0) - 'a';
if(str.charAt(1) >= str.charAt(0)) res = res - minus;
else res = res - minus + 1;
System.out.println(res);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 5a2ebfdf434327ca3a754c568fe357e7 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
public class Solution {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while(t-->0) {
char[] s = scan.next().toCharArray();
int result = (s[0] - 'a') * 25;
result += (s[0] <= s[1]) ? (s[1]-'a'):(s[1]-'a' + 1);
System.out.println(result);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | f0b69801617ef10d08b46fcc363863a9 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main {
public static void main(String [] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
StringBuilder sb = new StringBuilder();
int t = fs.nextInt();
while(t-- > 0) {
String s = fs.next();
int a = s.charAt(0)-'a';
int b = s.charAt(1)-'a';
sb.append((a > b) ? a*25+b+1 : a*25+b).append("\n");
}
out.print(sb);
out.close();
}
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());
}
long [] readArray(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());
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 5c9384602132e21c6c50d1299513e970 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Solution
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
String s=sc.next();
char first=s.charAt(0);
int initial=25*(first-97);
char second=s.charAt(1);
int finall=first>second?(second-97+1):(second-97);
System.out.println(initial+finall);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 45c59d9fed44b7d5c9fec02b80b488cc | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class B_Dictionary {
public static void main(String[] args) {
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
// try {
// out = new PrintWriter("output.txt");
// } catch (Exception e) {
// e.printStackTrace();
// }
int t = in.nextInt();
while (t-- > 0) {
char[] s = in.next().toCharArray();
int normal_index = ((s[0] - 'a') * 26) + s[1] - 'a' + 1;
int prev_same = s[0] - 'a';
int curr_same = 0;
if (s[1] > s[0]) curr_same++;
out.println(normal_index - prev_same - curr_same);
}
// first get the normal index, which is first char(0 based index) * 26 + second char (1 based)
// then we subtract prev values aa, bb,... for b series we have 1 such word i.e aa and for c we have 2 aa, bb
// if for curr word say ca, a < c so no change but for cd we have to subtract cc as well
out.flush();
out.close();
}
}
// For fast input output
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
// br = new BufferedReader(new InputStreamReader(System.in));
try {
br = new BufferedReader(new FileReader("input.txt"));
} catch(Exception e) {
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;
}
int[] readIntArray(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;
}
String[] readStringArray(int n) {
String[] a = new String[n];
for (int i = 0; i < n; i++) {
a[i] = next();
}
return a;
}
}
// end of fast i/o code | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | b4ddd3cb5ad7ad50c6c5b7e274bcbb10 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
Scanner sc = new Scanner(System.in);
//int test_case = Integer.parseInt(br.readLine());
long test_case = sc.nextLong();
for (long test = 0; test < test_case; test++) {
String s = sc.next();
HashMap<String,Integer> hm = new HashMap<>();
int cnt = 0;
for(int i=0;i<26;i++){
char one = (char) (i + 'a');
for(int j=0;j<26;j++){
if(i==j) continue;
char two = (char) (j+'a');
StringBuilder s1 = new StringBuilder();
s1.append(one);
s1.append(two);
hm.put(s1.toString(), cnt+1);
cnt++;
}
}
if(hm.containsKey(s)){
System.out.println(hm.get(s));
}else{
System.out.println(-1);
}
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | d8d8366bbf231d1ca636cf2159b28264 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(new BufferedInputStream(System.in));
int t = in.nextInt();
Map<String, Integer> mp = new HashMap<>();
int count = 0;
for (int i = 0; i < 26; i++) {
for (int j = 0; j < 26; j++) {
if (('a' + i) == ('a' + j))
continue;
count++;
String s = (char)('a' + i) + "" + (char)('a' + j);
mp.put(s, count);
}
}
for (int i = 0; i < t; i++) {
String s = in.next();
System.out.println(mp.get(s));
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 24db4570809ffd5391bed52eda4c099f | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Scanner;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.nio.charset.MalformedInputException;
import java.util.StringTokenizer;
import java.util.Arrays;
public class Cv {
public static void main(String[] args) {
FastScanner in = new FastScanner();
Scanner sc = new Scanner(System.in);
PrintWriter o = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
String s = sc.next();
char c = s.charAt(0);
char f = s.charAt(1);
int y = (int)c - 97;
int d = (int)f - 96;
if (f > c) {
d--;
}
o.println(y * 25 + d);
}
o.close();
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
public String next() {
while (!st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
byte nextByte() {
return Byte.parseByte(next());
}
short nextShort() {
return Short.parseShort(next());
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return java.lang.Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 8a8fd286c628fac5d60a4453ddda2004 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.Arrays;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner scan = new Scanner(System.in);
int T = scan.nextInt();
for(int i = 0; i < T; i++) {
String s = scan.next();
char[] ch = s.toCharArray();
int first = ch[0] - 'a';
int second = ch[1] - 'a';
if(first<second) {
System.out.println(25*first + second);
}else{
System.out.println(25*first + second + 1);
}
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 97dbc83008c57b747b558514df3d48c0 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static class Reader {
BufferedReader reader= new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tokenizer = new StringTokenizer("");
String next() throws IOException {
while (!tokenizer.hasMoreTokens() ) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
double nextDouble() throws IOException {
return Double.parseDouble(next());
}
long nextLong() throws IOException {
return Long.parseLong(next());
}
void close() throws IOException {
reader.close();
}
}
public static void main(String[] args) throws IOException {
Reader sc = new Reader();
StringBuilder ans = new StringBuilder();
HashMap<String, Integer> map = new HashMap<>();
ans.append("aa");
int cnt = 1;
for (int i = 0; i < 26; i++) {
for (int j = 0; j < 25; j++) {
char t = (char) (ans.charAt(1) + 1);
ans.deleteCharAt(1);
ans.append(t);
if (t != ans.charAt(0)) map.put(ans.toString(), cnt++);
}
char t = (char) (ans.charAt(0) + 1);
ans = new StringBuilder();
ans.append(t).append('a');
map.put(ans.toString(), cnt++);
}
int TT = sc.nextInt();
while (TT-- > 0) {
System.out.println(map.get(sc.next()));
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | d23d37a9bede12c23fb6cb2d064a96a6 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
while (t-- > 0)
{
String s = sc.nextLine();
char[] ch = s.toCharArray();
int position1 = (int) ch[0] - 96;
int position2 = (int) ch[1] - 96;
int pos = position2;
if (position1 < position2)
{
pos = position2-1;
}
System.out.println(25*(position1-1)+pos);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | a0063657787e839f18cea068b5753189 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | 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.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static FastScanner scan = new FastScanner();
public static void main(String[] args) throws Exception
{
int n = scan.nextInt();
for(int i = 0; i < n; i++)
{
char[] x = scan.next().toCharArray();
int answer = (x[0] - 'a') * 25 + (x[1] - 'a' + 1);
if(x[1] > x[0])
answer--;
System.out.println(answer);
}
}
public static void print(int[] arr)
{
//for debugging only
for(int x: arr)
out.print(x+" ");
out.println();
}
//input shenanigans
/*
Random stuff to try when stuck:
-for combo/probability problems, expand the given form we're interested in
-make everything the same then build an answer (constructive, make everything 0 then do something)
-something appears in parts of 2 --> model as graph
-assume a greedy then try to show why it works
-find way to simplify into one variable if multiple exist
-treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
-find lower and upper bounds on answer
-figure out what ur trying to find and isolate it
-see what observations you have and come up with more continuations
-work backwards (in constructive, go from the goal to the start)
-turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
-instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
-draw something
-simulate a process
-dont implement something unless if ur fairly confident its correct
-after 3 bad submissions move on to next problem if applicable
-do something instead of nothing and stay organized
-write stuff down
Random stuff to check when wa:
-if code is way too long/cancer then reassess
-switched N/M
-int overflow
-switched variables
-wrong MOD
-hardcoded edge case incorrectly
Random stuff to check when tle:
-continue instead of break
-condition in for/while loop bad
Random stuff to check when rte:
-switched N/M
-long to int/int overflow
-division by 0
-edge case for empty list/data structure/N=1
*/
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 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 long totient(long n)
{
long result = n;
for (int p = 2; p*p <= n; ++p)
if (n % p == 0)
{
while(n%p == 0)
n /= p;
result -= result/p;
}
if (n > 1)
result -= result/n;
return result;
/*
find phi(i) from 1 to N fast
O(N*loglogN)
long[] arr = new long[N+1];
for(int i=1; i <= N; i++)
arr[i] = i;
for(int v=2; v <= N; v++)
if(arr[v] == v)
for(int a=v; a <= N; a+=v)
arr[a] -= arr[a]/v;
*/
}
public static ArrayList<Integer> findDiv(int N)
{
//gens all divisors of N
ArrayList<Integer> ls1 = new ArrayList<Integer>();
ArrayList<Integer> ls2 = new ArrayList<Integer>();
for(int i=1; i <= (int)(Math.sqrt(N)+0.00000001); i++)
if(N%i == 0)
{
ls1.add(i);
ls2.add(N/i);
}
Collections.reverse(ls2);
for(int b: ls2)
if(b != ls1.get(ls1.size()-1))
ls1.add(b);
return ls1;
}
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 long power(long x, long y, long p)
{
//0^0 = 1
long res = 1L;
x = x%p;
while(y > 0)
{
if((y&1)==1)
res = (res*x)%p;
y >>= 1;
x = (x*x)%p;
}
return res;
}
//custom multiset (replace with HashMap if needed)
public static void push(TreeMap<Integer, Integer> map, int k, int v)
{
//map[k] += v;
if(!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k)+v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v)
{
//assumes map[k] >= v
//map[k] -= v
int lol = map.get(k);
if(lol == v)
map.remove(k);
else
map.put(k, lol-v);
}
public static int[] compress(int[] arr)
{
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1; //min value
for(int x: ls)
if(!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for(int i=0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static long[][] multiply(long[][] left, long[][] right)
{
long MOD = 1000000007L;
int N = left.length;
int M = right[0].length;
long[][] res = new long[N][M];
for(int a=0; a < N; a++)
for(int b=0; b < M; b++)
for(int c=0; c < left[0].length; c++)
{
res[a][b] += (left[a][c]*right[c][b])%MOD;
if(res[a][b] >= MOD)
res[a][b] -= MOD;
}
return res;
}
public static long[][] power(long[][] grid, long pow)
{
long[][] res = new long[grid.length][grid[0].length];
for(int i=0; i < res.length; i++)
res[i][i] = 1L;
long[][] curr = grid.clone();
while(pow > 0)
{
if((pow&1L) == 1L)
res = multiply(curr, res);
pow >>= 1;
curr = multiply(curr, curr);
}
return res;
}
}
class DSU
{
public int[] dsu;
public int[] size;
public DSU(int N)
{
dsu = new int[N+1];
size = new int[N+1];
for(int i=0; i <= N; i++)
{
dsu[i] = i;
size[i] = 1;
}
}
//with path compression, no find by rank
public int find(int x)
{
return dsu[x] == x ? x : (dsu[x] = find(dsu[x]));
}
public void merge(int x, int y)
{
int fx = find(x);
int fy = find(y);
dsu[fx] = fy;
}
public void merge(int x, int y, boolean sized)
{
int fx = find(x);
int fy = find(y);
size[fy] += size[fx];
dsu[fx] = fy;
}
}
class FenwickTree
{
//Binary Indexed Tree
//1 indexed
public int[] tree;
public int size;
public FenwickTree(int size)
{
this.size = size;
tree = new int[size+5];
}
public void add(int i, int v)
{
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
}
class SegmentTree
{
//Tlatoani's segment tree
//iterative implementation = low constant runtime factor
//range query, non lazy
final int[] val;
final int treeFrom;
final int length;
public SegmentTree(int treeFrom, int treeTo)
{
this.treeFrom = treeFrom;
int length = treeTo - treeFrom + 1;
int l;
for (l = 0; (1 << l) < length; l++);
val = new int[1 << (l + 1)];
this.length = 1 << l;
}
public void update(int index, int delta)
{
//replaces value
int node = index - treeFrom + length;
val[node] = delta;
for (node >>= 1; node > 0; node >>= 1)
val[node] = comb(val[node << 1], val[(node << 1) + 1]);
}
public int query(int from, int to)
{
//inclusive bounds
if (to < from)
return 0; //0 or 1?
from += length - treeFrom;
to += length - treeFrom + 1;
//0 or 1?
int res = 0;
for (; from + (from & -from) <= to; from += from & -from)
res = comb(res, val[from / (from & -from)]);
for (; to - (to & -to) >= from; to -= to & -to)
res = comb(res, val[(to - (to & -to)) / (to & -to)]);
return res;
}
public int comb(int a, int b)
{
//change this
return Math.max(a,b);
}
}
class LazySegTree
{
//definitions
private int NULL = -1;
private int[] tree;
private int[] lazy;
private int length;
public LazySegTree(int N)
{
length = N; int b;
for(b=0; (1<<b) < length; b++);
tree = new int[1<<(b+1)];
lazy = new int[1<<(b+1)];
}
public int query(int left, int right)
{
//left and right are 0-indexed
return get(1, 0, length-1, left, right);
}
private int get(int v, int currL, int currR, int L, int R)
{
if(L > R)
return NULL;
if(L <= currL && currR <= R)
return tree[v];
propagate(v);
int mid = (currL+currR)/2;
return comb(get(v*2, currL, mid, L, Math.min(R, mid)),
get(v*2+1, mid+1, currR, Math.max(L, mid+1), R));
}
public void update(int left, int right, int delta)
{
add(1, 0, length-1, left, right, delta);
}
private void add(int v, int currL, int currR, int L, int R, int delta)
{
if(L > R)
return;
if(currL == L && currR == R)
{
//exact covering
tree[v] += delta;
lazy[v] += delta;
return;
}
propagate(v);
int mid = (currL+currR)/2;
add(v*2, currL, mid, L, Math.min(R, mid), delta);
add(v*2+1, mid+1, currR, Math.max(L, mid+1), R, delta);
tree[v] = comb(tree[v*2], tree[v*2+1]);
}
private void propagate(int v)
{
//tree[v] already has lazy[v]
if(lazy[v] == 0)
return;
tree[v*2] += lazy[v];
lazy[v*2] += lazy[v];
tree[v*2+1] += lazy[v];
lazy[v*2+1] += lazy[v];
lazy[v] = 0;
}
private int comb(int a, int b)
{
return Math.max(a,b);
}
}
class RangeBit
{
//FenwickTree and RangeBit are faster than LazySegTree by constant factor
final int[] value;
final int[] weightedVal;
public RangeBit(int treeTo)
{
value = new int[treeTo+2];
weightedVal = new int[treeTo+2];
}
private void updateHelper(int index, int delta)
{
int weightedDelta = index*delta;
for(int j = index; j < value.length; j += j & -j)
{
value[j] += delta;
weightedVal[j] += weightedDelta;
}
}
public void update(int from, int to, int delta)
{
updateHelper(from, delta);
updateHelper(to + 1, -delta);
}
private int query(int to)
{
int res = 0;
int weightedRes = 0;
for (int j = to; j > 0; j -= j & -j)
{
res += value[j];
weightedRes += weightedVal[j];
}
return ((to + 1)*res)-weightedRes;
}
public int query(int from, int to)
{
if (to < from)
return 0;
return query(to) - query(from - 1);
}
}
class SparseTable
{
public int[] log;
public int[][] table;
public int N; public int K;
public SparseTable(int N)
{
this.N = N;
log = new int[N+2];
K = Integer.numberOfTrailingZeros(Integer.highestOneBit(N));
table = new int[N][K+1];
sparsywarsy();
}
private void sparsywarsy()
{
log[1] = 0;
for(int i=2; i <= N+1; i++)
log[i] = log[i/2]+1;
}
public void lift(int[] arr)
{
int n = arr.length;
for(int i=0; i < n; i++)
table[i][0] = arr[i];
for(int j=1; j <= K; j++)
for(int i=0; i + (1 << j) <= n; i++)
table[i][j] = Math.min(table[i][j-1], table[i+(1 << (j - 1))][j-1]);
}
public int query(int L, int R)
{
//inclusive, 1 indexed
L--; R--;
int mexico = log[R-L+1];
return Math.min(table[L][mexico], table[R-(1 << mexico)+1][mexico]);
}
}
class LCA
{
public int N, root;
public ArrayDeque<Integer>[] edges;
private int[] enter;
private int[] exit;
private int LOG = 17; //change this
private int[][] dp;
public LCA(int n, ArrayDeque<Integer>[] edges, int r)
{
N = n; root = r;
enter = new int[N+1];
exit = new int[N+1];
dp = new int[N+1][LOG];
this.edges = edges;
int[] time = new int[1];
//change to iterative dfs if N is large
dfs(root, 0, time);
dp[root][0] = 1;
for(int b=1; b < LOG; b++)
for(int v=1; v <= N; v++)
dp[v][b] = dp[dp[v][b-1]][b-1];
}
private void dfs(int curr, int par, int[] time)
{
dp[curr][0] = par;
enter[curr] = ++time[0];
for(int next: edges[curr])
if(next != par)
dfs(next, curr, time);
exit[curr] = ++time[0];
}
public int lca(int x, int y)
{
if(isAnc(x, y))
return x;
if(isAnc(y, x))
return y;
int curr = x;
for(int b=LOG-1; b >= 0; b--)
{
int temp = dp[curr][b];
if(!isAnc(temp, y))
curr = temp;
}
return dp[curr][0];
}
private boolean isAnc(int anc, int curr)
{
return enter[anc] <= enter[curr] && exit[anc] >= exit[curr];
}
}
class BitSet
{
private int CONS = 62; //safe
public long[] sets;
public int size;
public BitSet(int N)
{
size = N;
if(N%CONS == 0)
sets = new long[N/CONS];
else
sets = new long[N/CONS+1];
}
public void add(int i)
{
int dex = i/CONS;
int thing = i%CONS;
sets[dex] |= (1L << thing);
}
public int and(BitSet oth)
{
int boof = Math.min(sets.length, oth.sets.length);
int res = 0;
for(int i=0; i < boof; i++)
res += Long.bitCount(sets[i] & oth.sets[i]);
return res;
}
public int xor(BitSet oth)
{
int boof = Math.min(sets.length, oth.sets.length);
int res = 0;
for(int i=0; i < boof; i++)
res += Long.bitCount(sets[i] ^ oth.sets[i]);
return res;
}
}
class MaxFlow
{
//Dinic with optimizations (see magic array in dfs function)
public int N, source, sink;
public ArrayList<Edge>[] edges;
private int[] depth;
public MaxFlow(int n, int x, int y)
{
N = n;
source = x;
sink = y;
edges = new ArrayList[N+1];
for(int i=0; i <= N; i++)
edges[i] = new ArrayList<Edge>();
depth = new int[N+1];
}
public void addEdge(int from, int to, long cap)
{
Edge forward = new Edge(from, to, cap);
Edge backward = new Edge(to, from, 0L);
forward.residual = backward;
backward.residual = forward;
edges[from].add(forward);
edges[to].add(backward);
}
public long mfmc()
{
long res = 0L;
int[] magic = new int[N+1];
while(assignDepths())
{
long flow = dfs(source, Long.MAX_VALUE/2, magic);
while(flow > 0)
{
res += flow;
flow = dfs(source, Long.MAX_VALUE/2, magic);
}
magic = new int[N+1];
}
return res;
}
private boolean assignDepths()
{
Arrays.fill(depth, -69);
ArrayDeque<Integer> q = new ArrayDeque<Integer>();
q.add(source);
depth[source] = 0;
while(q.size() > 0)
{
int curr = q.poll();
for(Edge e: edges[curr])
if(e.capacityLeft() > 0 && depth[e.to] == -69)
{
depth[e.to] = depth[curr]+1;
q.add(e.to);
}
}
return depth[sink] != -69;
}
private long dfs(int curr, long bottleneck, int[] magic)
{
if(curr == sink)
return bottleneck;
for(; magic[curr] < edges[curr].size(); magic[curr]++)
{
Edge e = edges[curr].get(magic[curr]);
if(e.capacityLeft() > 0 && depth[e.to]-depth[curr] == 1)
{
long val = dfs(e.to, Math.min(bottleneck, e.capacityLeft()), magic);
if(val > 0)
{
e.augment(val);
return val;
}
}
}
return 0L; //no flow
}
private class Edge
{
public int from, to;
public long flow, capacity;
public Edge residual;
public Edge(int f, int t, long cap)
{
from = f;
to = t;
capacity = cap;
}
public long capacityLeft()
{
return capacity-flow;
}
public void augment(long val)
{
flow += val;
residual.flow -= val;
}
}
}
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
public char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 0ea661e94c9c44009fa1608585baa1ec | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.io.*;
import java.util.*;
public final class Main {
//int 2e9 - long 9e18
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(0, 1), new Pair(1, 0), new Pair(0, -1)};
static int mod = (int) (1e9 + 7);
static int mod2 = 998244353;
public static void main(String[] args) {
int tt = i();
while (tt-- > 0) {
solve();
}
out.flush();
}
public static void solve() {
String s = s();
int ans = 0;
ans += (s.charAt(0) - 'a') * 25;
ans += s.charAt(1) - 'a' + 1;
if (s.charAt(0) < s.charAt(1)){
ans --;
}
out.println(ans);
}
// (10,5) = 2 ,(11,5) = 3
static long upperDiv(long a, long b) {
return (a / b) + ((a % b == 0) ? 0 : 1);
}
static long sum(int[] a) {
long sum = 0;
for (int x : a) {
sum += x;
}
return sum;
}
static int[] preint(int[] a) {
int[] pre = new int[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] pre(int[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] post(int[] a) {
long[] post = new long[a.length + 1];
post[0] = 0;
for (int i = 0; i < a.length; i++) {
post[i + 1] = post[i] + a[a.length - 1 - i];
}
return post;
}
static long[] pre(long[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static void print(char A[]) {
for (char c : A) {
out.print(c);
}
out.println();
}
static void print(boolean A[]) {
for (boolean c : A) {
out.print(c + " ");
}
out.println();
}
static void print(int A[]) {
for (int c : A) {
out.print(c + " ");
}
out.println();
}
static void print(long A[]) {
for (long i : A) {
out.print(i + " ");
}
out.println();
}
static void print(List<Integer> A) {
for (int a : A) {
out.print(a + " ");
}
}
static int i() {
return in.nextInt();
}
static long l() {
return in.nextLong();
}
static double d() {
return in.nextDouble();
}
static String s() {
return in.nextLine();
}
static String c() {
return in.next();
}
static int[][] inputWithIdx(int N) {
int A[][] = new int[N][2];
for (int i = 0; i < N; i++) {
A[i] = new int[]{i, in.nextInt()};
}
return A;
}
static int[] input(int N) {
int A[] = new int[N];
for (int i = 0; i < N; i++) {
A[i] = in.nextInt();
}
return A;
}
static long[] inputLong(int N) {
long A[] = new long[N];
for (int i = 0; i < A.length; i++) {
A[i] = in.nextLong();
}
return A;
}
static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long GCD(long a, long b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long LCM(int a, int b) {
return (long) a / GCD(a, b) * b;
}
static long LCM(long a, long b) {
return a / GCD(a, b) * b;
}
// find highest i which satisfy a[i]<=x
static int lowerbound(int[] a, int x) {
int l = 0;
int r = a.length - 1;
while (l < r) {
int m = (l + r + 1) / 2;
if (a[m] <= x) {
l = m;
} else {
r = m - 1;
}
}
return l;
}
static void shuffle(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
}
static void shuffleAndSort(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static void shuffleAndSort(int[][] arr, Comparator<? super int[]> comparator) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int[] temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr, comparator);
}
static void shuffleAndSort(long[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
long temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static boolean isPerfectSquare(double number) {
double sqrt = Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static void swap(int A[], int a, int b) {
int t = A[a];
A[a] = A[b];
A[b] = t;
}
static void swap(char A[], int a, int b) {
char t = A[a];
A[a] = A[b];
A[b] = t;
}
static long pow(long a, long b, int mod) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow = (pow * x) % mod;
}
x = (x * x) % mod;
b /= 2;
}
return pow;
}
static long pow(long a, long b) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow *= x;
}
x = x * x;
b /= 2;
}
return pow;
}
static long modInverse(long x, int mod) {
return pow(x, mod - 2, mod);
}
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 = i + 6) {
if (N % i == 0 || N % (i + 2) == 0) {
return false;
}
}
return true;
}
public static String reverse(String str) {
if (str == null) {
return null;
}
return new StringBuilder(str).reverse().toString();
}
public static void reverse(int[] arr) {
int l = 0;
int r = arr.length - 1;
while (l < r) {
swap(arr, l, r);
l++;
r--;
}
}
public static String repeat(char ch, int repeat) {
if (repeat <= 0) {
return "";
}
final char[] buf = new char[repeat];
for (int i = repeat - 1; i >= 0; i--) {
buf[i] = ch;
}
return new String(buf);
}
public static int[] manacher(String s) {
char[] chars = s.toCharArray();
int n = s.length();
int[] d1 = new int[n];
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : Math.min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && chars[i - k] == chars[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
return d1;
}
public static int[] kmp(String s) {
int n = s.length();
int[] res = new int[n];
for (int i = 1; i < n; ++i) {
int j = res[i - 1];
while (j > 0 && s.charAt(i) != s.charAt(j)) {
j = res[j - 1];
}
if (s.charAt(i) == s.charAt(j)) {
++j;
}
res[i] = j;
}
return res;
}
}
class Pair {
int i;
int j;
Pair(int i, int j) {
this.i = i;
this.j = j;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Pair pair = (Pair) o;
return i == pair.i && j == pair.j;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
class ThreePair {
int i;
int j;
int k;
ThreePair(int i, int j, int k) {
this.i = i;
this.j = j;
this.k = k;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
ThreePair pair = (ThreePair) o;
return i == pair.i && j == pair.j && k == pair.k;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
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 Node {
int val;
public Node(int val) {
this.val = val;
}
}
class ST {
int n;
Node[] st;
ST(int n) {
this.n = n;
st = new Node[4 * Integer.highestOneBit(n)];
}
void build(Node[] nodes) {
build(0, 0, n - 1, nodes);
}
private void build(int id, int l, int r, Node[] nodes) {
if (l == r) {
st[id] = nodes[l];
return;
}
int mid = (l + r) >> 1;
build((id << 1) + 1, l, mid, nodes);
build((id << 1) + 2, mid + 1, r, nodes);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
void update(int i, Node node) {
update(0, 0, n - 1, i, node);
}
private void update(int id, int l, int r, int i, Node node) {
if (i < l || r < i) {
return;
}
if (l == r) {
st[id] = node;
return;
}
int mid = (l + r) >> 1;
update((id << 1) + 1, l, mid, i, node);
update((id << 1) + 2, mid + 1, r, i, node);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
Node get(int x, int y) {
return get(0, 0, n - 1, x, y);
}
private Node get(int id, int l, int r, int x, int y) {
if (x > r || y < l) {
return new Node(0);
}
if (x <= l && r <= y) {
return st[id];
}
int mid = (l + r) >> 1;
return comb(get((id << 1) + 1, l, mid, x, y), get((id << 1) + 2, mid + 1, r, x, y));
}
Node comb(Node a, Node b) {
if (a == null) {
return b;
}
if (b == null) {
return a;
}
return new Node(GCD(a.val, b.val));
}
static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | ba730f363194a6449fefd12d775f4507 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 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.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
// B -> CodeForcesProblemSet
public final class B {
static FastReader fr = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static final int gigamod = 1000000007;
static int t = 1;
static double epsilon = 0.00000001;
static boolean[] isPrime;
static int[] smallestFactorOf;
@SuppressWarnings("unused")
public static void main(String[] args) {
t = fr.nextInt();
OUTER:
for (int tc = 0; tc < t; tc++) {
char[] s = fr.next().toCharArray();
int idx = 1;
for (char char1 = 'a'; char1 <= 'z'; char1++)
for (char char2 = 'a'; char2 <= 'z'; char2++) {
if (char1 == char2) continue;
if (char1 == s[0] && char2 == s[1]) {
out.println(idx);
continue OUTER;
}
idx++;
}
}
out.close();
}
static class UnionFind {
// Uses weighted quick-union with path compression.
private int[] parent; // parent[i] = parent of i
private int[] size; // size[i] = number of sites in tree rooted at i
// Note: not necessarily correct if i is not a root node
private int count; // number of components
public UnionFind(int n) {
count = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
// Number of connected components.
public int count() {
return count;
}
// Find the root of p.
public int find(int p) {
while (p != parent[p])
p = parent[p];
return p;
}
public boolean connected(int p, int q) {
return find(p) == find(q);
}
public int numConnectedTo(int node) {
return size[find(node)];
}
// Weighted union.
public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
if (rootP == rootQ) return;
// make smaller root point to larger one
if (size[rootP] < size[rootQ]) {
parent[rootP] = rootQ;
size[rootQ] += size[rootP];
}
else {
parent[rootQ] = rootP;
size[rootP] += size[rootQ];
}
count--;
}
public static int[] connectedComponents(UnionFind uf) {
// We can do this in nlogn.
int n = uf.size.length;
int[] compoColors = new int[n];
for (int i = 0; i < n; i++)
compoColors[i] = uf.find(i);
HashMap<Integer, Integer> oldToNew = new HashMap<>();
int newCtr = 0;
for (int i = 0; i < n; i++) {
int thisOldColor = compoColors[i];
Integer thisNewColor = oldToNew.get(thisOldColor);
if (thisNewColor == null)
thisNewColor = newCtr++;
oldToNew.put(thisOldColor, thisNewColor);
compoColors[i] = thisNewColor;
}
return compoColors;
}
}
static class UGraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
@SuppressWarnings("unchecked")
public UGraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
adj[to].add(from);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int degree(int v) {
return adj[v].size();
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static void dfsMark(int current, boolean[] marked, UGraph g) {
if (marked[current]) return;
marked[current] = true;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, marked, g);
}
public static void dfsMark(int current, int from, long[] distTo, boolean[] marked, UGraph g, ArrayList<Integer> endPoints) {
if (marked[current]) return;
marked[current] = true;
if (from != -1)
distTo[current] = distTo[from] + 1;
HashSet<Integer> adj = g.adj(current);
int alreadyMarkedCtr = 0;
for (int adjc : adj) {
if (marked[adjc]) alreadyMarkedCtr++;
dfsMark(adjc, current, distTo, marked, g, endPoints);
}
if (alreadyMarkedCtr == adj.size())
endPoints.add(current);
}
public static void bfsOrder(int current, UGraph g) {
}
public static void dfsMark(int current, int[] colorIds, int color, UGraph g) {
if (colorIds[current] != -1) return;
colorIds[current] = color;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, colorIds, color, g);
}
public static int[] connectedComponents(UGraph g) {
int n = g.V();
int[] componentId = new int[n];
Arrays.fill(componentId, -1);
int colorCtr = 0;
for (int i = 0; i < n; i++) {
if (componentId[i] != -1) continue;
dfsMark(i, componentId, colorCtr, g);
colorCtr++;
}
return componentId;
}
public static boolean hasCycle(UGraph ug) {
int n = ug.V();
boolean[] marked = new boolean[n];
boolean[] hasCycleFirst = new boolean[1];
for (int i = 0; i < n; i++) {
if (marked[i]) continue;
hcDfsMark(i, ug, marked, hasCycleFirst, -1);
}
return hasCycleFirst[0];
}
// Helper for hasCycle.
private static void hcDfsMark(int current, UGraph ug, boolean[] marked, boolean[] hasCycleFirst, int parent) {
if (marked[current]) return;
if (hasCycleFirst[0]) return;
marked[current] = true;
HashSet<Integer> adjc = ug.adj(current);
for (int adj : adjc) {
if (marked[adj] && adj != parent && parent != -1) {
hasCycleFirst[0] = true;
return;
}
hcDfsMark(adj, ug, marked, hasCycleFirst, current);
}
}
}
static class Digraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
@SuppressWarnings("unchecked")
public Digraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public Digraph reversed() {
Digraph dg = new Digraph(V());
for (int i = 0; i < V(); i++)
for (int adjVert : adj(i)) dg.addEdge(adjVert, i);
return dg;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static int[] KosarajuSharirSCC(Digraph dg) {
int[] id = new int[dg.V()];
Digraph reversed = dg.reversed();
// Gotta perform topological sort on this one to get the stack.
Stack<Integer> revStack = Digraph.topologicalSort(reversed);
// Initializing id and idCtr.
id = new int[dg.V()];
int idCtr = -1;
// Creating a 'marked' array.
boolean[] marked = new boolean[dg.V()];
while (!revStack.isEmpty()) {
int vertex = revStack.pop();
if (!marked[vertex])
sccDFS(dg, vertex, marked, ++idCtr, id);
}
return id;
}
private static void sccDFS(Digraph dg, int source, boolean[] marked, int idCtr, int[] id) {
marked[source] = true;
id[source] = idCtr;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex]) sccDFS(dg, adjVertex, marked, idCtr, id);
}
public static Stack<Integer> topologicalSort(Digraph dg) {
// dg has to be a directed acyclic graph.
// We'll have to run dfs on the digraph and push the deepest nodes on stack first.
// We'll need a Stack<Integer> and a int[] marked.
Stack<Integer> topologicalStack = new Stack<Integer>();
boolean[] marked = new boolean[dg.V()];
// Calling dfs
for (int i = 0; i < dg.V(); i++)
if (!marked[i])
runDfs(dg, topologicalStack, marked, i);
return topologicalStack;
}
static void runDfs(Digraph dg, Stack<Integer> topologicalStack, boolean[] marked, int source) {
marked[source] = true;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex])
runDfs(dg, topologicalStack, marked, adjVertex);
topologicalStack.add(source);
}
}
static class FastReader {
private BufferedReader bfr;
private StringTokenizer st;
public FastReader() {
bfr = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
if (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(bfr.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());
}
char nextChar() {
return next().toCharArray()[0];
}
String nextString() {
return next();
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
double[] nextDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextDouble();
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
int[][] nextIntGrid(int n, int m) {
int[][] grid = new int[n][m];
for (int i = 0; i < n; i++) {
char[] line = fr.next().toCharArray();
for (int j = 0; j < m; j++)
grid[i][j] = line[j] - 48;
}
return grid;
}
}
static class LCA {
int[] height, first, segtree;
ArrayList<Integer> euler;
boolean[] visited;
int n;
LCA(ArrayList<Point>[] outFrom, int root) {
n = outFrom.length;
height = new int[n];
first = new int[n];
euler = new ArrayList<>();
visited = new boolean[n];
dfs(outFrom, root, 0);
int m = euler.size();
segtree = new int[m * 4];
build(1, 0, m - 1);
}
void dfs(ArrayList<Point>[] outFrom, int node, int h) {
visited[node] = true;
height[node] = h;
first[node] = euler.size();
euler.add(node);
for (Point to : outFrom[node]) {
if (!visited[(int) to.x]) {
dfs(outFrom, (int) to.x, h + 1);
euler.add(node);
}
}
}
void build(int node, int b, int e) {
if (b == e) {
segtree[node] = euler.get(b);
} else {
int mid = (b + e) / 2;
build(node << 1, b, mid);
build(node << 1 | 1, mid + 1, e);
int l = segtree[node << 1], r = segtree[node << 1 | 1];
segtree[node] = (height[l] < height[r]) ? l : r;
}
}
int query(int node, int b, int e, int L, int R) {
if (b > R || e < L)
return -1;
if (b >= L && e <= R)
return segtree[node];
int mid = (b + e) >> 1;
int left = query(node << 1, b, mid, L, R);
int right = query(node << 1 | 1, mid + 1, e, L, R);
if (left == -1) return right;
if (right == -1) return left;
return height[left] < height[right] ? left : right;
}
int lca(int u, int v) {
int left = first[u], right = first[v];
if (left > right) {
int temp = left;
left = right;
right = temp;
}
return query(1, 0, euler.size() - 1, left, right);
}
}
static class FenwickTree {
long[] array; // 1-indexed array, In this array We save cumulative information to perform efficient range queries and updates
public FenwickTree(int size) {
array = new long[size + 1];
}
public long rsq(int ind) {
assert ind > 0;
long sum = 0;
while (ind > 0) {
sum += array[ind];
//Extracting the portion up to the first significant one of the binary representation of 'ind' and decrementing ind by that number
ind -= ind & (-ind);
}
return sum;
}
public long rsq(int a, int b) {
assert b >= a && a > 0 && b > 0;
return rsq(b) - rsq(a - 1);
}
public void update(int ind, long value) {
assert ind > 0;
while (ind < array.length) {
array[ind] += value;
//Extracting the portion up to the first significant one of the binary representation of 'ind' and incrementing ind by that number
ind += ind & (-ind);
}
}
public int size() {
return array.length - 1;
}
}
static final class RedBlackCountSet {
// Manual implementation of RB-BST for C++ PBDS functionality.
// Modification required according to requirements.
// Colors for Nodes:
private static final boolean RED = true;
private static final boolean BLACK = false;
// Pointer to the root:
private Node root;
// Public constructor:
public RedBlackCountSet() {
root = null;
}
// Node class for storing key value pairs:
private class Node {
long key;
long value;
Node left, right;
boolean color;
long size; // Size of the subtree rooted at that node.
public Node(long key, long value, boolean color) {
this.key = key;
this.value = value;
this.left = this.right = null;
this.color = color;
this.size = value;
}
}
/***** Invariant Maintenance functions: *****/
// When a temporary 4 node is formed:
private Node flipColors(Node node) {
node.left.color = node.right.color = BLACK;
node.color = RED;
return node;
}
// When there's a right leaning red link and it is not a 3 node:
private Node rotateLeft(Node node) {
Node rn = node.right;
node.right = rn.left;
rn.left = node;
rn.color = node.color;
node.color = RED;
return rn;
}
// When there are 2 red links in a row:
private Node rotateRight(Node node) {
Node ln = node.left;
node.left = ln.right;
ln.right = node;
ln.color = node.color;
node.color = RED;
return ln;
}
/***** Invariant Maintenance functions end *****/
// Public functions:
public void put(long key) {
root = put(root, key);
root.color = BLACK; // One of the invariants.
}
private Node put(Node node, long key) {
// Standard insertion procedure:
if (node == null)
return new Node(key, 1, RED);
// Recursing:
int cmp = Long.compare(key, node.key);
if (cmp < 0)
node.left = put(node.left, key);
else if (cmp > 0)
node.right = put(node.right, key);
else
node.value++;
// Invariant maintenance:
if (node.left != null && node.right != null) {
if (node.right.color = RED && node.left.color == BLACK)
node = rotateLeft(node);
if (node.left.color == RED && node.left.left.color == RED)
node = rotateRight(node);
if (node.left.color == RED && node.right.color == RED)
node = flipColors(node);
}
// Property maintenance:
node.size = node.value + size(node.left) + size(node.right);
return node;
}
private long size(Node node) {
if (node == null)
return 0;
else
return node.size;
}
public long rank(long key) {
return rank(root, key);
}
// Modify according to requirements.
private long rank(Node node, long key) {
if (node == null) return 0;
int cmp = Long.compare(key, node.key);
if (cmp < 0)
return rank(node.left, key);
else if (cmp > 0)
return node.value + size(node.left) + rank(node.right, key);
else
return node.value + size(node.left);
}
}
static class SegmentTree {
private Node[] heap;
private int[] array;
private int size;
public SegmentTree(int[] array) {
this.array = Arrays.copyOf(array, array.length);
//The max size of this array is about 2 * 2 ^ log2(n) + 1
size = (int) (2 * Math.pow(2.0, Math.floor((Math.log((double) array.length) / Math.log(2.0)) + 1)));
heap = new Node[size];
build(1, 0, array.length);
}
public int size() {
return array.length;
}
//Initialize the Nodes of the Segment tree
private void build(int v, int from, int size) {
heap[v] = new Node();
heap[v].from = from;
heap[v].to = from + size - 1;
if (size == 1) {
heap[v].sum = array[from];
heap[v].min = array[from];
} else {
//Build childs
build(2 * v, from, size / 2);
build(2 * v + 1, from + size / 2, size - size / 2);
heap[v].sum = heap[2 * v].sum + heap[2 * v + 1].sum;
//min = min of the children
heap[v].min = Math.min(heap[2 * v].min, heap[2 * v + 1].min);
}
}
public int rsq(int from, int to) {
return rsq(1, from, to);
}
private int rsq(int v, int from, int to) {
Node n = heap[v];
//If you did a range update that contained this node, you can infer the Sum without going down the tree
if (n.pendingVal != null && contains(n.from, n.to, from, to)) {
return (to - from + 1) * n.pendingVal;
}
if (contains(from, to, n.from, n.to)) {
return heap[v].sum;
}
if (intersects(from, to, n.from, n.to)) {
propagate(v);
int leftSum = rsq(2 * v, from, to);
int rightSum = rsq(2 * v + 1, from, to);
return leftSum + rightSum;
}
return 0;
}
public int rMinQ(int from, int to) {
return rMinQ(1, from, to);
}
private int rMinQ(int v, int from, int to) {
Node n = heap[v];
//If you did a range update that contained this node, you can infer the Min value without going down the tree
if (n.pendingVal != null && contains(n.from, n.to, from, to)) {
return n.pendingVal;
}
if (contains(from, to, n.from, n.to)) {
return heap[v].min;
}
if (intersects(from, to, n.from, n.to)) {
propagate(v);
int leftMin = rMinQ(2 * v, from, to);
int rightMin = rMinQ(2 * v + 1, from, to);
return Math.min(leftMin, rightMin);
}
return Integer.MAX_VALUE;
}
public void update(int from, int to, int value) {
update(1, from, to, value);
}
private void update(int v, int from, int to, int value) {
//The Node of the heap tree represents a range of the array with bounds: [n.from, n.to]
Node n = heap[v];
if (contains(from, to, n.from, n.to)) {
change(n, value);
}
if (n.size() == 1) return;
if (intersects(from, to, n.from, n.to)) {
propagate(v);
update(2 * v, from, to, value);
update(2 * v + 1, from, to, value);
n.sum = heap[2 * v].sum + heap[2 * v + 1].sum;
n.min = Math.min(heap[2 * v].min, heap[2 * v + 1].min);
}
}
//Propagate temporal values to children
private void propagate(int v) {
Node n = heap[v];
if (n.pendingVal != null) {
change(heap[2 * v], n.pendingVal);
change(heap[2 * v + 1], n.pendingVal);
n.pendingVal = null; //unset the pending propagation value
}
}
//Save the temporal values that will be propagated lazily
private void change(Node n, int value) {
n.pendingVal = value;
n.sum = n.size() * value;
n.min = value;
array[n.from] = value;
}
//Test if the range1 contains range2
private boolean contains(int from1, int to1, int from2, int to2) {
return from2 >= from1 && to2 <= to1;
}
//check inclusive intersection, test if range1[from1, to1] intersects range2[from2, to2]
private boolean intersects(int from1, int to1, int from2, int to2) {
return from1 <= from2 && to1 >= from2 // (.[..)..] or (.[...]..)
|| from1 >= from2 && from1 <= to2; // [.(..]..) or [..(..)..
}
//The Node class represents a partition range of the array.
static class Node {
int sum;
int min;
//Here We store the value that will be propagated lazily
Integer pendingVal = null;
int from;
int to;
int size() {
return to - from + 1;
}
}
}
@SuppressWarnings("serial")
static class CountMap<T> extends TreeMap<T, Integer>{
CountMap() {
}
CountMap(T[] arr) {
this.putCM(arr);
}
public Integer putCM(T key) {
if (super.containsKey(key)) {
return super.put(key, super.get(key) + 1);
} else {
return super.put(key, 1);
}
}
public Integer removeCM(T key) {
Integer count = super.get(key);
if (count == null) return -1;
if (count == 1)
return super.remove(key);
else
return super.put(key, super.get(key) - 1);
}
public Integer getCM(T key) {
Integer count = super.get(key);
if (count == null)
return 0;
return count;
}
public void putCM(T[] arr) {
for (T l : arr)
this.putCM(l);
}
}
static class Point implements Comparable<Point> {
long x;
long y;
long id;
Point() {
x = y = id = 0;
}
Point(Point p) {
this.x = p.x;
this.y = p.y;
this.id = p.id;
}
Point(long a, long b, long id) {
this.x = a;
this.y = b;
this.id = id;
}
Point(long a, long b) {
this.x = a;
this.y = b;
}
@Override
public int compareTo(Point o) {
if (this.x < o.x)
return -1;
if (this.x > o.x)
return 1;
if (this.y < o.y)
return -1;
if (this.y > o.y)
return 1;
return 0;
}
public boolean equals(Point that) {
return this.compareTo(that) == 0;
}
}
static int longestPrefixPalindromeLen(char[] s) {
int n = s.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++)
sb.append(s[i]);
StringBuilder sb2 = new StringBuilder(sb);
sb.append('^');
sb.append(sb2.reverse());
// out.println(sb.toString());
char[] newS = sb.toString().toCharArray();
int m = newS.length;
int[] pi = piCalcKMP(newS);
return pi[m - 1];
}
static int KMPNumOcc(char[] text, char[] pat) {
int n = text.length;
int m = pat.length;
char[] patPlusText = new char[n + m + 1];
for (int i = 0; i < m; i++)
patPlusText[i] = pat[i];
patPlusText[m] = '^'; // Seperator
for (int i = 0; i < n; i++)
patPlusText[m + i] = text[i];
int[] fullPi = piCalcKMP(patPlusText);
int answer = 0;
for (int i = 0; i < n + m + 1; i++)
if (fullPi[i] == m)
answer++;
return answer;
}
static int[] piCalcKMP(char[] s) {
int n = s.length;
int[] pi = new int[n];
for (int i = 1; i < n; i++) {
int j = pi[i - 1];
while (j > 0 && s[i] != s[j])
j = pi[j - 1];
if (s[i] == s[j])
j++;
pi[i] = j;
}
return pi;
}
static String toBinaryString(long num, int bits) {
StringBuilder sb = new StringBuilder(Long.toBinaryString(num));
sb.reverse();
for (int i = sb.length(); i < bits; i++)
sb.append('0');
return sb.reverse().toString();
}
static long modDiv(long a, long b) {
return mod(a * power(b, gigamod - 2, gigamod));
}
static void coprimeGenerator(int m, int n, ArrayList<Point> coprimes, int limit, int numCoprimes) {
if (m > limit) return;
if (m <= limit && n <= limit)
coprimes.add(new Point(m, n));
if (coprimes.size() > numCoprimes) return;
coprimeGenerator(2 * m - n, m, coprimes, limit, numCoprimes);
coprimeGenerator(2 * m + n, m, coprimes, limit, numCoprimes);
coprimeGenerator(m + 2 * n, n, coprimes, limit, numCoprimes);
}
static long hash(long i) {
return (i * 2654435761L % gigamod);
}
static long hash2(long key) {
long h = Long.hashCode(key);
h ^= (h >>> 20) ^ (h >>> 12) ^ (h >>> 7) ^ (h >>> 4);
return h & (gigamod-1);
}
static int mapTo1D(int row, int col, int n, int m) {
// Maps elements in a 2D matrix serially to elements in
// a 1D array.
return row * m + col;
}
static int[] mapTo2D(int idx, int n, int m) {
// Inverse of what the one above does.
int[] rnc = new int[2];
rnc[0] = idx / m;
rnc[1] = idx % m;
return rnc;
}
static double distance(Point p1, Point p2) {
return Math.sqrt(Math.pow(p2.y-p1.y, 2) + Math.pow(p2.x-p1.x, 2));
}
static Map<Integer, Integer> smolNumPrimeFactorization(int num) {
if (smallestFactorOf == null)
primeGenerator(200001);
CountMap<Integer> fnps = new CountMap<>();
while (num != 1) {
fnps.putCM(smallestFactorOf[num]);
num /= smallestFactorOf[num];
}
return fnps;
}
static HashMap<Long, Integer> primeFactorization(long num) {
// Returns map of factor and its power in the number.
HashMap<Long, Integer> map = new HashMap<>();
while (num % 2 == 0) {
num /= 2;
Integer pwrCnt = map.get(2L);
map.put(2L, pwrCnt != null ? pwrCnt + 1 : 1);
}
for (long i = 3; i * i <= num; i += 2) {
while (num % i == 0) {
num /= i;
Integer pwrCnt = map.get(i);
map.put(i, pwrCnt != null ? pwrCnt + 1 : 1);
}
}
// If the number is prime, we have to add it to the
// map.
if (num != 1)
map.put(num, 1);
return map;
}
static HashSet<Long> divisors(long num) {
HashSet<Long> divisors = new HashSet<Long>();
divisors.add(1L);
divisors.add(num);
for (long i = 2; i * i <= num; i++) {
if (num % i == 0) {
divisors.add(num/i);
divisors.add(i);
}
}
return divisors;
}
static int bsearch(int[] arr, int val, int lo, int hi) {
// Returns the index of the first element
// larger than or equal to val.
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
idx = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
return idx;
}
static int bsearch(long[] arr, long val, int lo, int hi) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
idx = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
return idx;
}
static int bsearch(long[] arr, long val, int lo, int hi, boolean sMode) {
// Returns the index of the last element
// smaller than or equal to val.
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
hi = mid - 1;
} else {
idx = mid;
lo = mid + 1;
}
}
return idx;
}
static int bsearch(int[] arr, long val, int lo, int hi, boolean sMode) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] > val) {
hi = mid - 1;
} else {
idx = mid;
lo = mid + 1;
}
}
return idx;
}
static long factorial(long n) {
if (n <= 1)
return 1;
long factorial = 1;
for (int i = 1; i <= n; i++)
factorial = mod(factorial * i);
return factorial;
}
static long factorialInDivision(long a, long b) {
if (a == b)
return 1;
if (b < a) {
long temp = a;
a = b;
b = temp;
}
long factorial = 1;
for (long i = a + 1; i <= b; i++)
factorial = mod(factorial * i);
return factorial;
}
static long nCr(long n, long r, long[] fac) {
long p = 998244353;
// Base case
if (r == 0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
/*long fac[] = new long[(int)n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i % p;*/
return (fac[(int)n] * modInverse(fac[(int)r], p) % p
* modInverse(fac[(int)n - (int)r], p) % p) % p;
}
static long modInverse(long n, long p) {
return power(n, p - 2, p);
}
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
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1)==1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static long nPr(long n, long r) {
return factorialInDivision(n, n - r);
}
static int logk(long n, long k) {
return (int)(Math.log(n) / Math.log(k));
}
static boolean[] primeGenerator(int upto) {
// Sieve of Eratosthenes:
isPrime = new boolean[upto + 1];
smallestFactorOf = new int[upto + 1];
Arrays.fill(smallestFactorOf, 1);
Arrays.fill(isPrime, true);
isPrime[1] = isPrime[0] = false;
for (long i = 2; i < upto + 1; i++)
if (isPrime[(int) i]) {
smallestFactorOf[(int) i] = (int) i;
// Mark all the multiples greater than or equal
// to the square of i to be false.
for (long j = i; j * i < upto + 1; j++) {
if (isPrime[(int) j * (int) i]) {
isPrime[(int) j * (int) i] = false;
smallestFactorOf[(int) j * (int) i] = (int) i;
}
}
}
return isPrime;
}
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 gcd(long[] arr) {
int n = arr.length;
long gcd = arr[0];
for (int i = 1; i < n; i++) {
gcd = gcd(gcd, arr[i]);
}
return gcd;
}
static int gcd(int[] arr) {
int n = arr.length;
int gcd = arr[0];
for (int i = 1; i < n; i++) {
gcd = gcd(gcd, arr[i]);
}
return gcd;
}
static long lcm(long[] arr) {
long lcm = arr[0];
int n = arr.length;
for (int i = 1; i < n; i++) {
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
}
return lcm;
}
static long lcm(long a, long b) {
return (a * b)/gcd(a, b);
}
static boolean less(int a, int b) {
return a < b ? true : false;
}
static boolean isSorted(int[] a) {
for (int i = 1; i < a.length; i++) {
if (less(a[i], a[i - 1]))
return false;
}
return true;
}
static boolean isSorted(long[] a) {
for (int i = 1; i < a.length; i++) {
if (a[i] <= a[i - 1])
return false;
}
return true;
}
static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(long[] a, int i, int j) {
long temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(double[] a, int i, int j) {
double temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void sort(int[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(char[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(long[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(double[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void reverseSort(int[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(char[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverse(char[] arr) {
int n = arr.length;
for (int i = 0; i < n / 2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(long[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(double[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void shuffleArray(long[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
long tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(int[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
int tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(double[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
double tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(char[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
char tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
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 (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static String toString(int[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(boolean[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(long[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(char[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + "");
return sb.toString();
}
static String toString(int[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(long[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(double[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(char[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static long mod(long a, long m) {
return (a%m + m) % m;
}
static long mod(long num) {
return (num % gigamod + gigamod) % gigamod;
}
}
// NOTES:
// ASCII VALUE OF 'A': 65
// ASCII VALUE OF 'a': 97
// Range of long: 9 * 10^18
// ASCII VALUE OF '0': 48
// Primes upto 'n' can be given by (n / (logn)). | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 8ec4704540b388f65b14ba2632ce5399 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
import java.util.StringTokenizer;
public class Dict implements Runnable {
public static void main(String [] args) {
new Thread(null, new Dict(), "whatever", 1<<26).start();
}
public void run() {
FastScanner scanner = new FastScanner(System.in);
StringBuilder answers = new StringBuilder();
int tests = scanner.nextInt();
for (int t = 0; t < tests; t++) {
char[] input = scanner.next().toCharArray();
int offset = (input[0] < input[1]) ? 0 : 1;
answers.append(((input[0] - 'a') * 25) + (input[1] - 'a') + offset + "\n");
}
System.out.println(answers);
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(Reader in) {
br = new BufferedReader(in);
}
public FastScanner(InputStream in) {
this(new InputStreamReader(in));
}
public 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());}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 24942fd718ad7d59850faff5f1984277 | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public Main() {
FastScanner input = new FastScanner(System.in);
StringBuilder output = new StringBuilder();
int t = input.nextInt();
for (int i = 0; i < t; i++) {
String next = input.next();
char fChar = next.charAt(0);
char sChar = next.charAt(1);
int first = (fChar - 'a') * 26;
int second = sChar - 'a';
int subtract = fChar - 'a';
if (sChar < fChar) {
subtract--;
}
output.append(first + second - subtract + "\n");
}
System.out.println(output);
}
public static void main(String[] args) {
new Main();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(Reader in) {
br = new BufferedReader(in);
}
public FastScanner(InputStream in) {
this(new InputStreamReader(in));
}
public 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());}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | c4669f66b8cc262a7100d24eef87563a | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.io.*;
import java.util.*;
/*
*/
public class B{
static FastReader sc=null;
public static void main(String[] args) {
sc=new FastReader();
int t=sc.nextInt();
for(int tt=0;tt<t;tt++) {
char a[]=sc.next().toCharArray();
int pos=(a[0]-'a')*26+(a[1]-'a');
int sub=a[0]-'a'+(a[1]>a[0]?1:0);
System.out.println(pos-sub+1);
}
}
static int[] ruffleSort(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);
return a;
}
static void print(int a[]) {
for(int e:a) {
System.out.print(e+" ");
}
System.out.println();
}
static class FastReader{
StringTokenizer st=new StringTokenizer("");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
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());
}
long nextLong() {
return Long.parseLong(next());
}
int[] readArray(int n) {
int a[]=new int[n];
for(int i=0;i<n;i++)a[i]=sc.nextInt();
return a;
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | c50099c8c4b409c63a10122d974bd52f | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes |
import java.util.*;
import java.io.*;
public class B {
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;
}
}
public static void swap(long[] a, int i, int j) {
long temp = a[i];
a[i] = a[j];
a[j] = temp;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
FastReader t = new FastReader();
PrintWriter o = new PrintWriter(System.out);
int test = t.nextInt();
while (test-- > 0) {
String m = t.next();
long max = Integer.MIN_VALUE, min = Integer.MAX_VALUE;
// int n = t.nextInt();
// long[] a = new long[n];
// for (int i = 0; i < n; ++i) {
// a[i] = t.nextLong();
// }
long a = 0;
int i = m.charAt(0)-'a'+1;
int j = m.charAt(1)-'a'+1;
if (i>0)
a += (i-1)*25;
if (i<j)
{
a += (j-1);
}
else
a += j;
o.println(a);
}
o.flush();
o.close();
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | a36ce975e4e4bc0cb19435a6f5ee051f | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class Demo {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tc = sc.nextInt();
for (int itc=0; itc<tc; itc++) {
String s = sc.next();
char a = s.charAt(0);
char b = s.charAt(1);
int t = ((a-97)*26)+(b-97);
if (b>a && a>97) {
t--;
}
if (a>98) {
t-=(a-97-1);
}
System.out.println(t);
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | e452d8d3ebe986348e90a6a087a9888c | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.*;
public class dictionary{
public static void main(String [] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
while(t-->0){
String s = sc.nextLine();
int sum = (s.codePointAt(0)-97)*25;
if(s.codePointAt(1)<s.codePointAt(0)){
sum += s.codePointAt(1)-96;
}else{
sum += s.codePointAt(1)-97;
}
System.out.println(sum);
}
}
} | Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output | |
PASSED | 7114b28fdf9f1a30b54b2a8ee1cf60ad | train_107.jsonl | 1651502100 | The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word $$$a$$$ comes earlier than word $$$b$$$ in the dictionary if one of the following conditions hold: the first letter of $$$a$$$ is less than the first letter of $$$b$$$; the first letters of $$$a$$$ and $$$b$$$ are the same, and the second letter of $$$a$$$ is less than the second letter of $$$b$$$. So, the dictionary looks like that: Word $$$1$$$: ab Word $$$2$$$: ac ... Word $$$25$$$: az Word $$$26$$$: ba Word $$$27$$$: bc ... Word $$$649$$$: zx Word $$$650$$$: zy You are given a word $$$s$$$ from the Berland language. Your task is to find its index in the dictionary. | 512 megabytes | import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Dictionary {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
Map<Character, Integer> letters = new HashMap<>();
letters.put('a', 0);
letters.put('b', 1);
letters.put('c', 2);
letters.put('d', 3);
letters.put('e', 4);
letters.put('f', 5);
letters.put('g', 6);
letters.put('h', 7);
letters.put('i', 8);
letters.put('j', 9);
letters.put('k', 10);
letters.put('l', 11);
letters.put('m', 12);
letters.put('n', 13);
letters.put('o', 14);
letters.put('p', 15);
letters.put('q', 16);
letters.put('r', 17);
letters.put('s', 18);
letters.put('t', 19);
letters.put('u', 20);
letters.put('v', 21);
letters.put('w', 22);
letters.put('x', 23);
letters.put('y', 24);
letters.put('z', 25);
while(t != 0){
String word = sc.next();
Character first = word.charAt(0);
Character second = word.charAt(1);
int firstCharacterIndex = letters.get(first);
int secondCharacterIndex = letters.get(second);
int wordIndex = firstCharacterIndex * 25;
if(firstCharacterIndex < secondCharacterIndex){
wordIndex += secondCharacterIndex;
}else if(firstCharacterIndex > secondCharacterIndex){
wordIndex += secondCharacterIndex + 1;
}
System.out.println(wordIndex);
t -= 1;
}
}
}
| Java | ["7\n\nab\n\nac\n\naz\n\nba\n\nbc\n\nzx\n\nzy"] | 2 seconds | ["1\n2\n25\n26\n27\n649\n650"] | null | Java 8 | standard input | [
"combinatorics",
"math"
] | 2e3006d663a3c7ad3781aba1e37be3ca | The first line contains one integer $$$t$$$ ($$$1 \le t \le 650$$$) — the number of test cases. Each test case consists of one line containing $$$s$$$ — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language). | 800 | For each test case, print one integer — the index of the word $$$s$$$ in the dictionary. | standard output |
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