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 | 56ca7d6b26115533bca1903b6ae14fed | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.Arrays;
public class B {
static BufferedReader ins = new BufferedReader(new InputStreamReader(System.in));
static StreamTokenizer in = new StreamTokenizer(ins);
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
static int[] a = new int[200000];
static int n,m;
public static void main(String[] args)throws IOException {
int t =Integer.parseInt(ins.readLine());
while(t-->0){
Arrays.fill(a,0);
int n = Integer.parseInt(ins.readLine());
String ss = ins.readLine();
int res=0,ai=0;
int l=-1;
a[0]=1;
for (int i = 1; i <n; i++) {
if(ss.charAt(i)==ss.charAt(i-1))a[ai]++;
else{
ai++;
a[ai]=1;
}
}
for (int i = 0; i <= ai; i++) {
if(a[i]%2!=0){
if(l!=-1){
res = res+i-l;
l=-1;
}
else{
l=i;
}
}
}
out.println(res);
}
out.flush();
}
public static int in()throws IOException{
in.nextToken();
return (int)in.nval;
}
public static long inl()throws IOException{
in.nextToken();
return (long)in.nval;
}
public static double ind()throws IOException{
in.nextToken();
return in.nval;
}
public static String ins()throws IOException{
in.nextToken();
return in.sval;
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 8d47413d8d13d0a1bad20121bacaf550 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B1_Tokitsukaze_and_Good_01_String_easy_version {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int testCases, n;
static char a[];
/*
10
1110011000
-> 11 00 11 11 00
*/
static void solve(int index, ArrayList<Integer> list, int len) {
if (index >= n) {
return;
}
len = 1;
while (index + 1 < n && a[index] == a[index + 1]) {
++len;
++index;
}
list.add(len);
solve(index + 1, list, len);
}
static void solve(int t) {
ArrayList<Integer> list = new ArrayList<>();
solve(0, list, 1);
long arr[] = new long[list.size()];
int index = 0;
while (!list.isEmpty()) {
arr[index++] = list.get(0);
list.popFront();
}
int ans1 = 0, m = arr.length;
for (int i = 0; i < m - 1; i++) {
if (arr[i] % 2 == arr[i + 1] % 2 && arr[i] % 2 == 1) {
arr[i]--;
arr[i + 1]++;
ans1++;
} else if (arr[i] % 2 != arr[i + 1] % 2 && arr[i + 1] % 2 == 0) {
arr[i]--;
arr[i + 1]++;
ans1++;
}
}
ans.append(ans1);
if (t != testCases) {
ans.append("\n");
}
}
public static void main(String[] amit) throws IOException {
testCases = in.nextInt();
for (int t = 0; t < testCases; ++t) {
n = in.nextInt();
a = in.next().toCharArray();
solve(t + 1);
}
out.print(ans.toString());
out.flush();
in.close();
}
static boolean isSmaller(String str1, String str2) {
int n1 = str1.length(), n2 = str2.length();
if (n1 < n2) {
return true;
}
if (n2 < n1) {
return false;
}
for (int i = 0; i < n1; i++) {
if (str1.charAt(i) < str2.charAt(i)) {
return true;
} else if (str1.charAt(i) > str2.charAt(i)) {
return false;
}
}
return false;
}
static String sub(String str1, String str2) {
if (isSmaller(str1, str2)) {
String t = str1;
str1 = str2;
str2 = t;
}
String str = "";
int n1 = str1.length(), n2 = str2.length();
int diff = n1 - n2;
int carry = 0;
for (int i = n2 - 1; i >= 0; i--) {
int sub
= (((int) str1.charAt(i + diff) - (int) '0')
- ((int) str2.charAt(i) - (int) '0')
- carry);
if (sub < 0) {
sub = sub + 10;
carry = 1;
} else {
carry = 0;
}
str += String.valueOf(sub);
}
for (int i = n1 - n2 - 1; i >= 0; i--) {
if (str1.charAt(i) == '0' && carry > 0) {
str += "9";
continue;
}
int sub = (((int) str1.charAt(i) - (int) '0')
- carry);
if (i > 0 || sub > 0) {
str += String.valueOf(sub);
}
carry = 0;
}
return new StringBuilder(str).reverse().toString();
}
static String sum(String str1, String str2) {
if (str1.length() > str2.length()) {
String t = str1;
str1 = str2;
str2 = t;
}
String str = "";
int n1 = str1.length(), n2 = str2.length();
int diff = n2 - n1;
int carry = 0;
for (int i = n1 - 1; i >= 0; i--) {
int sum = ((int) (str1.charAt(i) - '0')
+ (int) (str2.charAt(i + diff) - '0') + carry);
str += (char) (sum % 10 + '0');
carry = sum / 10;
}
for (int i = n2 - n1 - 1; i >= 0; i--) {
int sum = ((int) (str2.charAt(i) - '0') + carry);
str += (char) (sum % 10 + '0');
carry = sum / 10;
}
if (carry > 0) {
str += (char) (carry + '0');
}
return new StringBuilder(str).reverse().toString();
}
static long detect_sum(int i, long a[], long sum) {
if (i >= a.length) {
return sum;
}
return detect_sum(i + 1, a, sum + a[i]);
}
static String mul(String num1, String num2) {
int len1 = num1.length();
int len2 = num2.length();
if (len1 == 0 || len2 == 0) {
return "0";
}
int result[] = new int[len1 + len2];
int i_n1 = 0;
int i_n2 = 0;
for (int i = len1 - 1; i >= 0; i--) {
int carry = 0;
int n1 = num1.charAt(i) - '0';
i_n2 = 0;
for (int j = len2 - 1; j >= 0; j--) {
int n2 = num2.charAt(j) - '0';
int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
carry = sum / 10;
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
if (carry > 0) {
result[i_n1 + i_n2] += carry;
}
i_n1++;
}
int i = result.length - 1;
while (i >= 0 && result[i] == 0) {
i--;
}
if (i == -1) {
return "0";
}
String s = "";
while (i >= 0) {
s += (result[i--]);
}
return s;
}
static class Node<T> {
T data;
Node<T> next;
public Node() {
this.next = null;
}
public Node(T data) {
this.data = data;
this.next = null;
}
public T getData() {
return data;
}
public void setData(T data) {
this.data = data;
}
public Node<T> getNext() {
return next;
}
public void setNext(Node<T> next) {
this.next = next;
}
@Override
public String toString() {
return this.getData().toString() + " ";
}
}
static class ArrayList<T> {
Node<T> head, tail;
int len;
public ArrayList() {
this.head = null;
this.tail = null;
this.len = 0;
}
int size() {
return len;
}
boolean isEmpty() {
return len == 0;
}
int indexOf(T data) {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
int index = -1, i = 0;
while (temp != null) {
if (temp.getData() == data) {
index = i;
}
i++;
temp = temp.getNext();
}
return index;
}
void add(T data) {
Node<T> newNode = new Node<>(data);
if (isEmpty()) {
head = newNode;
tail = newNode;
len++;
} else {
tail.setNext(newNode);
tail = newNode;
len++;
}
}
void see() {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
while (temp != null) {
out.print(temp.getData().toString() + " ");
out.flush();
temp = temp.getNext();
}
out.println();
out.flush();
}
void inserFirst(T data) {
Node<T> newNode = new Node<>(data);
Node<T> temp = head;
if (isEmpty()) {
head = newNode;
tail = newNode;
len++;
} else {
newNode.setNext(temp);
head = newNode;
len++;
}
}
T get(int index) {
if (isEmpty() || index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
if(index == 0) {
return head.getData();
}
Node<T> temp = head;
int i = 0;
T data = null;
while (temp != null) {
if (i == index) {
data = temp.getData();
}
i++;
temp = temp.getNext();
}
return data;
}
void addAt(T data, int index) {
if (index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> newNode = new Node<>(data);
int i = 0;
Node<T> temp = head;
while (temp.next != null) {
if (i == index) {
newNode.setNext(temp.next);
temp.next = newNode;
}
i++;
temp = temp.getNext();
}
// temp.setNext(temp);
len++;
}
void popFront() {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
if (head == tail) {
head = null;
tail = null;
} else {
head = head.getNext();
}
len--;
}
void removeAt(int index) {
if (index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
if (index == 0) {
this.popFront();
return;
}
Node<T> temp = head;
int i = 0;
Node<T> n = new Node<>();
while (temp != null) {
if (i == index) {
n.next = temp.next;
temp.next = n;
break;
}
i++;
n = temp;
temp = temp.getNext();
}
tail = n;
--len;
}
void clearAll() {
this.head = null;
this.tail = null;
}
}
static void merge(long a[], int left, int right, int mid) {
int n1 = mid - left + 1, n2 = right - mid;
long L[] = new long[n1];
long R[] = new long[n2];
for (int i = 0; i < n1; i++) {
L[i] = a[left + i];
}
for (int i = 0; i < n2; i++) {
R[i] = a[mid + 1 + i];
}
int i = 0, j = 0, k1 = left;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
a[k1] = L[i];
i++;
} else {
a[k1] = R[j];
j++;
}
k1++;
}
while (i < n1) {
a[k1] = L[i];
i++;
k1++;
}
while (j < n2) {
a[k1] = R[j];
j++;
k1++;
}
}
static void sort(long a[], int left, int right) {
if (left >= right) {
return;
}
int mid = (left + right) / 2;
sort(a, left, mid);
sort(a, mid + 1, right);
merge(a, left, right, mid);
}
static class Scanner {
BufferedReader in;
StringTokenizer st;
public Scanner() {
in = new BufferedReader(new InputStreamReader(System.in));
}
String next() throws IOException {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
String nextLine() throws IOException {
return in.readLine();
}
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 {
in.close();
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 7e29430449b976b3dfb01c0d534c0a27 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution{
public static void main(String[] args) throws IOException {
BufferedReader re = new BufferedReader(new InputStreamReader(System.in));
int testCase = Integer.parseInt(re.readLine());
StringBuilder ans = new StringBuilder();
for(int t = 1; t <= testCase; t++){
int n = Integer.parseInt(re.readLine());
String s = re.readLine();
char prev = s.charAt(0);
int answer = 0;
int count = 1;
for(int i = 1; i < n; i++){
if(prev != s.charAt(i)){
if(count % 2 == 0){
count = 1;
}else{
answer++;
count = 2;
}
}else {
count++;
}
prev = s.charAt(i);
}
ans.append(answer).append('\n');
}
System.out.print(ans);
re.close();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | f10b37b391bc1dbd7932e1e03319b932 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.*;
import java.util.*;
public class Main {
static long cr[][]=new long[1001][1001];
//static double EPS = 1e-7;
static long mod=998244353;
static long val=0;
static boolean flag=true;
public static void main(String[] args)
{
FScanner sc = new FScanner();
//Arrays.fill(prime, true);
//sieve();
//ncr();
int t=sc.nextInt();
StringBuilder sb = new StringBuilder();
while(t-->0)
{
int n=sc.nextInt();
String s=sc.next();
s=s+"#";
char c[]=s.toCharArray();
int count=1;
int ans=0;
ArrayList<Integer> A=new ArrayList<>();
for(int i=0;i<n;i++)
{
if(c[i]!=c[i+1])
{
if(count%2==1)
A.add(ans);
ans++;
count=1;
}
else
count++;
}
int res=0;
for(int i=1;i<A.size();i=i+2)
{
res=res+A.get(i)-A.get(i-1);
}
sb.append(res);
sb.append("\n");
}
System.out.println(sb.toString());
}
public static int solve(char c[],int i,int n,char prev,int dp[][])
{
if(i>=n)
return 0;
if(dp[i][prev-'a']!=-1)
return dp[i][prev-'a'];
int res=0;
if(c[i]==prev)
{ if(i+1>=n)
{res=2;
}
else
res= 2+solve(c,i+2,n,c[i+1],dp);
}
else
{
res= Math.max(solve(c,i+1,n,prev,dp),solve(c,i+1,n,c[i],dp));
}
dp[i][prev-'a']=res;
return res;
}
public static long powerLL(long x, long n)
{
long result = 1;
while (n > 0)
{
if (n % 2 == 1)
{
result = result * x % mod;
}
n = n / 2;
x = x * x % mod;
}
return result;
}
public static long gcd(long a,long b)
{
return b==0 ? a:gcd(b,a%b);
}
/* public static void sieve()
{ prime[0]=prime[1]=false;
int n=1000000;
for(int p = 2; p*p <=n; p++)
{
if(prime[p] == true)
{
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
*/
public static void ncr()
{
cr[0][0]=1;
for(int i=1;i<=1000;i++)
{
cr[i][0]=1;
for(int j=1;j<i;j++)
{
cr[i][j]=(cr[i-1][j-1]+cr[i-1][j])%mod;
}
cr[i][i]=1;
}
}
}
class pair
//implements Comparable<pair>
{
long a;long b;
pair(long a,long b)
{
this.b=b;
this.a=a;
}
}
class FScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer sb = new StringTokenizer("");
String next(){
while(!sb.hasMoreTokens()){
try{
sb = new StringTokenizer(br.readLine());
} catch(IOException e){ }
}
return sb.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
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] = nextInt();
return a;
}
float nextFloat(){
return Float.parseFloat(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | acde84c7ef989fcadb04268da7521ef0 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.function.BinaryOperator;
import java.util.stream.Collectors;
public class Main {
private final static long mod = 1000000007;
private final static int MAXN = 1000001;
private static long power(long x, long y, long m) {
long temp;
if (y == 0)
return 1;
temp = power(x, y / 2, m);
temp = (temp * temp) % m;
if (y % 2 == 0)
return temp;
else
return ((x % m) * temp) % m;
}
private static long power(long x, long y) {
return power(x, y, mod);
}
private static long gcd(long a, long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
static int nextPowerOf2(int a) {
return 1 << nextLog2(a);
}
static int nextLog2(int a) {
return (a == 0 ? 0 : 32 - Integer.numberOfLeadingZeros(a - 1));
}
private static long modInverse(long a, long m) {
long m0 = m;
long y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1) {
long q = a / m;
long t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
private static int[] getLogArr(int n) {
int arr[] = new int[n + 1];
for (int i = 1; i < n + 1; i++) {
arr[i] = (int) (Math.log(i) / Math.log(2) + 1e-10);
}
return arr;
}
private static int log[] = getLogArr(MAXN);
private static int getLRSpt(int st[][], int L, int R, BinaryOperator<Integer> binaryOperator) {
int j = log[R - L + 1];
return binaryOperator.apply(st[L][j], st[R - (1 << j) + 1][j]);
}
private static int[][] getSparseTable(int array[], BinaryOperator<Integer> binaryOperator) {
int k = log[array.length + 1] + 1;
int st[][] = new int[array.length + 1][k + 1];
for (int i = 0; i < array.length; i++)
st[i][0] = array[i];
for (int j = 1; j <= k; j++) {
for (int i = 0; i + (1 << j) <= array.length; i++) {
st[i][j] = binaryOperator.apply(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
}
return st;
}
private static int[][] getSparseTable(List<Integer> list, BinaryOperator<Integer> binaryOperator) {
int n=list.size();
int k = log[n + 1] + 1;
int st[][] = new int[n + 1][k + 1];
for (int i = 0; i < n; i++)
st[i][0] = list.get(i);
for (int j = 1; j <= k; j++) {
for (int i = 0; i + (1 << j) <= n; i++) {
st[i][j] = binaryOperator.apply(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
}
return st;
}
static class Subset {
int parent;
int rank;
@Override
public String toString() {
return "" + parent;
}
}
static int find(Subset[] subsets, int i) {
if (subsets[i].parent != i)
subsets[i].parent = find(subsets, subsets[i].parent);
return subsets[i].parent;
}
static void union(Subset[] Subsets, int x, int y) {
int xroot = find(Subsets, x);
int yroot = find(Subsets, y);
if (Subsets[xroot].rank < Subsets[yroot].rank)
Subsets[xroot].parent = yroot;
else if (Subsets[yroot].rank < Subsets[xroot].rank)
Subsets[yroot].parent = xroot;
else {
Subsets[xroot].parent = yroot;
Subsets[yroot].rank++;
}
}
private static int maxx(Integer... a) {
return Collections.max(Arrays.asList(a));
}
private static int minn(Integer... a) {
return Collections.min(Arrays.asList(a));
}
private static long maxx(Long... a) {
return Collections.max(Arrays.asList(a));
}
private static long minn(Long... a) {
return Collections.min(Arrays.asList(a));
}
private static class Pair<T extends Comparable<T>, U extends Comparable<U>> implements Comparable<Pair<T, U>> {
T a;
U b;
public Pair(T a, U b) {
this.a = a;
this.b = b;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair pair = (Pair) o;
return a.equals(pair.a) &&
b.equals(pair.b);
}
@Override
public int hashCode() {
return Objects.hash(a, b);
}
@Override
public String toString() {
return "(" + a +
"," + b +
')';
}
@Override
public int compareTo(Pair<T, U> o) {
return (this.a.equals(o.a) ? this.b.equals(o.b) ? 0 : this.b.compareTo(o.b) : this.a.compareTo(o.a));
}
}
public static int upperBound(List<Integer> list, int value) {
int low = 0;
int high = list.size();
while (low < high) {
final int mid = (low + high) / 2;
if (value >= list.get(mid)) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
private static int[] getLPSArray(String pattern) {
int i, j, n = pattern.length();
int[] lps = new int[pattern.length()];
lps[0] = 0;
for (i = 1, j = 0; i < n; ) {
if (pattern.charAt(i) == pattern.charAt(j)) {
lps[i++] = ++j;
} else if (j > 0) {
j = lps[j - 1];
} else {
lps[i++] = 0;
}
}
return lps;
}
private static List<Integer> findPattern(String text, String pattern) {
List<Integer> matchedIndexes = new ArrayList<>();
if (pattern.length() == 0) {
return matchedIndexes;
}
int[] lps = getLPSArray(pattern);
int i = 0, j = 0, n = text.length(), m = pattern.length();
while (i < n) {
if (text.charAt(i) == pattern.charAt(j)) {
i++;
j++;
}
if (j == m) {
matchedIndexes.add(i - j);
j = lps[j - 1];
}
if (i < n && text.charAt(i) != pattern.charAt(j)) {
if (j > 0) {
j = lps[j - 1];
} else {
i++;
}
}
}
return matchedIndexes;
}
private static Set<Long> getDivisors(long n) {
Set<Long> divisors = new HashSet<>();
divisors.add(1L);
divisors.add(n);
for (long i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
divisors.add(i);
divisors.add(n / i);
}
}
return divisors;
}
private static long getLCM(long a, long b) {
return (Math.max(a, b) / gcd(a, b)) * Math.min(a, b);
}
static long fac[] = new long[2000005];
static long ifac[] = new long[2000005];
private static void preCompute(int n) {
fac = new long[n + 1];
ifac = new long[n + 1];
fac[0] = ifac[0] = fac[1] = ifac[1] = 1;
int i;
for (i = 2; i < n + 1; i++) {
fac[i] = (i * fac[i - 1]) % mod;
ifac[i] = (power(i, mod - 2) * ifac[i - 1]) % mod;
}
}
private static long C(int n, int r) {
if (n < 0 || r < 0) return 1;
if (r > n) return 1;
return (fac[n] * ((ifac[r] * ifac[n - r]) % mod)) % mod;
}
static long getSum(long BITree[], int index) {
long sum = 0;
index = index + 1;
while (index > 0) {
index -= index & (-index);
}
return sum;
}
public static void updateBIT(long BITree[], int index, long val) {
index = index + 1;
while (index <= BITree.length - 1) {
BITree[index] += val;
index += index & (-index);
}
}
long[] constructBITree(int arr[], int m) {
int n = arr.length;
long BITree[] = new long[m + 1];
for (int i = 1; i <= n; i++)
BITree[i] = 0;
for (int i = 0; i < n; i++)
updateBIT(BITree, i, arr[i]);
return BITree;
}
private static int upperBound(int a[], int l, int r, int x) {
if (l > r) return -1;
if (l == r) {
if (a[l] <= x) {
return -1;
} else {
return a[l];
}
}
int m = (l + r) / 2;
if (a[m] <= x) {
return upperBound(a, m + 1, r, x);
} else {
return upperBound(a, l, m, x);
}
}
private static void mul(long A[][], long B[][]) {
int i, j, k, n = A.length;
long C[][] = new long[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
for (k = 0; k < n; k++) {
C[i][j] = (C[i][j] + (A[i][k] * B[k][j]) % mod) % mod;
}
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
A[i][j] = C[i][j];
}
}
}
private static void power(long A[][], long base[][], long n) {
if (n < 2) {
return;
}
power(A, base, n / 2);
mul(A, A);
if (n % 2 == 1) {
mul(A, base);
}
}
static void reverse(int a[], int l, int r) {
while (l < r) {
int t = a[l];
a[l] = a[r];
a[r] = t;
l++;
r--;
}
}
public void rotate(int[] nums, int k) {
k = k % nums.length;
reverse(nums, 0, nums.length - k - 1);
reverse(nums, nums.length - k, nums.length - 1);
reverse(nums, 0, nums.length - 1);
}
private 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 void main(String[] args) throws Exception {
long START_TIME = System.currentTimeMillis();
try (FastReader in = new FastReader();
FastWriter out = new FastWriter()) {
int n, t, i, j, f, k, m, ti, tidx, gm, q, g, l, r;
for (t = in.nextInt(), tidx = 1; tidx <= t; tidx++)
{
long x = 0, y = 0, z = 0, sum = 0, bhc = 0, ans = 0;
//out.print(String.format("Case #%d: ", tidx));
n=in.nextInt();
String s=in.next();
List<Integer>list=new ArrayList<>();
int cnt=0;
char p=s.charAt(0);
for(i=0;i<s.length();i++){
if(s.charAt(i)==p){
cnt++;
} else {
p=s.charAt(i);
list.add(cnt);
cnt=1;
}
}
list.add(cnt);
Integer a[]=list.toArray(new Integer[0]);
for(i=0;i<a.length-1;i++){
if(a[i]%2==1){
a[i]--;
a[i+1]++;
ans++;
}
}
out.println(ans);
}
/*if (args.length > 0 && "ex_time".equals(args[0])) {
out.print("\nTime taken: ");
out.println(System.currentTimeMillis() - START_TIME);
}*/
out.commit();
}
}
public static class FastReader implements Closeable {
private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer st;
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception 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[] nextIntArr(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
Integer[] nextIntegerArr(int n) {
Integer[] arr = new Integer[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
double[] nextDoubleArr(int n) {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = nextDouble();
}
return arr;
}
long[] nextLongArr(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextLong();
}
return arr;
}
List<Long> nextLongList(int n) {
List<Long> retList = new ArrayList<>();
for (int i = 0; i < n; i++) {
retList.add(nextLong());
}
return retList;
}
String[] nextStrArr(int n) {
String[] arr = new String[n];
for (int i = 0; i < n; i++) {
arr[i] = next();
}
return arr;
}
int[][] nextIntArr2(int n, int m) {
int[][] arr = new int[n][m];
for (int i = 0; i < n; i++) {
arr[i] = nextIntArr(m);
}
return arr;
}
long[][] nextLongArr2(int n, int m) {
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++) {
arr[i] = nextLongArr(m);
}
return arr;
}
@Override
public void close() throws IOException {
br.close();
}
}
public static class FastWriter implements Closeable {
BufferedWriter bw;
StringBuilder sb = new StringBuilder();
List<String> list = new ArrayList<>();
Set<String> set = new HashSet<>();
public FastWriter() {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
<T> void commit() throws IOException {
bw.write(sb.toString());
bw.flush();
sb = new StringBuilder();
}
public <T> void print(T obj) throws IOException {
sb.append(obj.toString());
commit();
}
public void println() throws IOException {
print("\n");
}
public <T> void println(T obj) throws IOException {
print(obj.toString() + "\n");
}
<T> void printArrLn(T[] arr) throws IOException {
for (int i = 0; i < arr.length - 1; i++) {
print(arr[i] + " ");
}
println(arr[arr.length - 1]);
}
<T> void printArr2(T[][] arr) throws IOException {
for (int j = 0; j < arr.length; j++) {
for (int i = 0; i < arr[j].length - 1; i++) {
print(arr[j][i] + " ");
}
println(arr[j][arr[j].length - 1]);
}
}
<T> void printColl(Collection<T> coll) throws IOException {
List<String> stringList = coll.stream().map(e -> ""+e).collect(Collectors.toList());
println(String.format("{%s}",String.join(",", stringList)));
}
void printCharN(char c, int n) throws IOException {
for (int i = 0; i < n; i++) {
print(c);
}
}
void printIntArr(int []arr) throws IOException {
for (int i = 0; i < arr.length - 1; i++) {
print(arr[i] + " ");
}
println(arr[arr.length - 1]);
}
void printIntArr2(int[][] arr) throws IOException {
for (int j = 0; j < arr.length; j++) {
for (int i = 0; i < arr[j].length - 1; i++) {
print(arr[j][i] + " ");
}
println(arr[j][arr[j].length - 1]);
}
}
@Override
public void close() throws IOException {
bw.close();
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 0937055a4b6f4a98b0ad5203bf5e61e2 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | // Online Java Compiler
// Use this editor to write, compile and run your Java code online
import java.util.*;
public class HelloWorld {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t--!=0){
int n=sc.nextInt();
sc.nextLine();
String val = sc.nextLine();
char ch='*';
int ans=0,cnt=0;
for(int i=0;i<n;i++){
if(val.charAt(i)!=ch){
if(ch=='*'){
ch=val.charAt(i);
cnt++;
}else{
if(cnt%2!=0){
ans++;
ch='*';
cnt=0;
}else{
cnt=1;
ch=val.charAt(i);
}
}
}else{
cnt++;
}
}
System.out.println(ans);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 1ec11a0c0404fcd05576f2b3e42cf086 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.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++) {
int n = fr.nextInt();
char[] s = fr.next().toCharArray();
char[] sorg = s.clone();
// determine two things:
//
// a. Min opers required to make 's' good
//
// b. Min segments that can be formed in
// these opers
// if it's a single element, we'd prefer
// to convert it to the other
int numOpers = 0;
for (int i = 0; i < n; i++) {
int j = i;
while (j < n - 1 && s[j + 1] == s[i])
j++;
// [i, j] is the segment
int len = j + 1 - i;
if (len % 2 == 0) {
i = j;
continue;
}
assert len % 2 != 0;
// we'll prefer changing the last element
s[j] -= '0';
s[j] ^= 1;
s[j] += '0';
// [i, j - 1] is the last segment,
// which has even length
numOpers++;
i = j - 1;
}
// in these opers,
out.println(numOpers);
}
out.close();
}
static class Query {
int type, idx;
long val;
Query(int tt, int ii, long xx) {
type = tt;
idx = ii;
val = xx;
}
Query(int tt, long xx) {
type = tt;
val = xx;
}
}
static class Tuple {
int val, idx, numPairs;
Tuple(int vv, int ii, int np) {
val = vv;
idx = ii;
numPairs = np;
}
}
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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | ab454c594abfefd319cad0876c6aaed8 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 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 Pair[] movesDiagonal = new Pair[]{new Pair(-1, -1), new Pair(-1, 1), new Pair(1, -1), new Pair(1, 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() {
int n = i();
String s = s();
int ans = 0;
for (int i = 0; i < n; i += 2) {
if (s.charAt(i) != s.charAt(i + 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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | d145ec06eac06d170942e2301cc477e1 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //import java.io.IOException;
import java.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.List;
public class TokitsukazeandGood {
static InputReader inputReader=new InputReader(System.in);
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static void solve() throws IOException
{
int count=0;
int n=inputReader.nextInt();
String str=inputReader.readString();
for (int i=1;i<n;i+=2)
{
if (str.charAt(i)!=str.charAt(i-1))
{
count++;
}
}out.println(count);
}
static PrintWriter out=new PrintWriter((System.out));
static void SortDec(long arr[])
{
List<Long>list=new ArrayList<>();
for(long ele:arr) {
list.add(ele);
}
Collections.sort(list,Collections.reverseOrder());
for (int i=0;i<list.size();i++)
{
arr[i]=list.get(i);
}
}
static void Sort(long arr[])
{
List<Long>list=new ArrayList<>();
for(long ele:arr) {
list.add(ele);
}
Collections.sort(list);
for (int i=0;i<list.size();i++)
{
arr[i]=list.get(i);
}
}
public static void main(String args[])throws IOException
{
int t=inputReader.nextInt();
while (t-->0) {
solve();
}
long s = System.currentTimeMillis();
// out.println(System.currentTimeMillis()-s+"ms");
out.close();
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c))
c = snext();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c))
c = snext();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | fbc7c40136af3ea32c6b31828662565c | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.io.ByteArrayOutputStream;
import java.io.BufferedWriter;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.FileNotFoundException;
import java.io.Writer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author tauros
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
RealFastReader in = new RealFastReader(inputStream);
RealFastWriter out = new RealFastWriter(outputStream);
CF789B1 solver = new CF789B1();
solver.solve(1, in, out);
out.close();
}
static class CF789B1 {
public void solve(int testNumber, RealFastReader in, RealFastWriter out) {
int cases = in.ni();
while (cases-- > 0) {
int n = in.ni();
char[] chars = in.ns(n);
int ans = 0;
for (int i = 0; i < n; i += 2) {
if (chars[i] != chars[i + 1]) {
ans++;
}
}
out.println(ans);
}
out.flush();
}
}
static class RealFastReader {
InputStream is;
private byte[] inbuf = new byte[1024];
public int lenbuf = 0;
public int ptrbuf = 0;
public RealFastReader(final 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++];
}
private boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b))
;
return b;
}
public char[] ns(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 int ni() {
return (int) nl();
}
public long nl() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
}
static class RealFastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private OutputStream out;
private Writer writer;
private int ptr = 0;
private RealFastWriter() {
out = null;
}
public RealFastWriter(Writer writer) {
this.writer = new BufferedWriter(writer);
out = new ByteArrayOutputStream();
}
public RealFastWriter(OutputStream os) {
this.out = os;
}
public RealFastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public RealFastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE) {
innerflush();
}
return this;
}
public RealFastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) {
innerflush();
}
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000) {
return 10;
}
if (l >= 100000000) {
return 9;
}
if (l >= 10000000) {
return 8;
}
if (l >= 1000000) {
return 7;
}
if (l >= 100000) {
return 6;
}
if (l >= 10000) {
return 5;
}
if (l >= 1000) {
return 4;
}
if (l >= 100) {
return 3;
}
if (l >= 10) {
return 2;
}
return 1;
}
public RealFastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE) {
innerflush();
}
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) {
return 19;
}
if (l >= 100000000000000000L) {
return 18;
}
if (l >= 10000000000000000L) {
return 17;
}
if (l >= 1000000000000000L) {
return 16;
}
if (l >= 100000000000000L) {
return 15;
}
if (l >= 10000000000000L) {
return 14;
}
if (l >= 1000000000000L) {
return 13;
}
if (l >= 100000000000L) {
return 12;
}
if (l >= 10000000000L) {
return 11;
}
if (l >= 1000000000L) {
return 10;
}
if (l >= 100000000L) {
return 9;
}
if (l >= 10000000L) {
return 8;
}
if (l >= 1000000L) {
return 7;
}
if (l >= 100000L) {
return 6;
}
if (l >= 10000L) {
return 5;
}
if (l >= 1000L) {
return 4;
}
if (l >= 100L) {
return 3;
}
if (l >= 10L) {
return 2;
}
return 1;
}
public RealFastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE) {
innerflush();
}
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public RealFastWriter writeln(int x) {
return write(x).writeln();
}
public RealFastWriter writeln() {
return write((byte) '\n');
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
if (writer != null) {
writer.write(((ByteArrayOutputStream) out).toString());
out = new ByteArrayOutputStream();
writer.flush();
} else {
out.flush();
}
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public RealFastWriter println(int x) {
return writeln(x);
}
public void close() {
flush();
try {
out.close();
} catch (Exception e) {
}
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 29014fbf5f339cbd2956e67c486568a8 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.io.ByteArrayOutputStream;
import java.io.BufferedWriter;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.FileNotFoundException;
import java.io.Writer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author tauros
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
RealFastReader in = new RealFastReader(inputStream);
RealFastWriter out = new RealFastWriter(outputStream);
CF789B1 solver = new CF789B1();
solver.solve(1, in, out);
out.close();
}
static class CF789B1 {
public void solve(int testNumber, RealFastReader in, RealFastWriter out) {
int cases = in.ni();
while (cases-- > 0) {
int n = in.ni();
char[] chars = in.ns(n);
int len = 0, eCnt = 0;
boolean hasO = false;
int ans = 0;
for (int i = 0; i < n; i++) {
len++;
if (i == n - 1 || chars[i + 1] != chars[i]) {
if (len % 2 == 1) {
if (!hasO) {
eCnt = 0;
hasO = true;
} else {
ans += eCnt + 1;
hasO = false;
}
} else {
eCnt++;
}
len = 0;
}
}
out.println(ans);
}
out.flush();
}
}
static class RealFastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private OutputStream out;
private Writer writer;
private int ptr = 0;
private RealFastWriter() {
out = null;
}
public RealFastWriter(Writer writer) {
this.writer = new BufferedWriter(writer);
out = new ByteArrayOutputStream();
}
public RealFastWriter(OutputStream os) {
this.out = os;
}
public RealFastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public RealFastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE) {
innerflush();
}
return this;
}
public RealFastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) {
innerflush();
}
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000) {
return 10;
}
if (l >= 100000000) {
return 9;
}
if (l >= 10000000) {
return 8;
}
if (l >= 1000000) {
return 7;
}
if (l >= 100000) {
return 6;
}
if (l >= 10000) {
return 5;
}
if (l >= 1000) {
return 4;
}
if (l >= 100) {
return 3;
}
if (l >= 10) {
return 2;
}
return 1;
}
public RealFastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE) {
innerflush();
}
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) {
return 19;
}
if (l >= 100000000000000000L) {
return 18;
}
if (l >= 10000000000000000L) {
return 17;
}
if (l >= 1000000000000000L) {
return 16;
}
if (l >= 100000000000000L) {
return 15;
}
if (l >= 10000000000000L) {
return 14;
}
if (l >= 1000000000000L) {
return 13;
}
if (l >= 100000000000L) {
return 12;
}
if (l >= 10000000000L) {
return 11;
}
if (l >= 1000000000L) {
return 10;
}
if (l >= 100000000L) {
return 9;
}
if (l >= 10000000L) {
return 8;
}
if (l >= 1000000L) {
return 7;
}
if (l >= 100000L) {
return 6;
}
if (l >= 10000L) {
return 5;
}
if (l >= 1000L) {
return 4;
}
if (l >= 100L) {
return 3;
}
if (l >= 10L) {
return 2;
}
return 1;
}
public RealFastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE) {
innerflush();
}
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public RealFastWriter writeln(int x) {
return write(x).writeln();
}
public RealFastWriter writeln() {
return write((byte) '\n');
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
if (writer != null) {
writer.write(((ByteArrayOutputStream) out).toString());
out = new ByteArrayOutputStream();
writer.flush();
} else {
out.flush();
}
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public RealFastWriter println(int x) {
return writeln(x);
}
public void close() {
flush();
try {
out.close();
} catch (Exception e) {
}
}
}
static class RealFastReader {
InputStream is;
private byte[] inbuf = new byte[1024];
public int lenbuf = 0;
public int ptrbuf = 0;
public RealFastReader(final 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++];
}
private boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b))
;
return b;
}
public char[] ns(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 int ni() {
return (int) nl();
}
public long nl() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 56901968c25d07c0f81fa774f364b399 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class k {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int res = 0;
String s = sc.next();
for (int i = 0;i<n;i+=2) {
if (s.charAt(i) != s.charAt(i+1)) {
res+=1;
}
}out.println(res);
}
out.close();
}
static boolean inc(int[] a) {
for (int i = 0; i < a.length-1; i++)
if (a[i] >= a[i+1]) return false;
return true;
}
static boolean decOrEq(int[] a) {
for (int i = 0; i < a.length-1; i++)
if (a[i] < a[i+1]) return false;
return true;
}
static final Random random = new Random();
static void shuffleSort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int r = random.nextInt(n), temp = a[r];
a[r] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static class 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();}
int[] readArray(int n) throws IOException {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 9c367bffe7bd6aa56a3ab1214ec80a82 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class main2{
public static void main(String [] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int count = 0;count<t;count++){
int n = sc.nextInt();
sc.nextLine();
String s = sc.nextLine();
boolean h = true;
int length = 1;
char c = s.charAt(0);
int total = 0;
for(int i = 1;i<s.length();i++){
if(c == s.charAt(i)){
length++;
}else{
if(length%2!=0){
if(total==0){
length = 2;
h = false;
c = s.charAt(i);
}else{
length++;
}
total++;
}else{
c = s.charAt(i);
length = 1;
}
}
}
if(h==true){
System.out.println(0);
}else{
System.out.println(total);
}
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | f1410578f60ef129877a7af601304adb | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
static PrintWriter pw;
static Scanner sc;
static pair[][] memo;
static int end;
static ArrayList<Integer> size;
static int inf = (int) 1e9;
static int get(int st) {
return st == 2 ? -1 : st;
}
static pair dp(int st, int i) {
if (i == end) {
return st == 0 ? new pair(0, 0) : new pair(inf, inf);
}
if (memo[st][i] != null)
return new pair(memo[st][i].x, memo[st][i].y);
int cur = size.get(i) + get(st);
if (cur % 2 == 0) {
pair p = dp(0, i + 1);
pair ans=new pair(p.x, p.y);
if (cur == 0)
ans.y--;
else
ans.y++;
// pw.println(st+" "+i+" "+ p);
return memo[st][i] = ans;
} else {
// pw.println(st + " " + i + " " + cur);
int y = cur - 1;
if (y == 0) {
pair p1 = dp(1, i + 1);
pair p2 = dp(2, i + 1);
p2.y++;
p1.x++;
p2.x++;
// pw.println(st+" "+i+" "+p1 + " " + p2);
return memo[st][i] = p1.merge(p2);
} else {
pair p1 = dp(1, i + 1);
pair p2 = dp(2, i + 1);
p1.y++;
p2.y++;
p1.x++;
p2.x++;
// pw.println(st+" "+i+" "+ p1 + " " + p2);
return memo[st][i] = p1.merge(p2);
}
}
}
public static void main(String[] args) throws Exception {
sc = new Scanner(System.in);
pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
char[] arr = sc.next().toCharArray();
char first = arr[0];
int cnt = 0;
size = new ArrayList<>();
for (int i = 0; i < n;) {
while (i < n && arr[i] == first) {
i++;
cnt++;
}
size.add(cnt);
if (i == n)
break;
first = arr[i];
cnt = 0;
}
end = size.size();
memo = new pair[3][end];
pw.println(dp(0, 0).x);
// pw.println(Arrays.deepToString(memo));
}
pw.flush();
}
static class pair {
int x, y;
public pair(int a, int b) {
x = a;
y = b;
}
public pair merge(pair p) {
if (x < p.x) {
return new pair(x, y);
} else if (p.x < x) {
return new pair(p.x, p.y);
} else if (y < p.y) {
return new pair(x, y);
}
return new pair(p.x, p.y);
}
public String toString() {
return x + " " + y;
}
}
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 Scanner(String s) throws IOException {
br = new BufferedReader(new FileReader(new File(s)));
}
public String readAllLines(BufferedReader reader) throws IOException {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line);
content.append(System.lineSeparator());
}
return content.toString();
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 3e6184c2869814e7cebcbf6af6159b1e | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//--------------------------INPUT READER---------------------------------//
static class fs {
public BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public fs() { this(System.in); }
public fs(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String ns() { return next(); }
int[] na(long nn) {
int n = (int) nn;
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
long[] nal(long nn) {
int n = (int) nn;
long[] l = new long[n];
for(int i = 0; i < n; i++) l[i] = nl();
return l;
}
}
//-----------------------------------------------------------------------//
//---------------------------PRINTER-------------------------------------//
static class Printer {
static PrintWriter w;
public Printer() {this(System.out);}
public Printer(OutputStream os) {
w = new PrintWriter(os);
}
public void p(int i) {w.println(i);}
public void p(long l) {w.println(l);}
public void p(double d) {w.println(d);}
public void p(String s) { w.println(s);}
public void pr(int i) {w.print(i);}
public void pr(long l) {w.print(l);}
public void pr(double d) {w.print(d);}
public void pr(String s) { w.print(s);}
public void pl() {w.println();}
public void close() {w.close();}
}
//-----------------------------------------------------------------------//
//--------------------------VARIABLES------------------------------------//
static fs sc = new fs();
static OutputStream outputStream = System.out;
static Printer w = new Printer(outputStream);
static long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
static int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
static long mod = 1000000007;
//-----------------------------------------------------------------------//
//--------------------------ADMIN_MODE-----------------------------------//
private static void ADMIN_MODE() throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
w = new Printer(new FileOutputStream("output.txt"));
sc = new fs(new FileInputStream("input.txt"));
}
}
//-----------------------------------------------------------------------//
//----------------------------START--------------------------------------//
public static void main(String[] args)
throws IOException {
ADMIN_MODE();
int t = sc.ni();while(t-->0)
solve();
w.close();
}
static void solve() throws IOException {
int n = sc.ni();
String str = sc.ns();
char[] strr = str.toCharArray();
int ct = 1;
char prev = strr[0];
int moves = 0;
for(int i = 1; i < n; i++) {
if(strr[i] == prev) ct++;
else {
if(ct%2==0) {
ct = 1;
prev = strr[i];
continue;
} else {
moves++;
ct = 0;
continue;
}
}
}
w.p(moves);
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | e4d1547a130727c7bb3fde9fd29b49f1 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//--------------------------INPUT READER---------------------------------//
static class fs {
public BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public fs() { this(System.in); }
public fs(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String ns() { return next(); }
int[] na(long nn) {
int n = (int) nn;
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
long[] nal(long nn) {
int n = (int) nn;
long[] l = new long[n];
for(int i = 0; i < n; i++) l[i] = nl();
return l;
}
}
//-----------------------------------------------------------------------//
//---------------------------PRINTER-------------------------------------//
static class Printer {
static PrintWriter w;
public Printer() {this(System.out);}
public Printer(OutputStream os) {
w = new PrintWriter(os);
}
public void p(int i) {w.println(i);}
public void p(long l) {w.println(l);}
public void p(double d) {w.println(d);}
public void p(String s) { w.println(s);}
public void pr(int i) {w.print(i);}
public void pr(long l) {w.print(l);}
public void pr(double d) {w.print(d);}
public void pr(String s) { w.print(s);}
public void pl() {w.println();}
public void close() {w.close();}
}
//-----------------------------------------------------------------------//
//--------------------------VARIABLES------------------------------------//
static fs sc = new fs();
static OutputStream outputStream = System.out;
static Printer w = new Printer(outputStream);
static long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
static int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
static long mod = 1000000007;
//-----------------------------------------------------------------------//
//--------------------------ADMIN_MODE-----------------------------------//
private static void ADMIN_MODE() throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
w = new Printer(new FileOutputStream("output.txt"));
sc = new fs(new FileInputStream("input.txt"));
}
}
//-----------------------------------------------------------------------//
//----------------------------START--------------------------------------//
public static void main(String[] args)
throws IOException {
ADMIN_MODE();
int t = sc.ni();while(t-->0)
solve();
w.close();
}
static void solve() throws IOException {
int n = sc.ni();
String str = sc.ns();
char[] strr = str.toCharArray();
int ct = 1;
char prev = strr[0];
int odd = 0;
int moves = 0;
for(int i = 1; i < n; i++) {
if(strr[i] == prev) {
ct++;
} else {
if(ct%2==0) {
prev = strr[i];
ct = 1;
continue;
}
moves++;
ct = 0;
}
}
if(ct%2==1) {
moves++;
}
w.p(moves);
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | e753860d84ea1a105125f91fda2b1777 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.*;
public class Codeforces {
static long mod= 10000_0000_7;
static int dp[][];
public static void main(String[] args) throws Exception {
PrintWriter out=new PrintWriter(System.out);
FastScanner fs=new FastScanner();
// DecimalFormat formatter= new DecimalFormat("#0.000000");
int t=fs.nextInt();
// int t=1;
// int n=1000000000;
outer:for(int time=1;time<=t;time++) {
int n=fs.nextInt();
char arr[]=fs.next().toCharArray();
List<Integer> list=new ArrayList<>();
int cnta[]=new int[n+1], cntb[]=new int[n+1];
for(int i=1;i<=n;i++) {
cnta[i]=cnta[i-1];
cntb[i]=cntb[i-1];
if(arr[i-1]=='0') {
cnta[i]++;
}
else cntb[i]++;
}
int x=0;
while(x<n) {
int j=x+1;
int cnt=1;
while(j<n&&arr[j]==arr[x]) {
j++;
cnt++;
}
list.add(cnt);
x=j;
}
// List<Integer> ind=new ArrayList<>();
// for(int i=0;i<list.size();i++) {
// int ele=list.get(i);
// if(ele%2==0) {
// }
// ind.add(i);
// }
int ans=0;
int i=0;
int sum=0;
// System.out.println(list);
while(i<list.size()) {
while(i<list.size()&&list.get(i)%2==0) {
sum+=list.get(i);
i++;
}
if(i==list.size()) break;
sum+=list.get(i);
i++;
int cnt=0;
int a=sum;
while(i<list.size()&&list.get(i)%2==0) {
sum+=list.get(i);
i++;
cnt++;
}
int b=sum;
sum+=list.get(i);
i++;
// System.out.println(a+" "+b);
ans+= cnt+1;
}
out.println(ans);
}
out.close();
}
static int find(int i,int j,int cnta[],int cntb[]) {
return 1+Math.min(cnta[j+1]-cnta[i], cntb[j+1]-cntb[i]);
}
static long pow(long a,long b) {
if(b<0) return 1;
long res=1;
while(b!=0) {
if((b&1)!=0) {
res*=a;
res%=mod;
}
a*=a;
a%=mod;
b=b>>1;
}
return res;
}
static long gcd(long a,long b) {
if(b==0) return a;
return gcd(b,a%b);
}
static long nck(int n,int k) {
if(k>n) return 0;
long res=1;
res*=fact(n);
res%=mod;
res*=modInv(fact(k));
res%=mod;
res*=modInv(fact(n-k));
res%=mod;
return res;
}
static long fact(long n) {
// return fact[(int)n];
long res=1;
for(int i=2;i<=n;i++) {
res*=i;
res%=mod;
}
return res;
}
static long modInv(long n) {
return pow(n,mod-2);
}
static void sort(int[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
int temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
static void sort(long[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
long temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
// Use this to input code since it is faster than a Scanner
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();
}
String nextLine() {
String str="";
try {
str= (br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
long[] readArrayL(int n) {
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=nextLong();
return a;
}
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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | cfdad86591e613156b16d3012d894833 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 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) {
boolean ok = true;
int cnt = 0;
int n = in.nextInt();
a = in.next().toCharArray();
for(int i = 1; i < n;i+=2) {
if(a[i] != a[i - 1]) {
cnt++;
}
}
System.out.println(cnt);
}
}
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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 355ff5d6cd3221e05cbf4217936329ca | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class pre54 {
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 main(String args[]){
FastReader obj = new FastReader();
int tc = obj.nextInt();
while(tc--!=0){
int n = obj.nextInt();
char str[] = obj.next().toCharArray();
int count = 0;
ArrayList<Integer> con = new ArrayList<>();
char c = str[0];
count = 1;
for(int i=1;i<n;i++){
if(c!=str[i]){
con.add(count);
count = 1;
c = str[i];
}else count++;
}
con.add(count);
// out.println(con);
int ans = 0;
for(int i=0;i<con.size()-1;i++){
if(con.get(i)%2!=0){
con.set(i,con.get(i)-1);
con.set(i+1,con.get(i+1)+1);
ans++;
}
}
out.println(ans);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | af0fa38d2fa2406a977e870eac235462 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.lang.*;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.File;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
/* 10^(7) = 1s.
* ceilVal = (a+b-1) / b */
static final int mod = 1000000007;
static final long temp = 998244353;
static final long MOD = 1000000007;
static final long M = (long)1e9+7;
static class Pair implements Comparable<Pair>
{
int first, second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
public int compareTo(Pair ob)
{
return (int)(first - ob.first);
}
}
static class Tuple implements Comparable<Tuple>
{
long first, second,third;
public Tuple(long first, long second, long third)
{
this.first = first;
this.second = second;
this.third = third;
}
public int compareTo(Tuple o)
{
return (int)(o.third - this.third);
}
}
public static class DSU
{
int count = 0;
int[] parent;
int[] rank;
public DSU(int n)
{
count = n;
parent = new int[n];
rank = new int[n];
Arrays.fill(parent, -1);
Arrays.fill(rank, 1);
}
public int find(int i)
{
return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
}
public void union(int a, int b) //Union Find by Rank
{
a = find(a);
b = find(b);
if(a == b) return;
if(rank[a] < rank[b])
{
parent[a] = b;
}
else if(rank[a] > rank[b])
{
parent[b] = a;
}
else
{
parent[b] = a;
rank[a] = 1 + rank[a];
}
count--;
}
public int countConnected()
{
return count;
}
}
static class Reader
{
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) throws IOException {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] longReadArray(int n) throws IOException {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
public static int gcd(int a, int b)
{
if(b == 0)
return a;
else
return gcd(b,a%b);
}
public static long lcm(long a, long b)
{
return (a / LongGCD(a, b)) * b;
}
public static long LongGCD(long a, long b)
{
if(b == 0)
return a;
else
return LongGCD(b,a%b);
}
public static long LongLCM(long a, long b)
{
return (a / LongGCD(a, b)) * b;
}
//Count the number of coprime's upto N
public static long phi(long n) //euler totient/phi function
{
long ans = n;
// for(long i = 2;i*i<=n;i++)
// {
// if(n%i == 0)
// {
// while(n%i == 0) n/=i;
// ans -= (ans/i);
// }
// }
//
// if(n > 1)
// {
// ans -= (ans/n);
// }
for(long i = 2;i<=n;i++)
{
if(isPrime(i))
{
ans -= (ans/i);
}
}
return ans;
}
public static long fastPow(long x, long n)
{
if(n == 0)
return 1;
else if(n%2 == 0)
return fastPow(x*x,n/2);
else
return x*fastPow(x*x,(n-1)/2);
}
public static long powMod(long x, long y, long p)
{
long res = 1;
x = x % p;
while (y > 0)
{
if (y % 2 == 1)
{
res = (res * x) % p;
}
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long modInverse(long n, long p)
{
return powMod(n, p - 2, p);
}
// Returns nCr % p using Fermat's little theorem.
public static long nCrModP(long n, long r,long p)
{
if (n<r)
return 0;
if (r == 0)
return 1;
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 - r)], p)
% p)
% p;
}
public static long fact(long n)
{
long[] fac = new long[(int)(n) + 1];
fac[0] = 1;
for (long i = 1; i <= n; i++)
fac[(int)(i)] = fac[(int)(i - 1)] * i;
return fac[(int)(n)];
}
public static long nCr(long n, long k)
{
long ans = 1;
for(long i = 0;i<k;i++)
{
ans *= (n-i);
ans /= (i+1);
}
return ans;
}
//Modular Operations for Addition and Multiplication.
public static long perfomMod(long x)
{
return ((x%M + M)%M);
}
public static long addMod(long a, long b)
{
return perfomMod(perfomMod(a)+perfomMod(b));
}
public static long subMod(long a, long b)
{
return perfomMod(perfomMod(a)-perfomMod(b));
}
public static long mulMod(long a, long b)
{
return perfomMod(perfomMod(a)*perfomMod(b));
}
public static boolean isPrime(long n)
{
if(n == 1)
{
return false;
}
//check only for sqrt of the number as the divisors
//keep repeating so only half of them are required. So,sqrt.
for(int i = 2;i*i<=n;i++)
{
if(n%i == 0)
{
return false;
}
}
return true;
}
public static List<Long> SieveList(int n)
{
boolean prime[] = new boolean[(int)(n+1)];
Arrays.fill(prime, true);
List<Long> l = new ArrayList<>();
for (long p = 2; p*p<=n; p++)
{
if (prime[(int)(p)] == true)
{
for(long i = p*p; i<=n; i += p)
{
prime[(int)(i)] = false;
}
}
}
for (long p = 2; p<=n; p++)
{
if (prime[(int)(p)] == true)
{
l.add(p);
}
}
return l;
}
public static int countDivisors(int x)
{
int c = 0;
for(int i = 1;i*i<=x;i++)
{
if(x%i == 0)
{
if(x/i != i)
{
c+=2;
}
else
{
c++;
}
}
}
return c;
}
public static long log2(long n)
{
long ans = (long)(log(n)/log(2));
return ans;
}
public static boolean isPow2(long n)
{
return (n != 0 && ((n & (n-1))) == 0);
}
public static boolean isSq(int x)
{
long s = (long)Math.round(Math.sqrt(x));
return s*s==x;
}
/*
*
* >= <=
0 1 2 3 4 5 6 7
5 5 5 6 6 6 7 7
lower_bound for 6 at index 3 (>=)
upper_bound for 6 at index 6(To get six reduce by one) (<=)
*/
public static int LowerBound(int a[], int x)
{
int l=-1,r=a.length;
while(l+1<r)
{
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
public static int UpperBound(int a[], int x)
{
int l=-1, r=a.length;
while(l+1<r)
{
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
public static void Sort(int[] a)
{
List<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
// Collections.reverse(l); //Use to Sort decreasingly
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
public static void ssort(char[] a)
{
List<Character> l = new ArrayList<>();
for (char i : a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
public static void main(String[] args) throws Exception
{
Reader sc = new Reader();
PrintWriter fout = new PrintWriter(System.out);
int tt = sc.nextInt();
fr:while(tt-- > 0)
{
int n = sc.nextInt();
char[] s = sc.next().toCharArray();
long ans = 0;
for(int i = 0;i<n;i+=2)
{
if(s[i] != s[i+1]) ans++;
}
fout.println(ans);
}
fout.close();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | c043fe4375b6569549b93d00d2cfe4a7 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.lang.*;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.File;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
/* 10^(7) = 1s.
* ceilVal = (a+b-1) / b */
static final int mod = 1000000007;
static final long temp = 998244353;
static final long MOD = 1000000007;
static final long M = (long)1e9+7;
static class Pair implements Comparable<Pair>
{
int first, second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
public int compareTo(Pair ob)
{
return (int)(first - ob.first);
}
}
static class Tuple implements Comparable<Tuple>
{
long first, second,third;
public Tuple(long first, long second, long third)
{
this.first = first;
this.second = second;
this.third = third;
}
public int compareTo(Tuple o)
{
return (int)(o.third - this.third);
}
}
public static class DSU
{
int count = 0;
int[] parent;
int[] rank;
public DSU(int n)
{
count = n;
parent = new int[n];
rank = new int[n];
Arrays.fill(parent, -1);
Arrays.fill(rank, 1);
}
public int find(int i)
{
return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
}
public void union(int a, int b) //Union Find by Rank
{
a = find(a);
b = find(b);
if(a == b) return;
if(rank[a] < rank[b])
{
parent[a] = b;
}
else if(rank[a] > rank[b])
{
parent[b] = a;
}
else
{
parent[b] = a;
rank[a] = 1 + rank[a];
}
count--;
}
public int countConnected()
{
return count;
}
}
static class Reader
{
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) throws IOException {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] longReadArray(int n) throws IOException {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
public static int gcd(int a, int b)
{
if(b == 0)
return a;
else
return gcd(b,a%b);
}
public static long lcm(long a, long b)
{
return (a / LongGCD(a, b)) * b;
}
public static long LongGCD(long a, long b)
{
if(b == 0)
return a;
else
return LongGCD(b,a%b);
}
public static long LongLCM(long a, long b)
{
return (a / LongGCD(a, b)) * b;
}
//Count the number of coprime's upto N
public static long phi(long n) //euler totient/phi function
{
long ans = n;
// for(long i = 2;i*i<=n;i++)
// {
// if(n%i == 0)
// {
// while(n%i == 0) n/=i;
// ans -= (ans/i);
// }
// }
//
// if(n > 1)
// {
// ans -= (ans/n);
// }
for(long i = 2;i<=n;i++)
{
if(isPrime(i))
{
ans -= (ans/i);
}
}
return ans;
}
public static long fastPow(long x, long n)
{
if(n == 0)
return 1;
else if(n%2 == 0)
return fastPow(x*x,n/2);
else
return x*fastPow(x*x,(n-1)/2);
}
public static long powMod(long x, long y, long p)
{
long res = 1;
x = x % p;
while (y > 0)
{
if (y % 2 == 1)
{
res = (res * x) % p;
}
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long modInverse(long n, long p)
{
return powMod(n, p - 2, p);
}
// Returns nCr % p using Fermat's little theorem.
public static long nCrModP(long n, long r,long p)
{
if (n<r)
return 0;
if (r == 0)
return 1;
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 - r)], p)
% p)
% p;
}
public static long fact(long n)
{
long[] fac = new long[(int)(n) + 1];
fac[0] = 1;
for (long i = 1; i <= n; i++)
fac[(int)(i)] = fac[(int)(i - 1)] * i;
return fac[(int)(n)];
}
public static long nCr(long n, long k)
{
long ans = 1;
for(long i = 0;i<k;i++)
{
ans *= (n-i);
ans /= (i+1);
}
return ans;
}
//Modular Operations for Addition and Multiplication.
public static long perfomMod(long x)
{
return ((x%M + M)%M);
}
public static long addMod(long a, long b)
{
return perfomMod(perfomMod(a)+perfomMod(b));
}
public static long subMod(long a, long b)
{
return perfomMod(perfomMod(a)-perfomMod(b));
}
public static long mulMod(long a, long b)
{
return perfomMod(perfomMod(a)*perfomMod(b));
}
public static boolean isPrime(long n)
{
if(n == 1)
{
return false;
}
//check only for sqrt of the number as the divisors
//keep repeating so only half of them are required. So,sqrt.
for(int i = 2;i*i<=n;i++)
{
if(n%i == 0)
{
return false;
}
}
return true;
}
public static List<Long> SieveList(int n)
{
boolean prime[] = new boolean[(int)(n+1)];
Arrays.fill(prime, true);
List<Long> l = new ArrayList<>();
for (long p = 2; p*p<=n; p++)
{
if (prime[(int)(p)] == true)
{
for(long i = p*p; i<=n; i += p)
{
prime[(int)(i)] = false;
}
}
}
for (long p = 2; p<=n; p++)
{
if (prime[(int)(p)] == true)
{
l.add(p);
}
}
return l;
}
public static int countDivisors(int x)
{
int c = 0;
for(int i = 1;i*i<=x;i++)
{
if(x%i == 0)
{
if(x/i != i)
{
c+=2;
}
else
{
c++;
}
}
}
return c;
}
public static long log2(long n)
{
long ans = (long)(log(n)/log(2));
return ans;
}
public static boolean isPow2(long n)
{
return (n != 0 && ((n & (n-1))) == 0);
}
public static boolean isSq(int x)
{
long s = (long)Math.round(Math.sqrt(x));
return s*s==x;
}
/*
*
* >= <=
0 1 2 3 4 5 6 7
5 5 5 6 6 6 7 7
lower_bound for 6 at index 3 (>=)
upper_bound for 6 at index 6(To get six reduce by one) (<=)
*/
public static int LowerBound(int a[], int x)
{
int l=-1,r=a.length;
while(l+1<r)
{
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
public static int UpperBound(int a[], int x)
{
int l=-1, r=a.length;
while(l+1<r)
{
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
public static void Sort(int[] a)
{
List<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
// Collections.reverse(l); //Use to Sort decreasingly
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
public static void ssort(char[] a)
{
List<Character> l = new ArrayList<>();
for (char i : a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
public static void main(String[] args) throws Exception
{
Reader sc = new Reader();
PrintWriter fout = new PrintWriter(System.out);
int tt = sc.nextInt();
fr:while(tt-- > 0)
{
int n = sc.nextInt();
char[] s = sc.next().toCharArray();
long ans = 0, c = 1;
for(int i = 1;i<n;i++)
{
if(s[i] != s[i-1])
{
if(c%2 == 0)
{
c = 1;
}
else if(c % 2 != 0)
{
ans++;
c = 0;
}
}
else
{
c++;
}
}
fout.println(ans);
}
fout.close();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 9a11af0c12b3cfbf2da69bcce5edd5aa | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
public class Template {
private static StringTokenizer st;
private static BufferedReader br;
private static PrintWriter pw;
public static void main(String[] args) throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
int test = nextInt();
while (test -- > 0)
{
int n = nextInt();
char a[] = next().toCharArray();
int cnt = 0;
boolean zeroOrOne = false;
char start = a[0];
List<Integer> part = new ArrayList<>();
for(int i = 0 ; i < n;i ++){
if(start == a[i])
cnt++;
else{
part.add(cnt);
cnt = 1;
start = a[i];
}
}
part.add(cnt);
int ans = 0;
for(int i = 0 ; i < part.size() -1 ;i ++){
if(part.get(i) % 2 == 0)
continue;
if(part.get(i + 1) >= 1){
part.set(i,part.get(i) + 1);
part.set(i + 1,part.get(i + 1) - 1);
ans ++;
}
}
System.out.println(ans);
}
}
private static int nextInt() throws IOException {
return Integer.parseInt(next());
}
private static long nextLong() throws IOException {
return Long.parseLong(next());
}
private static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
private static String next() throws IOException {
while (st==null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 4003c42343dab03d063dc6b08ac3adcb | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class B1_1678 {
public static void main(String[] args)
{
/*
* 1
14
10011011001001
*/
FastReader sc = new FastReader();
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
char[] ch=sc.next().toCharArray();
int[] arr=new int[n];
int c=1;
int ind=0;
for(int i=0;i<n-1;i++) {
if(ch[i]==ch[i+1]) {
c++;
continue;
}
arr[ind++]=c;
c=1;
}
arr[ind]=c;
int no=0;
int add=0;
int ans=0;
boolean check=true;
for(int i=0;i<=ind;i++) {
if(no>0) {
add++;
if(arr[i]%2!=0) {
no=0;
ans+=add;
add=0;
check=false;
}
}
if(arr[i]%2!=0 && check) no++;
check=true;
}
System.out.println(ans);
}
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 872041138e6dc08fb8d8908ad1e15999 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class TokitsukazeAndGood01StringEasyVersion {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st;
public static void main(String[] args) throws IOException {
int t = readInt();
while (t-- > 0) {
int n = readInt();
String str = readLine();
int[] s = new int[n];
for (int i = 0; i < n; i ++) s[i] = Character.getNumericValue(str.charAt(i));
ArrayList<Integer> segments = new ArrayList<Integer>();
int cur = s[0], len = 1;
for (int i = 1; i < n; i ++) {
if (s[i] != cur) {
segments.add(len);
len = 1;
cur = s[i];
} else len ++;
}
segments.add(len);
int ans = 0, sum = 0;
boolean odd = false;
for (int i = 0; i < segments.size(); i ++) {
if (!odd) {
if (segments.get(i) % 2 == 1) { odd = true; }
} else {
if (segments.get(i) % 2 == 1) { odd = false; ans ++; sum = 0; }
else { sum += segments.get(i); ans ++; }
}
}
if (odd) System.out.println(ans + sum);
else System.out.println(ans);
}
}
static String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine().trim());
return st.nextToken();
}
static long readLong() throws IOException {
return Long.parseLong(next());
}
static int readInt() throws IOException {
return Integer.parseInt(next());
}
static double readDouble() throws IOException {
return Double.parseDouble(next());
}
static char readCharacter() throws IOException {
return next().charAt(0);
}
static String readLine() throws IOException {
return br.readLine().trim();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 76decf1812b0b139b6ce722609d72a4e | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | // letsgoooooooooooooooooooooooooooooooo
import java.util.*;
import java.io.*;
public class Solution{
static int MOD=1000000007;
static PrintWriter pw;
static FastReader sc;
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());}
public char nextChar() throws IOException {return next().charAt(0);}
String nextLine(){
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int[] intArr(int a,int b,int n) throws IOException {int[]in=new int[n];for(int i=a;i<b;i++)in[i]=nextInt();return in;}
}
static void Solve() throws Exception{
int n =sc.nextInt();
String s =sc.next();
StringBuilder sb = new StringBuilder(s);
int count=0,ans=0;
for(int i=0;i<s.length();i++){
count=0;
char ch = s.charAt(i);
while(i<n && s.charAt(i)==ch){
i++;
count++;
}
if(count%2==1){
ans++;
}else{
i--;
}
// pw.println(s);
}
pw.println(ans);
}
public static void main(String[] args) throws Exception{
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
} catch (Exception e) {
System.err.println("Error");
}
sc= new FastReader();
pw = new PrintWriter(System.out);
int t=1;
t=sc.nextInt();
for(int ii=1;ii<=t;ii++) {
Solve();
}
pw.flush();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 77a7753571f78f6c780fb88d80e1cc6f | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
Scanner scn = new Scanner(System.in);
int tc = scn.nextInt();
while (tc != 0) {
int n = scn.nextInt();
String s = scn.next();
solve(n, s);
tc--;
}
}
public static void solve(int n, String s) {
if (n == 2) {
if (s.charAt(0) == s.charAt(1)) {
System.out.println("0");
return;
}
else {
System.out.println("1");
return;
}
}
ArrayList<Integer> dividedSegmentsSizeList = new ArrayList<>();
for (int i = 0 ; i < n ; ) {
int currentContinuousSegementSize = 1;
while (i < n - 1 && s.charAt(i) == s.charAt(i + 1)) {
currentContinuousSegementSize++;
i++;
}
i++;
dividedSegmentsSizeList.add(currentContinuousSegementSize);
}
int[] arr = new int[dividedSegmentsSizeList.size()];
for (int i = 0 ; i < arr.length ; i++) {
arr[i] = dividedSegmentsSizeList.get(i);
}
int minimumCount = 0;
for (int i = 0 ; i < arr.length - 1 ; i++) {
if (arr[i] % 2 != 0 && arr[i + 1] % 2 != 0) {
arr[i]--;
arr[i + 1]++;
minimumCount++;
}
else if (arr[i] % 2 != 0 && arr[i + 1] % 2 == 0) {
arr[i]--;
arr[i + 1]++;
minimumCount++;
}
}
System.out.println(minimumCount);
return;
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | bbffb6173a69d7323914f3a958e0c183 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //package com.company;
import java.util.*;
public class Practice {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
String s = sc.next();
char[] arr = s.toCharArray();
int cnt = 1;
int flag = 0;
for(int i = 0;i < n - 1;i++) {
if(s.charAt(i) == s.charAt(i + 1))
cnt++;
else {
if(cnt % 2 != 0)
flag = 1;
else
cnt = 1;
}
}
if(flag == 0)
System.out.println("0");
else {
int count = 0;
cnt = 1;
for(int i = 0;i < n - 1;i++) {
if(arr[i] == arr[i + 1])
cnt++;
if(arr[i] != arr[i + 1] && cnt % 2 == 0)
cnt = 1;
else if(arr[i] != arr[i + 1]) {
count++;
arr[i + 1] = arr[i];
i++;
cnt = 1;
}
}
System.out.println(count);
}
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 3b2f46293126959609bb315142358ef2 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class B2 {
public static void main(String[] args) throws IOException {
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw=new PrintWriter(System.out);
int t=Integer.parseInt(bf.readLine());
for(int i=0;i<t;i++) {
int n=Integer.parseInt(bf.readLine());
String s=bf.readLine();
int sol=0;
for(int j=0;j<n;j+=2) {
char x1=s.charAt(j);
char x2=s.charAt(j+1);
if((x1=='0'&&x2=='0')||(x1=='1'&&x2=='1')) {
continue;
}else {
sol++;
}
}
pw.println(sol);
}
pw.close();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 30b6996f1eee12a07a971b65839f2f1c | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
import static java.lang.System.out;
public class app {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n=sc.nextInt();
String s=sc.next();
ArrayList<Integer> aa=new ArrayList<>();
char a1=s.charAt(0);
int con=1;
for(int i=1;i<n;i++) {
if(a1==s.charAt(i)){
con++;
if(i==n-1&&a1==s.charAt(i)){
aa.add(con);
}
}
else{
aa.add(con);
con=1;
if(i==n-1&&a1!=s.charAt(i)){
aa.add(1);
}
a1=s.charAt(i);
}
}
int cons=0;
if(aa.size()==1) {
out.println(0);
}
if(aa.size()>1) {
int sss = aa.get(0);
for (int i = 1; i < aa.size(); i++) {
if (sss % 2 == 1) {
cons++;
sss = aa.get(i) - 1;
}
else{
sss=aa.get(i);
}
}
out.println(cons);
}
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 408d5585c2d44039aeb6360eb8c39803 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
public class tr1 {
static PrintWriter out;
static StringBuilder sb;
static long mod = 998244353;
static int m, id, c;
static ArrayList<int[]>[] ad;
static HashMap<Integer, Integer>[] going;
static long inf = Long.MAX_VALUE;
static char[][] g;
static boolean[][] vis;
static int[] ans, arr[], per;
static int h, w;
static HashMap<Integer, int[]> hmC, hmG;
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
char[] s = sc.nextLine().toCharArray();
ArrayList<Integer> seg = new ArrayList<>();
ArrayList<Integer> type = new ArrayList<>();
for (int i = 0; i < n; i++) {
int c = 1;
while (i < n - 1 && s[i] == s[i + 1]) {
i++;
c++;
}
seg.add(c);
type.add(s[i] - '0');
}
int ans = 0;
int carry = 0;
for (int i = 0; i < seg.size(); i++) {
int sum = carry + seg.get(i);
if (sum % 2 == 1)
ans++;
carry = sum % 2;
}
out.println(ans);
}
out.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream system) {
br = new BufferedReader(new InputStreamReader(system));
}
public Scanner(String file) throws Exception {
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 String nextLine() throws IOException {
return br.readLine();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public char nextChar() throws IOException {
return next().charAt(0);
}
public Long nextLong() throws IOException {
return Long.parseLong(next());
}
public int[] nextArrInt(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextArrLong(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextArrIntSorted(int n) throws IOException {
int[] a = new int[n];
Integer[] a1 = new Integer[n];
for (int i = 0; i < n; i++)
a1[i] = nextInt();
Arrays.sort(a1);
for (int i = 0; i < n; i++)
a[i] = a1[i].intValue();
return a;
}
public long[] nextArrLongSorted(int n) throws IOException {
long[] a = new long[n];
Long[] a1 = new Long[n];
for (int i = 0; i < n; i++)
a1[i] = nextLong();
Arrays.sort(a1);
for (int i = 0; i < n; i++)
a[i] = a1[i].longValue();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
public void waitForInput() throws InterruptedException {
Thread.sleep(3000);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 07866ea355803194bb9cb2b037a2e05b | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
public class Practice1 {
//
// static class Pair{
// int val;
// int data;
// Pair(int val,int data){
// this.val=val;
// this.data=data;
// }
// }
//
//
public static void printarr(int[] arr) {
int n=arr.length;
for(int i=0;i<n;i++) {
System.out.print(arr[i]+" ");
}
System.out.println();
}
// /** Code for Dijkstra's algorithm **/
public static class ListNode {
int vertex, weight;
ListNode(int v,int w) {
vertex = v;
weight = w;
}
int getVertex() { return vertex; }
int getWeight() { return weight; }
}
public static int[] dijkstra(
int V, HashMap<Integer,ArrayList<ListNode> > graph,
int source) {
int[] distance = new int[V];
for (int i = 0; i < V; i++)
distance[i] = Integer.MAX_VALUE;
distance[0] = 0;
PriorityQueue<ListNode> pq = new PriorityQueue<>(
(v1, v2) -> v1.getWeight() - v2.getWeight());
pq.add(new ListNode(source, 0));
while (pq.size() > 0) {
ListNode current = pq.poll();
for (ListNode n :
graph.get(current.getVertex())) {
if (distance[current.getVertex()]
+ n.getWeight()
< distance[n.getVertex()]) {
distance[n.getVertex()]
= n.getWeight()
+ distance[current.getVertex()];
pq.add(new ListNode(
n.getVertex(),
distance[n.getVertex()]));
}
}
}
// If you want to calculate distance from source to
// a particular target, you can return
// distance[target]
return distance;
}
//Methos to return all divisor of a number
static ArrayList<Long> allDivisors(long n) {
ArrayList<Long> al=new ArrayList<>();
long i=2;
while(i*i<=n) {
if(n%i==0) al.add(i);
if(n%i==0&&i*i!=n) al.add(n/i);
i++;
}
return al;
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static int find(boolean[] vis,int x,int y,int n,int m) {
if(x<0||y<0||x>=n||y>=m) return 0;
// if(vis[i][j]==true) return
return 0;
}
static long power(long N,long R)
{
long x=998244353;
if(R==0) return 1;
if(R==1) return N;
long temp= power(N,R/2)%x; // (a*b)%p = (a%p*b%p)*p
temp=(temp*temp)%x;
if(R%2==0){
return temp%x;
}else{
return (N*temp)%x;
}
}
static int[] sort(int[] arr) {
int n=arr.length;
ArrayList<Integer> al=new ArrayList<>();
for(int i=0;i<n;i++) {
al.add(arr[i]);
}
Collections.sort(al);
for(int i=0;i<n;i++) {
arr[i]=al.get(i);
}
return arr;
}
static int[] sortrev(int[] arr) {
int n=arr.length;
ArrayList<Integer> al=new ArrayList<>();
for(int i=0;i<n;i++) {
al.add(arr[i]);
}
Collections.sort(al,Collections.reverseOrder());
for(int i=0;i<n;i++) {
arr[i]=al.get(i);
}
return arr;
}
static long[] sort(long[] arr) {
long n=arr.length;
ArrayList<Long> al=new ArrayList<>();
for(int i=0;i<n;i++) {
al.add(arr[i]);
}
Collections.sort(al);
for(int i=0;i<n;i++) {
arr[i]=al.get(i);
}
return arr;
}
static class Pair{
int val;
int res;
Pair(int val,int res){
this.val=val;
this.res=res;
}
}
public static long find(ArrayList<Long> al,ArrayList<Integer> li,int ind) {
int n=al.size();
if(n==0) return 0;
//System.out.println("al : "+al);
ArrayList<Long> one=new ArrayList<>();
ArrayList<Long> zero=new ArrayList<>();
int a=li.get(ind);
for(Long i: al) {
if((i&(1<<a))==0) {
zero.add(i);
}else {
one.add(i);
}
}
//System.out.println("al : "+al);
if(ind==li.size()-1) {
long size1=one.size();
long size2=zero.size();
return size1*size1+size2*size2;
}else {
return find(one,li,ind+1)+find(zero,li,ind+1);
}
}
public static void main (String[] args) {
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
// out.print();
//out.println();
FastReader sc=new FastReader();
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
String str=sc.nextLine();
int count=1;
ArrayList<Integer> al=new ArrayList<>();
int one=0,zero=0;
for(int i=0;i<n;i++) {
if(str.charAt(i)=='1') {
one++;
}else {
zero++;
}
if(i==n-1) {
al.add(count);
continue;
}
if(str.charAt(i)!=str.charAt(i+1)) {
al.add(count);
count=1;
}else {
count++;
}
}
int n1=al.size();
int[] arr=new int[n1];
for(int i=0;i<n1;i++) {
arr[i]=al.get(i);
}
count=0;
for(int i=0;i<n1-1;i++) {
if(arr[i]==0||arr[i]%2==0) {
continue;
}
count++;
arr[i+1]--;
}
if(arr[n1-1]%2==0) {
out.println(count);
}else {
out.println(Math.min(one,zero));
}
}
out.close();
}
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;
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | d44cc55d11d09efe6e8f03a4802ddbcc | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class Main implements Runnable {
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
public static void main(String[] args) {
new Thread(null, new Main(), "", 256 * (1L << 20)).start();
}
public void run() {
try {
long t1 = System.currentTimeMillis();
if (System.getProperty("ONLINE_JUDGE") != null) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
} else {
in = new BufferedReader(new FileReader("input.txt"));
out = new PrintWriter("output.txt");
}
Locale.setDefault(Locale.US);
solve();
in.close();
out.close();
long t2 = System.currentTimeMillis();
System.err.println("Time = " + (t2 - t1));
} catch (Throwable t) {
t.printStackTrace(System.err);
System.exit(-1);
}
}
String readString() throws IOException {
while (!tok.hasMoreTokens()) {
tok = new StringTokenizer(in.readLine());
}
return tok.nextToken();
}
int readInt() throws IOException {
return Integer.parseInt(readString());
}
long readLong() throws IOException {
return Long.parseLong(readString());
}
double readDouble() throws IOException {
return Double.parseDouble(readString());
}
int[] readIntArray(int n) throws IOException {
int [] a = new int[n];
for(int i = 0; i < n; i++) {
a[i] = readInt();
}
return a;
}
long[] readLongArray(int n) throws IOException {
long [] a = new long[n];
for(int i = 0; i < n; i++) {
a[i] = readLong();
}
return a;
}
static final Random random=new Random();
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n);
long temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
void solveTest() throws IOException {
int n = readInt();
String s = readString();
char curChar = s.charAt(0);
int curCharCount = 1;
int ops = 0;
for(int i=1;i<n;i++) {
char c = s.charAt(i);
if(c == curChar) {
curCharCount++;
continue;
}
if(curCharCount % 2 == 1) {
ops++;
curChar = curChar == '0' ? '1' : '0';
curCharCount = 2;
} else {
curChar = curChar == '0' ? '1' : '0';
curCharCount = 1;
}
}
if(curCharCount % 2 == 1) {
ops++;
}
out.println(ops);
}
void solve() throws IOException {
int numTests = readInt();
while(numTests-->0) {
solveTest();
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 5d6aea65adca80bb8edb619b466fce72 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Map.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
import static java.lang.System.*;
public class Main
{
public void tq() throws Exception
{
st=new StringTokenizer(bq.readLine());
int tq=i();
sb=new StringBuilder(2000000);
o:
while(tq-->0)
{
int n=i();
char ar[]=s().toCharArray();
LinkedList<Integer> l=new LinkedList<>();
char a=ar[0];
int c=0;
for(char aa:ar)
{
if(aa==a)c++;
else
{
l.add(c);
c=1;
a=aa;
}
}
l.add(c);
c=0;
int p=l.remove();
while(l.size()>0)
{
int e=l.remove();
if(p%2!=0)
{
if(e%2!=0)
{
c++;
if(l.size()==0)break;
p=l.remove();
}
else
{
c++;
p=e-1;
}
continue;
}
p=e;
}
sl(c);
}
p(sb);
}
long mod=1000000007l;
int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
BufferedReader bq = new BufferedReader(new InputStreamReader(in));StringTokenizer st;StringBuilder sb;
public static void main(String[] a)throws Exception{new Main().tq();}
int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
void f1(int ar[],int v){fill(ar,v);}
void f2(int ar[][],int v){for(int a[]:ar)fill(a,v);}
void f3(int ar[][][],int v){for(int a[][]:ar)for(int b[]:a)fill(b,v);}
void f1(long ar[],long v){fill(ar,v);}
void f2(long ar[][],int v){for(long a[]:ar)fill(a,v);}
void f3(long ar[][][],int v){for(long a[][]:ar)for(long b[]:a)fill(b,v);}
void f(){out.flush();}
int p(int i,int p[]){return p[i]<0?i:(p[i]=p(p[i],p));}
boolean c(int x,int y,int n,int m){return x>=0&&x<n&&y>=0&&y<m;}
int[] so(int ar[])
{
Integer r[] = new Integer[ar.length];
for (int x = 0; x < ar.length; x++) r[x] = ar[x];
sort(r);
for (int x = 0; x < ar.length; x++) ar[x] = r[x];
return ar;
}
long[] so(long ar[])
{
Long r[] = new Long[ar.length];
for (int x = 0; x < ar.length; x++) r[x] = ar[x];
sort(r);
for (int x = 0; x < ar.length; x++) ar[x] = r[x];
return ar;
}
char[] so(char ar[])
{
Character r[] = new Character[ar.length];
for (int x = 0; x < ar.length; x++) r[x] = ar[x];
sort(r);
for (int x = 0; x < ar.length; x++) ar[x] = r[x];
return ar;
}
void p(Object p) {out.print(p);}void pl(Object p) {out.println(p);}void pl() {out.println();}
void s(String s) {sb.append(s);}
void s(int s) {sb.append(s);}
void s(long s) {sb.append(s);}
void s(char s) {sb.append(s);}
void s(double s) {sb.append(s);}
void ss() {sb.append(' ');}
void sl(String s) {s(s);sb.append("\n");}
void sl(int s) {s(s);sb.append("\n");}
void sl(long s) {s(s);sb.append("\n");}
void sl(char s) {s(s);sb.append("\n");}
void sl(double s) {s(s);sb.append("\n");}
void sl() {sb.append("\n");}
int l(int v) {return 31 - Integer.numberOfLeadingZeros(v);}
long l(long v) {return 63 - Long.numberOfLeadingZeros(v);}
int sq(int a) {return (int) sqrt(a);}
long sq(long a) {return (long) sqrt(a);}
int gcd(int a, int b)
{
while (b > 0)
{
int c = a % b;
a = b;
b = c;
}
return a;
}
long gcd(long a, long b)
{
while (b > 0l)
{
long c = a % b;
a = b;
b = c;
}
return a;
}
boolean p(String s, int i, int j)
{
while (i < j) if (s.charAt(i++) != s.charAt(j--)) return false;
return true;
}
boolean[] si(int n)
{
boolean bo[] = new boolean[n + 1];
bo[0] = bo[1] = true;
for (int x = 4; x <= n; x += 2) bo[x] = true;
for (int x = 3; x * x <= n; x += 2)
{
if (!bo[x])
{
int vv = (x << 1);
for (int y = x * x; y <= n; y += vv) bo[y] = true;
}
}
return bo;
}
long mul(long a, long b, long m)
{
long r = 1l;
a %= m;
while (b > 0)
{
if ((b & 1) == 1) r = (r * a) % m;
b >>= 1;
a = (a * a) % m;
}
return r;
}
int i() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return Integer.parseInt(st.nextToken());
}
long l() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return Long.parseLong(st.nextToken());
}
String s() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return st.nextToken();
}
double d() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return Double.parseDouble(st.nextToken());
}
void s(int a[])
{
for (int e : a){sb.append(e);sb.append(' ');}
sb.append("\n");
}
void s(long a[])
{
for (long e : a){sb.append(e);sb.append(' ');}
sb.append("\n");
}
void s(char a[])
{
for (char e : a){sb.append(e);sb.append(' ');}
sb.append("\n");
}
void s(int ar[][]) {for (int a[] : ar) s(a);}
void s(long ar[][]) {for (long a[] : ar) s(a);}
void s(char ar[][]) {for (char a[] : ar) s(a);}
int[] ari(int n) throws IOException
{
int ar[] = new int[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = Integer.parseInt(st.nextToken());
return ar;
}
long[] arl(int n) throws IOException
{
long ar[] = new long[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = Long.parseLong(st.nextToken());
return ar;
}
char[] arc(int n) throws IOException
{
char ar[] = new char[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = st.nextToken().charAt(0);
return ar;
}
double[] ard(int n) throws IOException
{
double ar[] = new double[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = Double.parseDouble(st.nextToken());
return ar;
}
String[] ars(int n) throws IOException
{
String ar[] = new String[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = st.nextToken();
return ar;
}
int[][] ari(int n, int m) throws IOException
{
int ar[][] = new int[n][m];
for (int x = 0; x < n; x++)ar[x]=ari(m);
return ar;
}
long[][] arl(int n, int m) throws IOException
{
long ar[][] = new long[n][m];
for (int x = 0; x < n; x++)ar[x]=arl(m);
return ar;
}
char[][] arc(int n, int m) throws IOException
{
char ar[][] = new char[n][m];
for (int x = 0; x < n; x++)ar[x]=arc(m);
return ar;
}
double[][] ard(int n, int m) throws IOException
{
double ar[][] = new double[n][m];
for (int x = 0; x < n; x++)ar[x]=ard(m);
return ar;
}
void p(int ar[])
{
sb = new StringBuilder(11 * ar.length);
for (int a : ar) {sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(long ar[])
{
StringBuilder sb = new StringBuilder(20 * ar.length);
for (long a : ar){sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(double ar[])
{
StringBuilder sb = new StringBuilder(22 * ar.length);
for (double a : ar){sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(char ar[])
{
StringBuilder sb = new StringBuilder(2 * ar.length);
for (char aa : ar){sb.append(aa);sb.append(' ');}
out.println(sb);
}
void p(String ar[])
{
int c = 0;
for (String s : ar) c += s.length() + 1;
StringBuilder sb = new StringBuilder(c);
for (String a : ar){sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(int ar[][])
{
StringBuilder sb = new StringBuilder(11 * ar.length * ar[0].length);
for (int a[] : ar)
{
for (int aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void p(long ar[][])
{
StringBuilder sb = new StringBuilder(20 * ar.length * ar[0].length);
for (long a[] : ar)
{
for (long aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void p(double ar[][])
{
StringBuilder sb = new StringBuilder(22 * ar.length * ar[0].length);
for (double a[] : ar)
{
for (double aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void p(char ar[][])
{
StringBuilder sb = new StringBuilder(2 * ar.length * ar[0].length);
for (char a[] : ar)
{
for (char aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void pl(Object... ar) {for (Object e : ar) p(e + " ");pl();}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | ab31cb0ac2eacc667f9f6ab6694bf358 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
static BufferedReader bf;
static PrintWriter out;
public static void main (String[] args)throws IOException {
bf = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
int t = Integer.parseInt(bf.readLine());
while(t-->0){
solve();
}
}
public static void solve()throws IOException{
int n = nextInt();
String[]arr = bf.readLine().split("");
String[]arr2 = arr.clone();
int count = 1;
int ans = 0;
for(int i =1;i<n;i++){
if(arr[i].equals(arr[i-1])){
count++;
}
else{
if(count%2 != 0){
arr[i] = arr[i-1].charAt(0)+"";
ans++;
count++;
}
else{
count = 1;
}
}
}
int ans2 = 0;
count = 1;
for(int i = n-2;i>=0;i--){
if(arr2[i].equals(arr2[i+1])){
count++;
}
else{
if(count%2 != 0){
arr2[i] = arr2[i+1].charAt(0)+"";
ans2++;
count++;
}
else{
count = 1;
}
}
}
println(Math.min(ans,ans2));
}
public static int check(long mid,int []arr,long count){
for(int i = 0;i<arr.length;i++){
if(arr[i] != 0){
count = count - (mid - arr[i]);
}
}
if(count == 0)return 0;
if(count < 0)return -1;
return 1;
}
// code for input
public static void print(String s ){
System.out.print(s);
}
public static void print(int num ){
System.out.print(num);
}
public static void print(long num ){
System.out.print(num);
}
public static void println(String s){
System.out.println(s);
}
public static void println(int num){
System.out.println(num);
}
public static void println(long num){
System.out.println(num);
}
public static void println(){
System.out.println();
}
public static int toInt(String s){
return Integer.parseInt(s);
}
public static long toLong(String s){
return Long.parseLong(s);
}
public static String[] nextStringArray()throws IOException{
return bf.readLine().split(" ");
}
public static int nextInt()throws IOException{
return Integer.parseInt(bf.readLine());
}
public static long nextLong()throws IOException{
return Long.parseLong(bf.readLine());
}
public static String nextString()throws IOException{
return bf.readLine();
}
public static int[] nextIntArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
int[]arr = new int[n];
for(int i =0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
return arr;
}
public static long[] nextLongArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
long[]arr = new long[n];
for(int i =0;i<n;i++){
arr[i] = Long.parseLong(str[i]);
}
return arr;
}
public static int[][] newIntMatrix(int r,int c)throws IOException{
int[][]arr = new int[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Integer.parseInt(str[j]);
}
}
return arr;
}
public static long[][] newLongMatrix(int r,int c)throws IOException{
long[][]arr = new long[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Long.parseLong(str[j]);
}
}
return arr;
}
static class pair{
int one;
int two;
pair(int one,int two){
this.one = one ;
this.two =two;
}
}
public static long gcd(long a,long b){
if(b == 0)return a;
return gcd(b,a%b);
}
public static long lcm(long a,long b){
return (a*b)/(gcd(a,b));
}
public static boolean isPalindrome(String s){
int i = 0;
int j = s.length()-1;
while(i<=j){
if(s.charAt(i) != s.charAt(j)){
return false;
}
i++;
j--;
}
return true;
}
}
// use some math tricks it might help
// sometimes just try to think in straightforward plan in A and B problems don't always complecate the questions with thinking too much differently
// always use long number to do 10^9+7 modulo
// if a problem is related to binary string it could also be related to parenthesis
// *****try to use binary search(it is a very beautiful thing it can work in some of the very unexpected problems ) in the question it might work******
// try sorting
// try to think in opposite direction of question it might work in your way
// if a problem is related to maths try to relate some of the continuous subarray with variables like - > a+b+c+d or a,b,c,d in general
// if the question is to much related to left and/or right side of any element in an array then try monotonic stack it could work.
// in range query sums try to do binary search it could work
// analyse the time complexity of program thoroughly
// anylyse the test cases properly
// if we divide any number by 2 till it gets 1 then there will be (number - 1) operation required
// try to do the opposite operation of what is given in the problem
//think about the base cases properly
//If a question is related to numbers try prime factorisation or something related to number theory
// keep in mind unique strings | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | d5d79a3163d28a01af5d2a65235c48bb | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.ArrayList;
import java.util.List;
public class MainB {
public static void main(String[] args) throws IOException {
// System.setIn(new FileInputStream("input.txt"));
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine());
for (int test_case = 0; test_case < T; test_case++) {
int n = Integer.parseInt(br.readLine());
String s = br.readLine();
int cnt = 1;
char cur = s.charAt(0);
List<Integer> list = new ArrayList<>();
for (int i = 1; i < n; i++) {
if(s.charAt(i) == cur){
cnt++;
}else{
cur = s.charAt(i);
list.add(cnt);
cnt = 1;
}
}
list.add(cnt);
int left = -1, ans = 0;
for (int i = 0; i < list.size(); i++) {
if(list.get(i) % 2 == 1){
if(left == -1) {
left = i;
}else{
ans += i - left;
left = -1;
}
}
}
System.out.println(ans);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 20b409eedeb86298bc4d921cde9a57a9 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(in.readLine());
while (t-- > 0) {
StringTokenizer st1 = new StringTokenizer(in.readLine());
String s = in.readLine();
int n = Integer.parseInt(st1.nextToken());
int[] a = new int[n];
for (int i=0; i<n; i++) {
char c = s.charAt(i);
if (c == '1')
a[i] = 1;
}
// List<Integer> temp = new ArrayList<>();
int zero = 0, one = 0, ans = 0;
int prev = a[0] == '0' ? 0 : 1;
for (int i=0; i<n; i++) {
if (prev != a[i]) {
// if (zero > 0)
// temp.add(zero);
// else
// temp.add(one);
if (zero%2 !=0 || one%2 != 0) {
ans++;
prev = a[i];
zero = 0;
one = 0;
if (a[i] == 0) zero++;
else one++;
}
}
if (a[i] == 0) zero++;
else one++;
prev = a[i];
}
out.println(ans);
}
in.close();
out.close();
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | b9a263b28d07d9da9453e504ae0cf1a1 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
// import java.lang.invoke.ConstantBootstraps;
// import java.math.BigInteger;
// import java.beans.IndexedPropertyChangeEvent;
import java.io.*;
@SuppressWarnings("unchecked")
public class Main implements Runnable {
static FastReader in;
static PrintWriter out;
static int bit(long n) {
return (n == 0) ? 0 : (1 + bit(n & (n - 1)));
}
static void p(Object o) {
out.print(o);
}
static void pn(Object o) {
out.println(o);
}
static void pni(Object o) {
out.println(o);
out.flush();
}
static String n() throws Exception {
return in.next();
}
static String nln() throws Exception {
return in.nextLine();
}
static int ni() throws Exception {
return Integer.parseInt(in.next());
}
static long nl() throws Exception {
return Long.parseLong(in.next());
}
static double nd() throws Exception {
return Double.parseDouble(in.next());
}
static class FastReader {
static BufferedReader br;
static StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception {
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
throw new Exception(e.toString());
}
return str;
}
}
static long power(long a, long b) {
if (a == 0L)
return 0L;
if (b == 0)
return 1;
long val = power(a, b / 2);
val = val * val;
if ((b % 2) != 0)
val = val * a;
return val;
}
static long power(long a, long b, long mod) {
if (a == 0L)
return 0L;
if (b == 0)
return 1;
long val = power(a, b / 2L, mod) % mod;
val = (val * val) % mod;
if ((b % 2) != 0)
val = (val * a) % mod;
return val;
}
static ArrayList<Long> prime_factors(long n) {
ArrayList<Long> ans = new ArrayList<Long>();
while (n % 2 == 0) {
ans.add(2L);
n /= 2L;
}
for (long i = 3; i * i <= n; i++) {
while (n % i == 0) {
ans.add(i);
n /= i;
}
}
if (n > 2) {
ans.add(n);
}
return ans;
}
static void sort(ArrayList<Long> a) {
Collections.sort(a);
}
static void reverse_sort(ArrayList<Long> a) {
Collections.sort(a, Collections.reverseOrder());
}
static void swap(long[] a, int i, int j) {
long temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(List<Long> a, int i, int j) {
long temp = a.get(i);
a.set(j, a.get(i));
a.set(j, temp);
}
static void sieve(boolean[] prime) {
int n = prime.length - 1;
Arrays.fill(prime, true);
for (int i = 2; i * i <= n; i++) {
if (prime[i]) {
for (int j = 2 * i; j <= n; j += i) {
prime[j] = false;
}
}
}
}
static long gcd(long a, long b) {
if (a < b) {
long temp = a;
a = b;
b = temp;
}
if (b == 0)
return a;
return gcd(b, a % b);
}
static HashMap<Long, Long> map_prime_factors(long n) {
HashMap<Long, Long> map = new HashMap<>();
while (n % 2 == 0) {
map.put(2L, map.getOrDefault(2L, 0L) + 1L);
n /= 2L;
}
for (long i = 3; i * i <= n; i++) {
while (n % i == 0) {
map.put(i, map.getOrDefault(i, 0L) + 1L);
n /= i;
}
}
if (n > 2) {
map.put(n, map.getOrDefault(n, 0L) + 1L);
}
return map;
}
static List<Long> divisor(long n) {
List<Long> ans = new ArrayList<>();
ans.add(1L);
long count = 0;
for (long i = 2L; i * i <= n; i++) {
if (n % i == 0) {
if (i == n / i)
ans.add(i);
else {
ans.add(i);
ans.add(n / i);
}
}
}
return ans;
}
static void sum_of_divisors(int n) {
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
dp[i] += i;
for (int j = i + i; j <= n; j += i) {
dp[j] += i;
}
}
}
static void prime_factorization_using_sieve(int n) {
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
for (int i = 2; i <= n; i++) {
dp[i] = Math.min(dp[i], i);// dp[i] stores smallest prime number which divides number i
for (int j = 2 * i; j <= n; j++) {// can calculate prime factorization in O(logn) time by dividing
// val/=dp[val]; till 1 is obtained
dp[j] = Math.min(dp[j], i);
}
}
}
/*
* ----------------------------------------------------Sorting------------------
* ------------------------------------------------
*/
public static void sort(long[] arr, int l, int r) {
if (l >= r)
return;
int mid = (l + r) / 2;
sort(arr, l, mid);
sort(arr, mid + 1, r);
merge(arr, l, mid, r);
}
public static void sort(int[] arr, int l, int r) {
if (l >= r)
return;
int mid = (l + r) / 2;
sort(arr, l, mid);
sort(arr, mid + 1, r);
merge(arr, l, mid, r);
}
static void merge(int[] arr, int l, int mid, int r) {
int[] left = new int[mid - l + 1];
int[] right = new int[r - mid];
for (int i = l; i <= mid; i++) {
left[i - l] = arr[i];
}
for (int i = mid + 1; i <= r; i++) {
right[i - (mid + 1)] = arr[i];
}
int left_start = 0;
int right_start = 0;
int left_length = mid - l + 1;
int right_length = r - mid;
int temp = l;
while (left_start < left_length && right_start < right_length) {
if (left[left_start] < right[right_start]) {
arr[temp] = left[left_start++];
} else {
arr[temp] = right[right_start++];
}
temp++;
}
while (left_start < left_length) {
arr[temp++] = left[left_start++];
}
while (right_start < right_length) {
arr[temp++] = right[right_start++];
}
}
static void merge(long[] arr, int l, int mid, int r) {
long[] left = new long[mid - l + 1];
long[] right = new long[r - mid];
for (int i = l; i <= mid; i++) {
left[i - l] = arr[i];
}
for (int i = mid + 1; i <= r; i++) {
right[i - (mid + 1)] = arr[i];
}
int left_start = 0;
int right_start = 0;
int left_length = mid - l + 1;
int right_length = r - mid;
int temp = l;
while (left_start < left_length && right_start < right_length) {
if (left[left_start] < right[right_start]) {
arr[temp] = left[left_start++];
} else {
arr[temp] = right[right_start++];
}
temp++;
}
while (left_start < left_length) {
arr[temp++] = left[left_start++];
}
while (right_start < right_length) {
arr[temp++] = right[right_start++];
}
}
// static int[] smallest_prime_factor;
// static int count = 1;
// static int[] p = new int[100002];
// static long[] flat_tree = new long[300002];
// static int[] in_time = new int[1000002];
// static int[] out_time = new int[1000002];
// static long[] subtree_gcd = new long[100002];
// static int w = 0;
// static boolean poss = true;
/*
* (a^b^c)%mod
* Using fermats Little theorem
* x^(mod-1)=1(mod)
* so b^c can be written as b^c=x*(mod-1)+y
* then (a^(x*(mod-1)+y))%mod=(a^(x*(mod-1))*a^(y))mod
* the term (a^(x*(mod-1)))%mod=a^(mod-1)*a^(mod-1)
*
*/
// ---------------------------------------------------Segment_Tree----------------------------------------------------------------//
// static class comparator implements Comparator<node> {
// public int compare(node a, node b) {
// return a.a - b.a > 0 ? 1 : -1;
// }
// }
static class Segment_Tree {
private long[] segment_tree;
public Segment_Tree(int n) {
this.segment_tree = new long[4 * n + 1];
}
void build(int index, int left, int right, int[] a) {
if (left == right) {
segment_tree[index] = a[left];
return;
}
int mid = (left + right) / 2;
build(2 * index + 1, left, mid, a);
build(2 * index + 2, mid + 1, right, a);
segment_tree[index] = segment_tree[2 * index + 1] + segment_tree[2 * index + 2];
}
long query(int index, int left, int right, int l, int r) {
if (left > right)
return 0;
if (left >= l && r >= right) {
return segment_tree[index];
}
if (l > right || left > r)
return 0;
int mid = (left + right) / 2;
return query(2 * index + 1, left, mid, l, r) + query(2 * index + 2, mid + 1, right, l, r);
}
void update(int index, int left, int right, int node, int val) {
if (left == right) {
segment_tree[index] += val;
return;
}
int mid = (left + right) / 2;
if (node <= mid)
update(2 * index + 1, left, mid, node, val);
else
update(2 * index + 2, mid + 1, right, node, val);
segment_tree[index] = segment_tree[2 * index + 1] + segment_tree[2 * index + 2];
}
}
static class min_Segment_Tree {
private long[] segment_tree;
public min_Segment_Tree(int n) {
this.segment_tree = new long[4 * n + 1];
}
void build(int index, int left, int right, int[] a) {
if (left == right) {
segment_tree[index] = a[left];
return;
}
int mid = (left + right) / 2;
build(2 * index + 1, left, mid, a);
build(2 * index + 2, mid + 1, right, a);
segment_tree[index] = Math.min(segment_tree[2 * index + 1], segment_tree[2 * index + 2]);
}
long query(int index, int left, int right, int l, int r) {
if (left > right)
return Integer.MAX_VALUE;
if (left >= l && r >= right) {
return segment_tree[index];
}
if (l > right || left > r)
return Integer.MAX_VALUE;
int mid = (left + right) / 2;
return Math.min(query(2 * index + 1, left, mid, l, r), query(2 * index + 2, mid + 1, right, l, r));
}
void update(int index, int left, int right, int node, int val) {
if (left == right) {
segment_tree[index] += val;
return;
}
int mid = (left + right) / 2;
if (node <= mid)
update(2 * index + 1, left, mid, node, val);
else
update(2 * index + 2, mid + 1, right, node, val);
segment_tree[index] = Math.min(segment_tree[2 * index + 1], segment_tree[2 * index + 2]);
}
}
static class max_Segment_Tree {
public long[] segment_tree;
public max_Segment_Tree(int n) {
this.segment_tree = new long[4 * n + 1];
}
void build(int index, int left, int right, long[] a) {
// pn(index+" "+left+" "+right);
if (left == right) {
segment_tree[index] = a[left];
return;
}
int mid = (left + right) / 2;
build(2 * index + 1, left, mid, a);
build(2 * index + 2, mid + 1, right, a);
segment_tree[index] = Math.max(segment_tree[2 * index + 1], segment_tree[2 * index + 2]);
}
long query(int index, int left, int right, int l, int r) {
if (left >= l && r >= right) {
return segment_tree[index];
}
if (l > right || left > r)
return Long.MIN_VALUE;
int mid = (left + right) / 2;
long ans1=query(2 * index + 1, left, mid, l, r);
long ans2=query(2 * index + 2, mid + 1, right, l, r);
long max=Math.max(ans1,ans2);
// pn(index+" "+left+" "+right+" "+max+" "+l+" "+r+" "+ans1+" "+ans2);
return max;
}
void update(int index, int left, int right, int node, int val) {
if (left == right) {
segment_tree[index] += val;
return;
}
int mid = (left + right) / 2;
if (node <= mid)
update(2 * index + 1, left, mid, node, val);
else
update(2 * index + 2, mid + 1, right, node, val);
segment_tree[index] = Math.max(segment_tree[2 * index + 1], segment_tree[2 * index + 2]);
}
}
// // ------------------------------------------------------ DSU
// // --------------------------------------------------------------------//
static class dsu {
private int[] parent;
private int[] rank;
private int[] size;
public dsu(int n) {
this.parent = new int[n + 1];
this.rank = new int[n + 1];
this.size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
rank[i] = 1;
size[i] = 1;
}
}
int findParent(int a) {
if (parent[a] == a)
return a;
else
return parent[a] = findParent(parent[a]);
}
void join(int a, int b) {
int parent_a = findParent(a);
int parent_b = findParent(b);
if (parent_a == parent_b)
return;
if (rank[parent_a] > rank[parent_b]) {
parent[parent_b] = parent_a;
size[parent_a] += size[parent_b];
} else if (rank[parent_a] < rank[parent_b]) {
parent[parent_a] = parent_b;
size[parent_b] += size[parent_a];
} else {
parent[parent_a] = parent_b;
size[parent_b] += size[parent_a];
rank[parent_b]++;
}
}
}
// ------------------------------------------------Comparable---------------------------------------------------------------------//
public static class rectangle {
int x1, x3, y1, y3;// lower left and upper rigth coordinates
int x2, y2, x4, y4;// remaining coordinates
/*
* (x4,y4) (x3,y3)
* ____________
* | |
* |____________|
*
* (x1,y1) (x2,y2)
*/
public rectangle(int x1, int y1, int x3, int y3) {
this.x1 = x1;
this.y1 = y1;
this.x3 = x3;
this.y3 = y3;
this.x2 = x3;
this.y2 = y1;
this.x4 = x1;
this.y4 = y3;
}
public long area() {
if (x3 < x1 || y3 < y1)
return 0;
return (long) Math.abs(x1 - x3) * (long) Math.abs(y1 - y3);
}
}
static long intersection(rectangle a, rectangle b) {
if (a.x3 < a.x1 || a.y3 < a.y1 || b.x3 < b.x1 || b.y3 < b.y1)
return 0;
long l1 = ((long) Math.min(a.x3, b.x3) - (long) Math.max(a.x1, b.x1));
long l2 = ((long) Math.min(a.y3, b.y3) - (long) Math.max(a.y1, b.y1));
if (l1 < 0 || l2 < 0)
return 0;
long area = ((long) Math.min(a.x3, b.x3) - (long) Math.max(a.x1, b.x1))
* ((long) Math.min(a.y3, b.y3) - (long) Math.max(a.y1, b.y1));
if (area < 0)
return 0;
return area;
}
// --------------------------------------------------------------Multiset---------------------------------------------------------------//
public static class multiset {
public TreeMap<Integer, Integer> map;
public int size = 0;
public multiset() {
map = new TreeMap<>();
}
public multiset(int[] a) {
map = new TreeMap<>();
size = a.length;
for (int i = 0; i < a.length; i++) {
map.put(a[i], map.getOrDefault(a[i], 0) + 1);
}
}
void add(int a) {
size++;
map.put(a, map.getOrDefault(a, 0) + 1);
}
void remove(int a) {
size--;
int val = map.get(a);
map.put(a, val - 1);
if (val == 1)
map.remove(a);
}
void removeAll(int a) {
if (map.containsKey(a)) {
size -= map.get(a);
map.remove(a);
}
}
int ceiling(int a) {
if (map.ceilingKey(a) != null) {
int find = map.ceilingKey(a);
return find;
} else
return Integer.MIN_VALUE;
}
int floor(int a) {
if (map.floorKey(a) != null) {
int find = map.floorKey(a);
return find;
} else
return Integer.MAX_VALUE;
}
int lower(int a) {
if (map.lowerKey(a) != null) {
int find = map.lowerKey(a);
return find;
} else
return Integer.MAX_VALUE;
}
int higher(int a) {
if (map.higherKey(a) != null) {
int find = map.higherKey(a);
return find;
} else
return Integer.MIN_VALUE;
}
int first() {
return map.firstKey();
}
int last() {
return map.lastKey();
}
boolean contains(int a) {
if (map.containsKey(a))
return true;
return false;
}
int size() {
return size;
}
void clear() {
map.clear();
}
int poll() {
if (map.size() == 0) {
return Integer.MAX_VALUE;
}
size--;
int first = map.firstKey();
if (map.get(first) == 1) {
map.pollFirstEntry();
} else
map.put(first, map.get(first) - 1);
return first;
}
int polllast() {
if (map.size() == 0) {
return Integer.MAX_VALUE;
}
size--;
int last = map.lastKey();
if (map.get(last) == 1) {
map.pollLastEntry();
} else
map.put(last, map.get(last) - 1);
return last;
}
}
static class pair implements Comparable<pair> {
int a;
int b;
int dir;
public pair(int a, int b, int dir) {
this.a = a;
this.b = b;
this.dir = dir;
}
public int compareTo(pair p) {
// if (this.b == Integer.MIN_VALUE || p.b == Integer.MIN_VALUE)
// return (int) (this.index - p.index);
return (int) (this.a - p.a);
}
}
static class pair2 implements Comparable<pair2> {
long a;
int index;
public pair2(long a, int index) {
this.a = a;
this.index = index;
}
public int compareTo(pair2 p) {
return (int) (this.a - p.a);
}
}
static class node implements Comparable<node> {
int l;
int r;
public node(int l, int r) {
this.l = l;
this.r = r;
}
public int compareTo(node a) {
if (this.l == a.l) {
return this.r - a.r;
}
return (int) (this.l - a.l);
}
}
static long ans = 0;
static int leaf = 0;
static boolean poss = true;
static long mod = 1000000007L;
static int[] dx = { -1, 0, 0, 1, -1, -1, 1, 1 };
static int[] dy = { 0, -1, 1, 0, -1, 1, -1, 1 };
static boolean cycle;
int count = 0;
public static void main(String[] args) throws Exception {
// new Thread(null,new Main(), "1", 1 << 26).start();
long start = System.nanoTime();
in = new FastReader();
out = new PrintWriter(System.out, false);
int tc = ni();
while (tc-- > 0) {
int n=ni();
String s=n();
int ans=0;
List<Integer> arr=new ArrayList<>();
int index=0;
int minus=0;
Map<Integer,Integer> map=new HashMap<>();
for(int i=0;i<n;i++){
char c=s.charAt(i);
int j=i;
while(j<n && s.charAt(j)==c){
j++;
}
if(((j-i)%2)==1){
arr.add(index);
map.put(index, j-i);
}
index++;
i=j-1;
}
for(int i=1;i<arr.size();i+=2){
ans+=arr.get(i)-arr.get(i-1);
if(map.get(arr.get(i))==1 || map.get(arr.get(i-1))==1)minus--;
}
pn(ans);
}
long end = System.nanoTime();
// pn((end-start)*1.0/1000000000);
out.flush();
out.close();
}
static boolean KMPSearch(String pat, char[] txt,int[] lps,int m)
{
int M = pat.length();
int N = m;
int j = 0;
int i = 0; // index for txt[]
while ((N - i) >= (M - j)) {
if (pat.charAt(j) == txt[i] || txt[i]=='#') {
j++;
i++;
}
if (j == M) {
return true;
// j = lps[j - 1];
}
// mismatch after j matches
else if (i < N && pat.charAt(j) != txt[i]) {
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
return false;
}
// static boolean find(int i, int length, int[] a, int n) {
// if (i == n) {
// if (length == 0)
// return true;
// return false;
// }
// if (i > n)
// return false;
// boolean ans = false;
// ans = find(i + 1, length + 1, a, n);
// if (length == 0) {
// ans = ans || find(i + a[i] + 1, 0, a, n);
// }
// if (a[i] == length) {
// ans = ans || (find(i + 1, 0, a, n));
// }
// return ans;
// }
static int find(int val, int[][] pre) {
int l = 0;
int r = pre.length - 1;
int ans = -1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (pre[mid][1] >= val) {
ans = mid;
l = mid + 1;
} else
r = mid - 1;
}
if (ans == -1) {
return -1;
} else
return pre[ans][0];
}
static long nc(long[] fact,int a,int b){
return (fact[a])/(fact[a-b]*fact[b]);
}
static long ncm(long[] fact, long[] fact_inv, int n, int m) {
if (n < m)
return 0L;
long a = fact[n];
long b = fact_inv[n - m];
long c = fact_inv[m];
a = (a * b) % mod;
return (a * c) % mod;
}
public void run() {
try {
in = new FastReader();
out = new PrintWriter(System.out);
int tc = ni();
while (tc-- > 0) {
}
out.flush();
} catch (Exception e) {
}
}
static void dfs(int i, int p, boolean[] visited, List<List<Integer>> arr, int length) {
visited[i] = true;
for (int nei : arr.get(i)) {
if (visited[nei] || nei == p)
continue;
dfs(nei, i, visited, arr, length + 1);
}
}
static boolean inside(int i, int j, int n, int m) {
if (i >= 0 && j >= 0 && i < n && j < m)
return true;
return false;
}
static int binary_search(int[] a, int val) {
int l = 0;
int r = a.length - 1;
int ans = 0;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] <= val) {
ans = mid;
l = mid + 1;
} else
r = mid - 1;
}
return ans;
}
static int[] longest_common_prefix(String s) {
int m = s.length();
int[] lcs = new int[m];
int len = 0;
int i = 1;
lcs[0] = 0;
while (i < m) {
if (s.charAt(i) == s.charAt(len)) {
lcs[i++] = ++len;
} else {
if (len == 0) {
lcs[i] = 0;
i++;
} else
len = lcs[len - 1];
}
}
return lcs;
}
static void swap(char[] a, char[] b, int i, int j) {
char temp = a[i];
a[i] = b[j];
b[j] = temp;
}
static void factorial(long[] fact, long[] fact_inv, int n, long mod) {
fact[0] = 1;
for (int i = 1; i < n; i++) {
fact[i] = (i * fact[i - 1]) % mod;
}
for (int i = 0; i < n; i++) {
fact_inv[i] = power(fact[i], mod - 2, mod);// (1/x)%m can be calculated by fermat's little theoram which is
// (x**(m-2))%m when m is prime
}
// (a^(b^c))%m is equal to, let res=(b^c)%(m-1) then (a^res)%m
// https://www.geeksforgeeks.org/find-power-power-mod-prime/?ref=rp
}
static void find(int i, int n, int[] row, int[] col, int[] d1, int[] d2) {
if (i >= n) {
ans++;
return;
}
for (int j = 0; j < n; j++) {
if (col[j] == 0 && d1[i - j + n - 1] == 0 && d2[i + j] == 0) {
col[j] = 1;
d1[i - j + n - 1] = 1;
d2[i + j] = 1;
find(i + 1, n, row, col, d1, d2);
col[j] = 0;
d1[i - j + n - 1] = 0;
d2[i + j] = 0;
}
}
}
static int answer(int l, int r, int[][] dp) {
if (l > r)
return 0;
if (l == r) {
dp[l][r] = 1;
return 1;
}
if (dp[l][r] != -1)
return dp[l][r];
int val = Integer.MIN_VALUE;
int mid = l + (r - l) / 2;
val = 1 + Math.max(answer(l, mid - 1, dp), answer(mid + 1, r, dp));
return dp[l][r] = val;
}
static void print(int[] a) {
for (int i = 0; i < a.length; i++)
p(a[i] + " ");
pn("");
}
static long count(long n) {
long count = 0;
while (n != 0) {
count += n % 10;
n /= 10;
}
return count;
}
static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static int LcsOfPrefix(String a, String b) {
int i = 0;
int j = 0;
int count = 0;
while (i < a.length() && j < b.length()) {
if (a.charAt(i) == b.charAt(j)) {
j++;
count++;
}
i++;
}
return a.length() + b.length() - 2 * count;
}
static void reverse(int[] a, int n) {
for (int i = 0; i < n / 2; i++) {
int temp = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = temp;
}
}
static char get_char(int a) {
return (char) (a + 'a');
}
static int find1(int[] a, int val) {
int ans = -1;
int l = 0;
int r = a.length - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] <= val) {
l = mid + 1;
ans = mid;
} else
r = mid - 1;
}
return ans;
}
static int find2(int[] a, int val) {
int l = 0;
int r = a.length - 1;
int ans = -1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] <= val) {
ans = mid;
l = mid + 1;
} else
r = mid - 1;
}
return ans;
}
// static void dfs(List<List<Integer>> arr, int node, int parent, long[] val) {
// p[node] = parent;
// in_time[node] = count;
// flat_tree[count] = val[node];
// subtree_gcd[node] = val[node];
// count++;
// for (int adj : arr.get(node)) {
// if (adj == parent)
// continue;
// dfs(arr, adj, node, val);
// subtree_gcd[node] = gcd(subtree_gcd[adj], subtree_gcd[node]);
// }
// out_time[node] = count;
// flat_tree[count] = val[node];
// count++;
// }
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 2df021ec0b2783bc4cd8b6a0c4371521 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static Scanner sc;
static PrintWriter pw;
public static void main(String[] args) throws IOException {
sc = new Scanner(System.in);
pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
char[] arr = sc.next().toCharArray();
int res = 0;
for (int i = 0; i < n; i++) {
char curr = arr[i];
int j = i,len = 0;
while(j < n && arr[j] == curr) {
j++;
len++;
}
if(len % 2 == 1) {
i = j;
res++;
}else {
i = j - 1;
}
}
pw.println(res);
}
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 readAllLines(BufferedReader reader) throws IOException {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line);
content.append(System.lineSeparator());
}
return content.toString();
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | b3f82bba206f20563a2777d6cca364ee | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class Vaibhav {
static long bit[];
static boolean prime[];
static class Pair implements Comparable<Pair> {
long x;
int y;
Pair(long x, int y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o){
return (int)(this.x-o.x);
}
}
static int st[]; //array to store segment tree
// A utility function to get minimum of two numbers
static int minVal(int x, int y) {
return (x < y) ? x : y;
}
// A utility function to get the middle index from corner
// indexes.
static int getMid(int s, int e) {
return s + (e - s) / 2;
}
/* A recursive function to get the minimum value in a given
range of array indexes. The following are parameters for
this function.
st --> Pointer to segment tree
index --> Index of current node in the segment tree. Initially
0 is passed as root is always at index 0
ss & se --> Starting and ending indexes of the segment
represented by current node, i.e., st[index]
qs & qe --> Starting and ending indexes of query range */
static int RMQUtil(int ss, int se, int qs, int qe, int index)
{
// If segment of this node is a part of given range, then
// return the min of the segment
if (qs <= ss && qe >= se)
return st[index];
// If segment of this node is outside the given range
if (se < qs || ss > qe)
return Integer.MAX_VALUE;
// If a part of this segment overlaps with the given range
int mid = getMid(ss, se);
return minVal(RMQUtil(ss, mid, qs, qe, 2 * index + 1),
RMQUtil(mid + 1, se, qs, qe, 2 * index + 2));
}
// Return minimum of elements in range from index qs (query
// start) to qe (query end). It mainly uses RMQUtil()
static int RMQ(int n, int qs, int qe)
{
// Check for erroneous input values
if (qs < 0 || qe > n - 1 || qs > qe) {
// System.out.println("Invalid Input");
return -1;
}
return RMQUtil(0, n - 1, qs, qe, 0);
}
// A recursive function that constructs Segment Tree for
// array[ss..se]. si is index of current node in segment tree st
static int constructSTUtil(int arr[], int ss, int se, int si)
{
// If there is one element in array, store it in current
// node of segment tree and return
if (ss == se) {
st[si] = arr[ss];
return arr[ss];
}
// If there are more than one elements, then recur for left and
// right subtrees and store the minimum of two values in this node
int mid = getMid(ss, se);
st[si] = minVal(constructSTUtil(arr, ss, mid, si * 2 + 1),
constructSTUtil(arr, mid + 1, se, si * 2 + 2));
return st[si];
}
/* Function to construct segment tree from given array. This function
allocates memory for segment tree and calls constructSTUtil() to
fill the allocated memory */
static void constructST(int arr[], int n)
{
// Allocate memory for segment tree
//Height of segment tree
int x = (int) (Math.ceil(Math.log(n) / Math.log(2)));
//Maximum size of segment tree
int max_size = 2 * (int) Math.pow(2, x) - 1;
st = new int[max_size]; // allocate memory
// Fill the allocated memory st
constructSTUtil(arr, 0, n - 1, 0);
}
static long power(long x, long y, long p) {
if (y == 0) return 1;
if (x == 0) return 0;
long res = 1l;
x = x % p;
while (y > 0) {
if (y % 2 == 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long power(long x, long y) {
if (y == 0) return 1;
if (x == 0) return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1) res = (res * x);
y = y >> 1;
x = (x * x);
}
return res;
}
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 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());
}
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;
}
long nextLong() {
return Long.parseLong(next());
}
}
static void sieveOfEratosthenes(int n) {
prime = new boolean[n + 1];
for (int i = 0; i <= n; i++) prime[i] = true;
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) prime[i] = false;
}
}
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
static int binomialCoeff(int n, int r)
{
if (r > n)
return 0;
long m = 1000000007;
long inv[] = new long[r + 1];
inv[0] = 1;
if(r+1>=2)
inv[1] = 1;
// Getting the modular inversion
// for all the numbers
// from 2 to r with respect to m
// here m = 1000000007
for (int i = 2; i <= r; i++) {
inv[i] = m - (m / i) * inv[(int) (m % i)] % m;
}
int ans = 1;
// for 1/(r!) part
for (int i = 2; i <= r; i++) {
ans = (int) (((ans % m) * (inv[i] % m)) % m);
}
// for (n)*(n-1)*(n-2)*...*(n-r+1) part
for (int i = n; i >= (n - r + 1); i--) {
ans = (int) (((ans % m) * (i % m)) % m);
}
return ans;
}
public static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static BigInteger bi(String str) {
return new BigInteger(str);
}
// Author - vaibhav_1710
static FastScanner fs = null;
static long ans;
static long mod=998244353;
static ArrayList<Integer> al[];
static int dp[][][];
static int cntbig;
static int cntless;
static HashSet<Long> h;
static int a[];
public static void main(String[] args) {
fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t =fs.nextInt();
outer:
while (t-- > 0 ){
int n = fs.nextInt();
String s = fs.next();
ArrayList<Integer> al = new ArrayList<>();
int i=0;
int cnt=0;
int cnto=0;
int cnt1=0;
int cnt0=0;
for(i=0;i<s.length();i++){
if(s.charAt(i)=='0') cnt0++;
else cnt1++;
}
i=0;
while(i<s.length()){
char c = s.charAt(i);
cnt=0;
while(i<n && s.charAt(i)==c){
cnt++;
i++;
}
if(cnt%2!=0) cnto++;
al.add(cnt);
}
if(cnto%2!=0){
out.println(Math.min(cnt0,cnt1));
continue outer;
}
int f=-1;
int ss=-1;
int ans=0;
for(i=0;i<al.size();i++){
if(al.get(i)%2!=0){
if(f==-1){
f = i;
}else{
ans += (i-f);
f = -1;
}
}
}
out.println(ans);
}
out.close();
}
public static long query(int a,int b){
System.out.println("? "+a+" "+b);
System.out.flush();
long v = fs.nextLong();
return v;
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | e2de77926808afb528767ba0d4caf170 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //package kg.my_algorithms.Codeforces;
import java.io.*;
import java.util.*;
/*
*/
public class Solution {
public static void main(String[] args) throws IOException {
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
FastReader fr = new FastReader();
StringBuilder sb = new StringBuilder();
int testCases = fr.nextInt();
for(int test=1;test<=testCases;test++){
List<Integer> groups = new ArrayList<>();
int n = fr.nextInt();
String s = fr.next();
char prev = s.charAt(0);
int groupSize = 1;
for(int i=1;i<s.length();i++){
if(prev!=s.charAt(i)){
groups.add(groupSize);
prev = s.charAt(i);
groupSize = 1;
}
else groupSize++;
}
groups.add(groupSize);
int changes = 0;
for(int i=0;i<groups.size()-1;i++){
if(groups.get(i)%2==1){
if(groups.get(i+1)%2==1){
changes++;
i++;
}
else{
changes++;
groups.set(i+1,groups.get(i+1)+1);
}
}
}
sb.append(changes).append("\n");
}
output.write(sb.toString());
output.flush();
}
}
//Fast Input
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 58c18afdc0051470041750c7a40612a0 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.StringTokenizer;
public class Tokitsukaze_and_Good_01String_easy_version
{
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 java.lang.Exception
{
try
{
Scanner obj = new Scanner (System.in);
int t = obj.nextInt();
while (t > 0)
{
t--;
int n = obj.nextInt();
String str = obj.next();
int i = 0;
ArrayList<Integer> arr = new ArrayList<Integer>();
while (i < n)
{
char chi = str.charAt(i);
int j = i;
while (j < n)
{
char chj = str.charAt(j);
if (chi == chj)
{
j++;
}
else
{
break;
}
}
if ((j - i) % 2 == 0)
{
arr.add (0);
}
else
{
arr.add (1);
}
i = j;
}
int l = arr.size(), c = 0;
for (i=0;i<l-1;i++)
{
if (arr.get (i) == 1)
{
c++;
int v = arr.get (i+1);
int ele = arr.set (i+1, (v ^ 1));
}
}
System.out.println (c);
}
}catch(Exception e){
return;
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 97bfe76565cdb1699f9eaa55d41599f4 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 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++) {
int N = scan.nextInt();
String s = scan.next();
char[] ch = s.toCharArray();
int ans = 0;
int count = 1;
for(int j = 1; j < N; j++){
if(ch[j] == ch[j-1]) {
count++;
}else{
if(count%2 == 0) {
count = 1;
}else{
ch[j] = ch[j-1];
ans++;
j++;
count = 1;
}
}
}
System.out.println(ans);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 57fa633ea63c55d557cba4ecc66b7b27 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static int[] arr = new int[1000000];
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t>0) {
int n = scanner.nextInt();
scanner.nextLine();
String s = scanner.nextLine();
char prev='2';
int count=0;
int ans=0;
//System.out.println(s);
for(int i=0;i<n;i++){
if(prev!='2'){
if(prev!=s.charAt(i)){
if(count%2 == 1){
ans++;
}
}
}
count++;
prev = s.charAt(i);
}
if(count%2==1){
ans++;
}
System.out.println(ans);
t--;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | a43a763f7973d5a9d56d23ecf3d4a4a3 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int testCount = Integer.parseInt(sc.nextLine());
int[] result = new int[testCount];
for (int i = 0; i < testCount; i++) {
int stringLen = Integer.parseInt(sc.nextLine());
String str = sc.nextLine();
int primMin = 0;
if (str.charAt(0) == str.charAt(1)) {
primMin = 0;
} else {
primMin = 1;
}
for (int j = 3; j < str.length(); j += 2) {
if (str.charAt(j - 1) == str.charAt(j)) {
primMin = primMin;
} else {
primMin = primMin + 1;
}
}
result[i] = primMin;
}
for (int i = 0; i < result.length; i++) {
System.out.println(result[i]);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 6c8a917119cdb0eb61f79950cd69cb89 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();//用例个数
while(n!=0){
int ans = 0;
int len=sc.nextInt();//s长度
String s=sc.next();
for (int i = 0; i < len-1; i++) {
if (s.charAt(i)!=s.charAt(i+1)){
ans++;
}
i++;
}
System.out.println(ans);
n--;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 18c8952c640152b823782b1659da68b1 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.util.Scanner;
public class GoodString {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-->0){
int n = in.nextInt();
String s = in.next();
int count = 0;
for (int i = 1; i < n; i+=2) {
if(s.charAt(i-1) != s.charAt(i))
count++;
}
System.out.println(count);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 565a5f2b775411e212b910d655f697c5 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1678B
{
static final int INF = Integer.MAX_VALUE/2;
public static void main(String[] followthekkathyoninsta) throws Exception
{
BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(infile.readLine());
int T = Integer.parseInt(st.nextToken());
StringBuilder sb = new StringBuilder();
while(T-->0)
{
st = new StringTokenizer(infile.readLine());
int N = Integer.parseInt(st.nextToken());
String line = infile.readLine().trim();
int[] arr = new int[N];
for(int i=0; i < N; i++)
arr[i] = line.charAt(i)-'0';
Pair[][] dp = new Pair[2][2];
fill(dp);
dp[1][arr[0]] = new Pair(0, 0);
dp[1][arr[0]^1] = new Pair(1, 0);
for(int t=1; t < N; t++)
{
Pair[][] next = new Pair[2][2];
fill(next);
for(int a=0; a < 2; a++)
for(int b=0; b < 2; b++)
if(dp[a][b].moves != INF)
{
//extend from block
int na = a^1;
int nb = b;
int cost = b^arr[t];
Pair temp = new Pair(dp[a][b].moves+cost, dp[a][b].blocks);
if(next[na][nb].compareTo(temp) > 0)
next[na][nb] = temp;
//create new block
if(a == 0)
{
na = 1;
nb = b^1;
cost ^= 1;
temp = new Pair(dp[a][b].moves+cost, dp[a][b].blocks+1);
if(next[na][nb].compareTo(temp) > 0)
next[na][nb] = temp;
}
}
dp = next;
}
Pair res = dp[0][0];
if(res.compareTo(dp[0][1]) > 0)
res = dp[0][1];
sb.append(res.moves);
sb.append("\n");
}
System.out.print(sb);
}
public static void fill(Pair[][] dp)
{
for(int a=0; a < 2; a++)
for(int b=0; b < 2; b++)
dp[a][b] = new Pair(INF, INF);
}
}
/*
dp[i][odd/even][tag] = minimum moves such that for first i elements, the current block = tag and it's length is odd or even
block variable won't include itself
can also store the hard version easily
*/
class Pair
{
public int moves;
public int blocks;
public Pair(int a, int b)
{
moves = a;
blocks = b;
}
public int compareTo(Pair oth)
{
if(moves != oth.moves)
return moves-oth.moves;
return blocks-oth.blocks;
}
}
/*
1
8
11001111
*/ | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 18ddf94e5dfeea972528ea8b6a77d47f | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
import java.io.*;
import java.lang.reflect.Array;
public class CodeForces {
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
try {
br = new BufferedReader(new FileReader("input.txt"));
PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
System.setOut(out);
} 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;
}
}
void shuffle(int a[], int n) {
for (int i = 0; i < n; i++) {
int t = (int) Math.random() * a.length;
int x = a[t];
a[t] = a[i];
a[i] = x;
}
}
long lcm(long a, long b) {
long val = a * b;
return val / gcd(a, b);
}
void swap(long a[], long[] b, int i) {
long temp = a[i];
a[i] = b[i];
b[i] = temp;
}
long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
long mod = (long) 1e9 + 7, c;
FastReader in;
PrintWriter o;
public static void main(String[] args) throws IOException {
new CodeForces().run();
}
void run() throws IOException {
in = new FastReader();
OutputStream op = System.out;
o = new PrintWriter(op);
int t;
t = 1;
t = in.nextInt();
for (int i = 1; i <= t; i++) {
helper();
o.println();
}
o.flush();
o.close();
}
public void helper() {
int n=in.nextInt();
char a[]=in.nextLine().toCharArray();
int ans=0;
for(int i=0;i<n;i++)
{
int j=i;
while(i+1<n && a[i+1]==a[j])
i++;
if((i-j+1)%2==0)continue;
ans++;
i++;
}
o.print(ans);
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 1967c93b8cd1bc001dea0e6f70ecc750 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args)throws Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw=new PrintWriter(System.out);
int t=Integer.parseInt(br.readLine());
while(t-->0)
{
int n=Integer.parseInt(br.readLine());
String s=br.readLine();
ArrayList<Integer> al=new ArrayList<>();
int ct=1;
for(int i=0;i<n-1;i++)
{
if(s.charAt(i+1)==s.charAt(i))
{
ct++;
}
else
{
al.add(ct);
ct=1;
}
}
al.add(ct);
int ans=0;
for(int i=0;i<al.size()-1;i++)
{
if(al.get(i)%2==1)
{
al.set(i,al.get(i)-1);
al.set(i+1,al.get(i+1)+1);
ans++;
}
}
pw.println(ans);
}
pw.flush();
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | ee197a80f6bf975773952da3cd81ca67 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main
{
public static void main(String args[])
{
FastScanner input = new FastScanner();
int tc = input.nextInt();
work:
while (tc-- > 0) {
int n = input.nextInt();
char a[] = input.next().toCharArray();
int ans=0;
for (int i = 0; i <n; i++) {
if(a[i]=='0')
{
int count=1;
while(i+1<n&&a[i+1]=='0')
{
count++;
i++;
}
if(count%2==1)
{
ans++;
i++;
}
}
else{
int count=1;
while(i+1<n&&a[i+1]=='1')
{
count++;
i++;
}
if(count%2==1)
{
ans++;
i++;
}
}
}
System.out.println(ans);
}
}
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 nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine() throws IOException
{
return br.readLine();
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 9c646fdbaa169e93adc295b655168d42 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Roy
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
B1TokitsukazeAndGood01StringEasyVersion solver = new B1TokitsukazeAndGood01StringEasyVersion();
solver.solve(1, in, out);
out.close();
}
static class B1TokitsukazeAndGood01StringEasyVersion {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int tCases = in.readInteger();
for (int cs = 1; cs <= tCases; ++cs) {
int n = in.readInteger();
String s = in.readString();
int segLength = 1;
List<Integer> segList = new ArrayList<>();
for (int i = 1; i < n; i++) {
if (s.charAt(i) == s.charAt(i - 1)) segLength++;
else {
segList.add(segLength);
segLength = 1;
}
}
segList.add(segLength);
int ans = 0;
for (int i = 0, sz = segList.size() - 1; i < sz; i++) {
int x = segList.get(i);
if (MathUtility.isOdd(x)) {
ans++;
segList.set(i, x + 1);
segList.set(i + 1, segList.get(i + 1) - 1);
}
}
out.printLine(ans);
}
}
}
static class InputReader {
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
private final InputStream stream;
private final byte[] buf = new byte[1024];
public InputReader(InputStream stream) {
this.stream = stream;
}
private long readWholeNumber(int c) {
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInteger() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = (int) readWholeNumber(c);
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
this.print(objects);
writer.println();
}
public void close() {
writer.flush();
writer.close();
}
}
static class MathUtility {
public static <T> boolean isOdd(T number) {
long longNumber = Long.parseLong(number.toString());
return ((longNumber & 1) > 0);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 9b36c7f60b0d811c38b89cf8af70e175 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //package com.company;
import java.util.*;
import java.io.*;
public class TokitsukazeEasy {
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(--t >= 0) {
int len = Integer.parseInt(br.readLine());
String binS = br.readLine();
int num = 0; int c = 0; ArrayList<Boolean> conS = new ArrayList<>();
for(int i = 0; i < len; i++) {
if(i == 0) {
num = binS.charAt(i) - '0'; c++;
}
else {
if(binS.charAt(i) - '0' == num) {
c++;
if(i == len - 1) {
conS.add(c % 2 == 1);
}
}
else {
conS.add(c % 2 == 1);
num = binS.charAt(i) - '0'; c = 1;
}
}
}
int ans = 0;
for(int i = 0; i < conS.size(); i++) {
if(conS.get(i)) {
ans++; conS.set(i, !conS.get(i));
if(i != conS.size() - 1) conS.set(i+1, !conS.get(i+1));
}
}
System.out.println(ans);
}
br.close();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 05e0e58ac8230e2c525d5dcb49e08781 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Sanket Makani
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
B1TokitsukazeAndGood01StringEasyVersion solver = new B1TokitsukazeAndGood01StringEasyVersion();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class B1TokitsukazeAndGood01StringEasyVersion {
public void solve(int testNumber, InputReader sc, PrintWriter w) {
int n = sc.nextInt();
char c[] = sc.next().toCharArray();
List<Integer> list = new ArrayList<>();
int current = 1;
for (int i = 1; i < n; i++) {
if (c[i] == c[i - 1]) {
current++;
} else {
list.add(current);
current = 1;
}
}
list.add(current);
int answer = 0;
for (int i = 0; i < list.size() - 1; i++) {
int currentElem = list.get(i);
int nextElem = list.get(i + 1);
if (currentElem % 2 == 0) {
continue;
}
nextElem--;
list.set(i + 1, nextElem);
answer++;
}
w.println(answer);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 1869bd5642a8f2eecef91b32e803ef57 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.io.*;
public class B {
private static FastReader fr;
private static OutputStream out;
private static int mod = (int)(1e9+7);
private void solve() {
int t = fr.nextInt();
while(t-- > 0) {
int n = fr.nextInt();
char arr[] = fr.next().toCharArray();
int ans = 0;
int ones = arr[0] == '1' ? 1 : 0;
int zeros = 1 - ones;
for(int i=1;i<n;i++){
if(arr[i] == arr[i-1]){
ones += arr[i] == '1' ? 1 : 0;
zeros += arr[i] == '0' ? 1 : 0;
}else if(arr[i-1] == '0'){
ones = 1;
if(zeros % 2 != 0){
ans ++;
ones ++;
}
zeros = 0;
}else{
zeros = 1;
if(ones % 2 != 0){
ans ++;
zeros ++;
}
ones = 0;
}
}
println(ans);
}
}
public static void main(String args[]) throws IOException{
new B().run();
}
private static int modInverse(int a) {
int m0 = mod;
int y = 0, x = 1;
if (mod == 1)
return 0;
while (a > 1) {
int q = (int)(a / mod);
int t = mod;
mod = a % mod;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
private ArrayList<Integer> factors(int n, boolean include){
ArrayList<Integer> factors = new ArrayList<>();
if(n < 0)
return factors;
if(include) {
factors.add(1);
if(n > 1)
factors.add(n);
}
int i = 2;
for(;i*i<n;i++) {
if(n % i == 0) {
factors.add(i);
factors.add(n / i);
}
}
if(i * i == n) {
factors.add(i);
}
return factors;
}
private ArrayList<Integer> PrimeFactors(int n) {
ArrayList<Integer> primes = new ArrayList<>();
int i = 2;
while (i * i <= n) {
if (n % i == 0) {
primes.add(i);
n /= i;
while (n % i == 0) {
primes.add(i);
n /= i;
}
}
i++;
}
if(n > 1)
primes.add(i);
return primes;
}
private boolean isPrime(int n) {
if(n == 0 || n == 1) {
return false;
}
if(n % 2 == 0) {
return false;
}
for(int i=3;i*i<=n;i+=2) {
if(n % i == 0) {
return false;
}
}
return true;
}
private ArrayList<Integer> Sieve(int n){
boolean bool[] = new boolean[n+1];
Arrays.fill(bool, true);
bool[0] = bool[1] = false;
for(int i=2;i*i<=n;i++) {
if(bool[i]) {
int j = 2;
while(i*j <= n) {
bool[i*j] = false;
j++;
}
}
}
ArrayList<Integer> primes = new ArrayList<>();
for(int i=2;i<=n;i++) {
if(bool[i])
primes.add(i);
}
return primes;
}
private HashMap<Integer, Integer> addToHashMap(HashMap<Integer, Integer> map, int arr[]){
for(int val: arr) {
if(!map.containsKey(val)) {
map.put(val, 1);
}else {
map.put(val, map.get(val) + 1);
}
}
return map;
}
private int factorial(int n) {
long fac = 1;
for(int i=2;i<=n;i++) {
fac *= i;
fac %= mod;
}
return (int)(fac % mod);
}
// private static int pow(int base,int exp){
// if(exp == 0){
// return 1;
// }else if(exp == 1){
// return base;
// }
// int a = pow(base,exp/2);
// a = ((a % mod) * (a % mod)) % mod;
// if(exp % 2 == 1) {
// a = ((a % mod) * (base % mod));
// }
// return a;
// }
private static int gcd(int a,int b){
if(a == 0){
return b;
}
return gcd(b%a,a);
}
private static int lcm(int a,int b){
return (a * b)/gcd(a,b);
}
private void run() throws IOException{
fr = new FastReader();
out = new BufferedOutputStream(System.out);
solve();
out.flush();
out.close();
}
private static class FastReader{
private static BufferedReader br;
private static StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
out = new BufferedOutputStream(System.out);
}
public String next() {
while (st == null || !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;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public char[] nextCharArray() {
return next().toCharArray();
}
public int[] nextIntArray(int n) {
int arr[] = new int[n];
for(int i=0;i<n;i++) {
arr[i] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
long arr[] = new long[n];
for(int i=0;i<n;i++) {
arr[i] = nextLong();
}
return arr;
}
public String[] nextStringArray(int n) {
String arr[] = new String[n];
for(int i=0;i<n;i++) {
arr[i] = next();
}
return arr;
}
}
public static void print(Object str) {
try {
out.write(str.toString().getBytes());
} catch (IOException e) {
e.printStackTrace();
}
}
public static void println(Object str) {
try {
out.write((str.toString() + "\n").getBytes());
} catch (IOException e) {
e.printStackTrace();
}
}
public static void println() {
println("");
}
public static void printArray(Object str[]){
for(Object s : str) {
try {
out.write(str.toString().getBytes());
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | ea40962a1489dfe06d00008e1ff8d3d5 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
public class GoodString{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.parseInt(sc.nextLine());
for(int i = 0 ; i < t ; i++) {
int len = Integer.parseInt(sc.nextLine());
String num = sc.nextLine();
int count = 1;
int moves = 0;
for(int j = 0 ; j < num.length() ; j++) {
if(j > 0) {
if(num.charAt(j) == num.charAt(j-1)) {
count ++;
}
else if(num.charAt(j) != num.charAt(j-1)) {
if(count % 2 ==1) {
moves ++;
count = 0;
}
else {
count ++;
}
}
}
}
System.out.println(moves);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 767641e19093fe6d34768f18fdb4b9a4 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 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) {
int n = sc.nextInt(), cnt = 0;
char c[] = sc.next().toCharArray();
for(int i=0; i<c.length; i+=2)
if(c[i] != c[i+1])
cnt++;
System.out.println(cnt);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | a2f008a7efba09fd28c496abde8fca66 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
//1。Tokitsukaze and Good 01-String (简易版)
//1110111000
//(3 1) (2 0) (2 1) (3 0)
public class Tokit {
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
int n=scan.nextInt();
for (int i = 0; i < n; i++) {
int x=scan.nextInt();
String s=scan.next();
int change=0;
int count=0;
for (int j = 0; j <s.length(); j++) {
if(count!=0) {
if(s.charAt(j)!=s.charAt(j-1)) {
if(count%2!=0) {
change++;
j++;
}
count=0;
}
}
count++;
}
System.out.println(change);
}
scan.close();
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 9b266579ea22c4d4a0926a4ed09c2d02 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class _1678B1 {
FastScanner scn;
PrintWriter w;
PrintStream fs;
int MOD = 1000000007;
int MAX = 200005;
long mul(long x, long y) {long res = x * y; return (res >= MOD ? res % MOD : res);}
long power(long x, long y) {if (y < 0) return 1; long res = 1; x %= MOD; while (y!=0) {if ((y & 1)==1)res = mul(res, x); y >>= 1; x = mul(x, x);} return res;}
void ruffleSort(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);}
void reverseSort(int[] arr){List<Integer> list = new ArrayList<>();for (int i=0; i<arr.length; i++){list.add(arr[i]);}Collections.sort(list, Collections.reverseOrder());for (int i = 0; i < arr.length; i++){arr[i] = list.get(i);}}
boolean LOCAL;
void debug(Object... o){if(LOCAL)System.err.println(Arrays.deepToString(o));}
//SPEED IS NOT THE CRITERIA, CODE SHOULD BE A NO BRAINER, CMP KILLS, MOCIM
void solve(){
int t=scn.nextInt();
while(t-->0)
{
int n= scn.nextInt();
String s = scn.next();
int ans = 0;
for(int i=n-1;i>=0;i--){
int j = i;
int val = s.charAt(i)-'0';
int ct= 0;
while(j>=0&&(s.charAt(j)-'0')==val){
ct++;
j--;
}
j++;
i=j;
if((ct&1)==0){
continue;
}else{
ans++;
i--;
}
}
w.println(ans);
}
}
void run() {
try {
long ct = System.currentTimeMillis();
scn = new FastScanner(new File("input.txt"));
w = new PrintWriter(new File("output.txt"));
fs=new PrintStream("error.txt");
System.setErr(fs);
LOCAL=true;
solve();
w.close();
System.err.println(System.currentTimeMillis() - ct);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
void runIO() {
scn = new FastScanner(System.in);
w = new PrintWriter(System.out);
LOCAL=false;
solve();
w.close();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return null;
st = new StringTokenizer(s);
}
return st.nextToken();
}
boolean hasMoreTokens() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return false;
st = new StringTokenizer(s);
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
long[] nextLongArray(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());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
int lowerBound(int[] arr, int x){int n = arr.length, si = 0, ei = n - 1;while(si <= ei){int mid = si + (ei - si)/2;if(arr[mid] == x){if(mid-1 >= 0 && arr[mid-1] == arr[mid]){ei = mid-1;}else{return mid;}}else if(arr[mid] > x){ei = mid - 1; }else{si = mid+1;}}return si; }
int upperBound(int[] arr, int x){int n = arr.length, si = 0, ei = n - 1;while(si <= ei){int mid = si + (ei - si)/2;if(arr[mid] == x){if(mid+1 < n && arr[mid+1] == arr[mid]){si = mid+1;}else{return mid + 1;}}else if(arr[mid] > x){ei = mid - 1; }else{si = mid+1;}}return si; }
int upperBound(ArrayList<Integer> list, int x){int n = list.size(), si = 0, ei = n - 1;while(si <= ei){int mid = si + (ei - si)/2;if(list.get(mid) == x){if(mid+1 < n && list.get(mid+1) == list.get(mid)){si = mid+1;}else{return mid + 1;}}else if(list.get(mid) > x){ei = mid - 1; }else{si = mid+1;}}return si; }
void swap(int[] arr, int i, int j){int temp = arr[i];arr[i] = arr[j];arr[j] = temp;}
long nextPowerOf2(long v){if (v == 0) return 1;v--;v |= v >> 1;v |= v >> 2;v |= v >> 4;v |= v >> 8;v |= v >> 16;v |= v >> 32;v++;return v;}
int gcd(int a, int b) {if(a == 0){return b;}return gcd(b%a, a);} // TC- O(logmax(a,b))
boolean nextPermutation(int[] arr) {if(arr == null || arr.length <= 1){return false;}int last = arr.length-2;while(last >= 0){if(arr[last] < arr[last+1]){break;}last--;}if (last < 0){return false;}if(last >= 0){int nextGreater = arr.length-1;for(int i=arr.length-1; i>last; i--){if(arr[i] > arr[last]){nextGreater = i;break;}}swap(arr, last, nextGreater);}int i = last + 1, j = arr.length - 1;while(i < j){swap(arr, i++, j--);}return true;}
public static void main(String[] args) {
new _1678B1().runIO();
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 6c770f6cd9cf3f8af1fe9e59b08b804b | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | // Problem: B1. Tokitsukaze and Good 01-String (easy version)
// Contest: Codeforces - Codeforces Round #789 (Div. 2)
// URL: https://codeforces.com/contest/1678/problem/B1
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
import java.util.*;
public class My_template {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
solve(sc);
}
}
private static void solve(Scanner sc) {
int n=sc.nextInt();
String s=sc.next();
int c=1;
int ans=0;
char[] ar=s.toCharArray();
int i=0;
char curr=ar[0];
for( i=1;i<s.length();i++){
if(ar[i]==curr){
c++;
continue;
}
if(c%2==0){
curr=ar[i];
c=1;
}else
{
ans++;
c++;
}
}
System.out.println(ans);
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | acbe5f60690d121f9fec715976ba6ba9 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
// * * * 2100 * * * //
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class A {
static class Pair {
int x;
int y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
static class Edge implements Comparable<Edge> {
int u;
int v;
int w;
Edge(int u, int v, int w) {
this.u = u;
this.v = v;
this.w = w;
}
public int compareTo(Edge o) {
return this.w - o.w;
}
}
public static void main(String[] args) {
FastScanner sc = new FastScanner();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
char ca[] = (sc.nextLine()).toCharArray();
boolean can = true;
int prev = 0;
int count = 0;
for (int i = 1; i < n; i++) {
if (ca[i] != ca[i - 1]) {
if ((i - prev) % 2 != 0) {
count++;
} else
prev = i;
}
}
System.out.println(count);
}
}
static void sort(int[] a) {
// ruffle
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;
}
// then sort
Arrays.sort(a);
}
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());
}
double nextDouble() {
return Double.parseDouble(next());
}
String str = "";
String nextLine() {
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
static long gcd(long n, long l) {
if (l == 0)
return n;
return gcd(l, n % l);
}
static void sieveOfEratosthenes(int n, ArrayList<Integer> al) {
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++) {
if (prime[i] == true)
al.get(i);
}
}
static final int mod = 100000000 + 7;
static final int max_val = 2147483647;
static final int min_val = max_val + 1;
static long fastPow(long base, long exp) {
if (exp == 0)
return 1;
long half = fastPow(base, exp / 2);
if (exp % 2 == 0)
return mul(half, half);
return mul(half, mul(half, base));
}
static long mul(long a, long b) {
return a * b % mod;
}
static int nCr(int n, int r) {
return fact(n) / (fact(r) *
fact(n - r));
}
static int fact(int n) {
int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// a -> z == 97 -> 122
// String.format("%.9f", ans) ,--> to get upto 9 decimal places , (ans is
// double)
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 76f81f6a0dc44b8800b59fc2c6459662 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class B {
public static void main(String[]args) throws IOException {
Scanner sc=new Scanner(System.in);
PrintWriter out=new PrintWriter(System.out);
int t=sc.nextInt();
a:while(t-->0) {
int n=sc.nextInt();
String s=sc.next();
ArrayList<Integer>arr=new ArrayList<Integer>();
int ans=0;
for(int i=0;i<n;) {
int cnt=0;
while(i<n&&s.charAt(i)=='0') {
cnt++;
i++;
}
if(cnt>0) {arr.add(cnt);cnt=0;}
while(i<n&&s.charAt(i)=='1') {
cnt++;
i++;
}
if(cnt>0) arr.add(cnt);
}
boolean odd=false;
int lst=-1;
for(int i=0;i<arr.size();i++) {
if(arr.get(i)%2==0)continue;
if(odd) {
ans+=i-lst;
odd=false;
}else {
odd=true;
lst=i;
}
}
out.println(ans);
}
out.close();
}
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 boolean hasNext() {return st.hasMoreTokens();}
public int nextInt() throws IOException {return Integer.parseInt(next());}
public double nextDouble() throws IOException {return Double.parseDouble(next());}
public long nextLong() throws IOException {return Long.parseLong(next());}
public String nextLine() throws IOException {return br.readLine();}
public boolean ready() throws IOException {return br.ready(); }
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 76eddbf16d5dee442b316cdde0cfee44 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static FastReader sc=new FastReader();
// static long dp[][];
// static boolean v[][][];
// static int mod=998244353;;
// static int mod=1000000007;
static long oset[];
static int oset_p;
static long mod=1000000007;
// static int max;
static int bit[];
//static long fact[];
//static HashMap<Long,Long> mp;
//static StringBuffer sb=new StringBuffer("");
//static HashMap<Integer,Integer> map;
static PrintWriter out=new PrintWriter(System.out);
public static void main(String[] args)
{
// StringBuffer sb=new StringBuffer("");
int ttt=1;
ttt =i();
outer :while (ttt-- > 0)
{
int n=i();
long cnt=0;
String s=s();
int r=0;
int l=0;
while(r<n)
{
int ch=s.charAt(r);
while(r<n&&s.charAt(r)==ch)
r++;
if((r-l)%2!=0)
{
cnt++;
r++;
}
l=r;
}
out.println(cnt);
}
out.close();
}
static class Pair implements Comparable<Pair>
{
int x;
int y;
// int z;
Pair(int x,int y){
this.x=x;
this.y=y;
// this.z=z;
}
@Override
public int compareTo(Pair o) {
if(this.x>o.x)
return 1;
else if(this.x<o.x)
return -1;
else {
if(this.y>o.y)
return 1;
else if(this.y<o.y)
return -1;
else
return 0;
}
}
public int hashCode()
{
final int temp = 14;
int ans = 1;
ans =x*31+y*13;
return ans;
}
@Override
public boolean equals(Object o)
{
if (this == o) {
return true;
}
if (o == null) {
return false;
}
if (this.getClass() != o.getClass()) {
return false;
}
Pair other = (Pair)o;
if (this.x != other.x || this.y!=other.y) {
return false;
}
return true;
}
//
/* FOR TREE MAP PAIR USE */
// public int compareTo(Pair o) {
// if (x > o.x) {
// return 1;
// }
// if (x < o.x) {
// return -1;
// }
// if (y > o.y) {
// return 1;
// }
// if (y < o.y) {
// return -1;
// }
// return 0;
// }
}
static int find(int A[],int a) {
if(A[a]==a)
return a;
return A[a]=find(A, A[a]);
}
//static int find(int A[],int a) {
// if(A[a]==a)
// return a;
// return find(A, A[a]);
//}
//FENWICK TREE
static void update(int i, int x){
for(; i < bit.length; i += (i&-i))
bit[i] += x;
}
static int sum(int i){
int ans = 0;
for(; i > 0; i -= (i&-i))
ans += bit[i];
return ans;
}
//END
static long summation(long A[],int si,int ei)
{
long ans=0;
for(int i=si;i<=ei;i++)
ans+=A[i];
return ans;
}
static void add(long v,Map<Long,Long>mp) {
if(!mp.containsKey(v)) {
mp.put(v, (long)1);
}
else {
mp.put(v, mp.get(v)+(long)1);
}
}
static void remove(long v,Map<Long,Long>mp) {
if(mp.containsKey(v)) {
mp.put(v, mp.get(v)-(long)1);
if(mp.get(v)==0)
mp.remove(v);
}
}
public static int upper(long A[],long k,int si,int ei)
{
int l=si;
int u=ei;
int ans=-1;
while(l<=u) {
int mid=(l+u)/2;
if(A[mid]<=k) {
ans=mid;
l=mid+1;
}
else {
u=mid-1;
}
}
return ans;
}
public static int lower(long A[],long k,int si,int ei)
{
int l=si;
int u=ei;
int ans=-1;
while(l<=u) {
int mid=(l+u)/2;
if(A[mid]<=k) {
l=mid+1;
}
else {
ans=mid;
u=mid-1;
}
}
return ans;
}
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 int[] input(int n) {
int A[]=new int[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextInt();
}
return A;
}
static long[] inputL(int n) {
long A[]=new long[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextLong();
}
return A;
}
static String[] inputS(int n) {
String A[]=new String[n];
for(int i=0;i<n;i++) {
A[i]=sc.next();
}
return A;
}
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 reverse(long A[]) {
int n=A.length;
long B[]=new long[n];
for(int i=0;i<n;i++) {
B[i]=A[n-i-1];
}
for(int i=0;i<n;i++)
A[i]=B[i];
}
static void reverse(int A[]) {
int n=A.length;
int B[]=new int[n];
for(int i=0;i<n;i++) {
B[i]=A[n-i-1];
}
for(int i=0;i<n;i++)
A[i]=B[i];
}
static void input(int A[],int B[]) {
for(int i=0;i<A.length;i++) {
A[i]=sc.nextInt();
B[i]=sc.nextInt();
}
}
static long[][] input(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]=l();
}
}
return A;
}
static char[][] charinput(int n,int m){
char A[][]=new char[n][m];
for(int i=0;i<n;i++) {
String s=s();
for(int j=0;j<m;j++) {
A[i][j]=s.charAt(j);
}
}
return A;
}
static int nextPowerOf2(int n)
{
if(n==0)
return 1;
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
static int highestPowerof2(int x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
static long highestPowerof2(long x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
static int max(int A[]) {
int max=Integer.MIN_VALUE;
for(int i=0;i<A.length;i++) {
max=Math.max(max, A[i]);
}
return max;
}
static int min(int A[]) {
int min=Integer.MAX_VALUE;
for(int i=0;i<A.length;i++) {
min=Math.min(min, A[i]);
}
return min;
}
static long max(long A[]) {
long max=Long.MIN_VALUE;
for(int i=0;i<A.length;i++) {
max=Math.max(max, A[i]);
}
return max;
}
static long min(long A[]) {
long min=Long.MAX_VALUE;
for(int i=0;i<A.length;i++) {
min=Math.min(min, A[i]);
}
return min;
}
static long [] prefix(long A[]) {
long p[]=new long[A.length];
p[0]=A[0];
for(int i=1;i<A.length;i++)
p[i]=p[i-1]+A[i];
return p;
}
static long [] prefix(int A[]) {
long p[]=new long[A.length];
p[0]=A[0];
for(int i=1;i<A.length;i++)
p[i]=p[i-1]+A[i];
return p;
}
static long [] suffix(long A[]) {
long p[]=new long[A.length];
p[A.length-1]=A[A.length-1];
for(int i=A.length-2;i>=0;i--)
p[i]=p[i+1]+A[i];
return p;
}
static long [] suffix(int A[]) {
long p[]=new long[A.length];
p[A.length-1]=A[A.length-1];
for(int i=A.length-2;i>=0;i--)
p[i]=p[i+1]+A[i];
return p;
}
static void fill(int dp[]) {
Arrays.fill(dp, -1);
}
static void fill(int dp[][]) {
for(int i=0;i<dp.length;i++)
Arrays.fill(dp[i], -1);
}
static void fill(int dp[][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
Arrays.fill(dp[i][j],-1);
}
}
}
static void fill(int dp[][][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
for(int k=0;k<dp[0][0].length;k++) {
Arrays.fill(dp[i][j][k],-1);
}
}
}
}
static void fill(long dp[]) {
Arrays.fill(dp, -1);
}
static void fill(long dp[][]) {
for(int i=0;i<dp.length;i++)
Arrays.fill(dp[i], -1);
}
static void fill(long dp[][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
Arrays.fill(dp[i][j],-1);
}
}
}
static void fill(long dp[][][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
for(int k=0;k<dp[0][0].length;k++) {
Arrays.fill(dp[i][j][k],-1);
}
}
}
}
static int min(int a,int b) {
return Math.min(a, b);
}
static int min(int a,int b,int c) {
return Math.min(a, Math.min(b, c));
}
static int min(int a,int b,int c,int d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static int max(int a,int b) {
return Math.max(a, b);
}
static int max(int a,int b,int c) {
return Math.max(a, Math.max(b, c));
}
static int max(int a,int b,int c,int d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long min(long a,long b) {
return Math.min(a, b);
}
static long min(long a,long b,long c) {
return Math.min(a, Math.min(b, c));
}
static long min(long a,long b,long c,long d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static long max(long a,long b) {
return Math.max(a, b);
}
static long max(long a,long b,long c) {
return Math.max(a, Math.max(b, c));
}
static long max(long a,long b,long c,long d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long power(long x, long y, long p)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long power(long x, long y)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x);
y = y >> 1;
x = (x * x);
}
return res;
}
static void print(int A[]) {
for(int i : A) {
out.print(i+" ");
}
out.println();
}
static void print(long A[]) {
for(long i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static long mod(long x) {
return ((x%mod + mod)%mod);
}
static String reverse(String s) {
StringBuffer p=new StringBuffer(s);
p.reverse();
return p.toString();
}
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
static void sort(int[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static void sort(long[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static String sort(String s) {
Character ch[]=new Character[s.length()];
for(int i=0;i<s.length();i++) {
ch[i]=s.charAt(i);
}
Arrays.sort(ch);
StringBuffer st=new StringBuffer("");
for(int i=0;i<s.length();i++) {
st.append(ch[i]);
}
return st.toString();
}
static HashMap<Integer,Long> hash(int A[]){
HashMap<Integer,Long> map=new HashMap<Integer, Long>();
for(int i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+(long)1);
}
else {
map.put(i, (long)1);
}
}
return map;
}
static HashMap<Long,Long> hash(long A[]){
HashMap<Long,Long> map=new HashMap<Long, Long>();
for(long i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+(long)1);
}
else {
map.put(i, (long)1);
}
}
return map;
}
static TreeMap<Integer,Integer> tree(int A[]){
TreeMap<Integer,Integer> map=new TreeMap<Integer, Integer>();
for(int i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static TreeMap<Long,Integer> tree(long A[]){
TreeMap<Long,Integer> map=new TreeMap<Long, Integer>();
for(long i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static boolean prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean prime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
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;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 2775928fb412b67ef6dbe009b69f5627 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
/**
*
* @Har_Har_Mahadev
*/
/**
* Main , Solution , Remove Public
*/
public class B {
public static void process() throws IOException {
int n = sc.nextInt();
String s = sc.next();
int prev = s.charAt(0);
int cc = 1;
int ans = 0;
for(int i = 1; i<n; i++) {
int curr = s.charAt(i);
if(curr == prev) {
cc++;
continue;
}
if(cc%2 == 0) {
prev = curr;
cc = 1;
continue;
}
cc++;
ans++;
}
System.out.println(ans);
}
//=============================================================================
//--------------------------The End---------------------------------
//=============================================================================
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
// google(TTT++);
process();
}
out.flush();
// tr(System.currentTimeMillis()-s+"ms");
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
/*
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof Pair)) return false;
Pair key = (Pair) o;
return x == key.x && y == key.y;
}
@Override
public int hashCode() {
int result = x;
result = 31 * result + y;
return result;
}
*/
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
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 lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
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 void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
//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);
}
// compress Big value to Time Limit
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;
}
// Fast Writer
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
// if(x < 0){ x = 0; }
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
// Fast Inputs
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[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = 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[] readArrayDouble(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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 765a1d582d45ab95e0dd032fa4f9be49 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.*;
public class Main {
public static void main(String[] args){
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
solve(in, out);
out.close();
}
static String reverse(String s) {
return (new StringBuilder(s)).reverse().toString();
}
static void sieveOfEratosthenes(int n, int factors[], ArrayList<Integer> ar)
{
factors[1]=1;
int p;
for(p = 2; p*p <=n; p++)
{
if(factors[p] == 0)
{
ar.add(p);
factors[p]=p;
for(int i = p*p; i <= n; i += p)
if(factors[i]==0)
factors[i] = p;
}
}
for(;p<=n;p++){
if(factors[p] == 0)
{
factors[p] = p;
ar.add(p);
}
}
}
static void sort(int ar[]) {
int n = ar.length;
ArrayList<Integer> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static void sort1(long ar[]) {
int n = ar.length;
ArrayList<Long> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static long ncr(long n, long r, long mod) {
if (r == 0)
return 1;
long val = ncr(n - 1, r - 1, mod);
val = (n * val) % mod;
val = (val * modInverse(r, mod)) % mod;
return val;
}
static int findMax(int a[], int n, int vis[], int i, int d){
if(i>=n)
return 0;
if(vis[i]==1)
return findMax(a, n, vis, i+1, d);
int max = 0;
for(int j=i+1;j<n;j++){
if(Math.abs(a[i]-a[j])>d||vis[j]==1)
continue;
vis[j] = 1;
max = Math.max(max, 1 + findMax(a, n, vis, i+1, d));
vis[j] = 0;
}
return max;
}
static void findSub(ArrayList<ArrayList<Integer>> ar, int n, ArrayList<Integer> a, int i){
if(i==n){
ArrayList<Integer> b = new ArrayList<Integer>();
for(int y:a){
b.add(y);
}
ar.add(b);
return;
}
for(int j=0;j<n;j++){
if(j==i)
continue;
a.set(i,j);
findSub(ar, n, a, i+1);
}
}
// *-------------------code starts here--------------------------------------------------*
static HashMap<Long,Integer>map;
public static void solve(InputReader sc, PrintWriter pw){
long mod=(long)1e9+7;
int t=sc.nextInt();
// int t=1;
L : while(--t>=0){
int n=sc.nextInt();
char ch[]=sc.next().toCharArray();
int cnt=0;
for(int i=0;i<n;i+=2){
if(ch[i]!=ch[i+1]){
cnt++;
}
}
pw.println(cnt);
}
}
// *-------------------code ends here-----------------------------------------------------*
static void assignAnc(ArrayList<Integer> ar[],int sz[], int pa[] ,int curr, int par){
sz[curr] = 1;
pa[curr] = par;
for(int v:ar[curr]){
if(par==v)
continue;
assignAnc(ar, sz, pa, v, curr);
sz[curr] += sz[v];
}
}
static int findLCA(int a, int b, int par[][], int depth[]){
if(depth[a]>depth[b]){
a = a^b;
b = a^b;
a = a^b;
}
int diff = depth[b] - depth[a];
for(int i=19;i>=0;i--){
if((diff&(1<<i))>0){
b = par[b][i];
}
}
if(a==b)
return a;
for(int i=19;i>=0;i--){
if(par[b][i]!=par[a][i]){
b = par[b][i];
a = par[a][i];
}
}
return par[a][0];
}
static void formArrayForBinaryLifting(int n, int par[][]){
for(int j=1;j<20;j++){
for(int i=0;i<n;i++){
if(par[i][j-1]==-1)
continue;
par[i][j] = par[par[i][j-1]][j-1];
}
}
}
static long lcm(int a, int b){
return a*b/gcd(a,b);
}
static class Pair1 {
long a;
long b;
Pair1(long a, long b) {
this.a = a;
this.b = b;
}
}
static class Pair implements Comparable<Pair> {
int a;
int b;
// int c;
Pair(int a, int b) {
this.a = a;
this.b = b;
// this.c = c;
}
public int compareTo(Pair p) {
return Integer.compare(a,p.a);
}
}
static boolean isPrime(int 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;
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long fast_pow(long base, long n, long M) {
if (n == 0)
return 1;
if (n == 1)
return base % M;
long halfn = fast_pow(base, n / 2, M);
if (n % 2 == 0)
return (halfn * halfn) % M;
else
return (((halfn * halfn) % M) * base) % M;
}
static long modInverse(long n, long M) {
return fast_pow(n, M - 2, M);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 9992768);
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();
}
public int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
public long[] readarray(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 8d4244912c49c8f6d1def41e0242b89d | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class GoodStringEasy {
public static void main(String[] args) {
FastReader in = new FastReader();
int t = in.nextInt();
for (int i = 0; i < t; i++) {
int n = in.nextInt();
String s = in.nextLine();
int c0 = 0, c1 = 0, total = 0;
for (int j = 0; j < n; j++) {
char c = s.charAt(j);
if (j + 1 == n) {
if (c == '0') c0++;
else c1++;
if (c0 % 2 != 0 || c1 % 2 != 0) total++;
break;
}
if (c == '0') {
if (c1 % 2 == 0) {
c0++;
}
else {
total++;
}
c1 = 0;
}
else {
if (c0 % 2 == 0) {
c1++;
}
else {
total++;
}
c0 = 0;
}
}
System.out.println(total);
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
if (st.hasMoreTokens()) {
str = st.nextToken("\n");
}
else {
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 79d19cb6ced429aac665f221215bf18f | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
/**
* @Author LWX
* @date 2022/5/11 21:43
*/
public class Tokitsukaze_and_Good_01_String_easy_version {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
for (int i = 0; i < n; i++) {
int len = input.nextInt();
String line = input.next();
System.out.println(minimumPos(len, line));
}
}
private static int minimumPos(int len, String line) {
if (len == 0) return 0;
int current = -1;
int count = 0;
int ans = 0;
for (char c : line.toCharArray()) {
int cur = c - '0';
if (current == -1) {
current = cur;
count = 1;
} else {
if (current == cur) {
count++;
} else {
if (count % 2 == 0) {
current = cur;
count = 1;
} else {
count++;
ans++;
}
}
}
}
return ans;
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 0c5c047ed129b8c28ca5f008071be695 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class B1_1678 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int t=0; t<T; t++) {
in.nextInt();
char[] S = (in.next() + ' ').toCharArray();
int answer = 0;
boolean unpairedOdd = false;
int count = 0;
char last = ' ';
for (char c : S) {
if (c == last) {
count++;
} else {
if (unpairedOdd) {
answer++;
}
if (count%2 == 1) {
unpairedOdd = !unpairedOdd;
}
count = 1;
last = c;
}
}
System.out.println(answer);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 3c6ec352e9ec93fec4775a96455c246d | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //package december;
import java.util.*;
import java.io.*;
public class codeforces1678B {
public static void main (String [] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int numCases = Integer.parseInt(br.readLine());
for (int rep = 0; rep<numCases;rep++)
{
int length = Integer.parseInt(br.readLine());
String str = br.readLine();
int count = 0;
for (int i = 0; i<length;i+=2)
{
if (str.charAt(i)!=str.charAt(i+1))
{
count++;
}
}
System.out.println(count);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 42d5d8efb4019766274a67643a9cc241 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
for(int i = 0; i < t; i++){
int n = scanner.nextInt();
String s = scanner.next();
System.out.println(solve(s));
}
}
public static int solve(String s){
int counter = 0;
for(int i = 0; i < s.length(); i+=2){
if(s.charAt(i) != s.charAt(i + 1))
counter++;
}
return counter;
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | a0cd5729beb992c833c3ee02147dbcd8 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
public class Five {
private static int find(String s)
{
List<Integer> res = new ArrayList<>();
int cnt=1;
for(int i=1;i<s.length();i++)
{
if(s.charAt(i)==s.charAt(i-1))
cnt++;
else
{
res.add(cnt);cnt=1;
}
}
cnt=0;
int arr[] = new int[res.size()];
for(int i=0;i< res.size();i++)
{
arr[i]=res.get(i);
}
for(int i=0;i< res.size();i++)
{
if(arr[i]%2!=0)
{
arr[i]-=1;
if(i+1< res.size()) arr[i+1]+=1;
cnt++;
}
}
return cnt;
}
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-->0) {
int n = sc.nextInt();
String input = sc.next();
System.out.println(find(input));
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 78dc027ff5783242b329823fa4c6e79e | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //package com.company;
import java.awt.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.net.InetAddress;
import java.text.CollationKey;
import java.util.*;
public class Main implements Runnable {
int size[];
static int e9 = 1000000007;
int max = Integer.MIN_VALUE;
long max1 = Long.MIN_VALUE;
long min1 = Long.MAX_VALUE;
long fib[][];
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;
}
}
int parent[];
int rank[];
int sie=1000000;
boolean sieve[]=new boolean[sie+1];
ArrayList<ArrayList<Integer>> arrayLists;
HashMap<String,Integer> hashMap;
long fib1[][];
long r[];
int n;
int m;
long co=0;
long v[];
int dp[][][];
boolean check=false;
int k;
int il[];
String s;
ArrayList<String> arrayList;
long ans=0;
Set<String> set;
public static void main(String[] args) throws IOException {
new Thread(null,new Main(),"Main",1<<28).start();
}
public void run() {
FastReader scanner = new FastReader();
// Scanner scanner = new Scanner(System.in);
// BufferedReader bufferedReader=new BufferedReader(new InputStreamReader(System.in));
PrintWriter printWriter = new PrintWriter(System.out);
// System.out.println(-3%23);
long t1=scanner.nextLong();
high:while (t1-- > 0) {
int n=scanner.nextInt();
String s=scanner.next();
char arr[]=s.toCharArray();
int c=1;
int ans=0;
for (int i = 1; i < n; i++) {
if(arr[i]==arr[i-1]){
c++;
}
else{
// System.out.println(c+" "+i);
if(c%2!=0){
ans++;
arr[i]=arr[i-1];
i++;
}
c=1;
}
}
printWriter.println(ans);
}
printWriter.flush();
}
static long mod=1_000_000_007;
public void sieve(){
for (long p = 2; p<=sie; p++)
{
if (sieve[(int)p] == true)
{
for (long i = p * p; i <=sie; i += p) {
// long o=i;
//// System.out.println(o+" "+p);
// if(o<0){
// return;
// }
sieve[(int) (i)] = false;
}
}
}
}
public long lcm(long a,long b){
return (a*b)/gcd(a,b);
}
public long mul(long a, long b) {
// System.out.println(a+" "+b);
return ((a)*(b))%mod;
}
public long fact(int x) {
long ans=1;
for (int i=2; i<=x; i++) ans=mul(ans, i);
return ans;
}
public long fastPow(long base, long exp) {
if (exp==0) return 1;
long half=fastPow(base, exp/2);
if (exp%2==0) return mul(half, half)%mod;
return mul(half, mul(half, base))%mod;
}
public long modInv(long x) {
return fastPow(x, mod-2);
}
public long nc2(int n,int k){
return n*(n-1)/2;
}
public long nCk(int n, int k) {
return mul(fact(n), mul(modInv(fact(k)), modInv(fact(n-k))));
}
public void parent(int n){
parent=new int[n+1];
size=new int[n+1];
rank=new int[n+1];
for (int i = 0; i <=n; i++) {
parent[i]=i;
rank[i]=0;
// size[i]=1;
}
}
public void union(int i,int j){
int root1=find(i);
int root2=find(j);
// if(root1 != root2) {
// parent[root2] = root1;
//// sz[a] += sz[b];
// }
if(root1==root2){
return;
}
if(rank[root1]>rank[root2]){
parent[root2]=root1;
size[root1]+=size[root2];
}
else if(rank[root1]<rank[root2]){
parent[root1]=root2;
size[root2]+=size[root1];
}
else{
parent[root2]=root1;
rank[root1]+=1;
size[root1]+=size[root2];
}
}
public int find(int p){
if(parent[p]!=p){
if(parent[p]==-1){
return parent[p];
}
parent[p]=find(parent[p]);
}
return parent[p];
}
public double dist(double x1,double y1,double x2,double y2){
double e=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);
double e1=Math.sqrt(e);
return e1;
}
public void make(int p){
parent[p]=p;
rank[p]=1;
}
Random rand = new Random();
public void sort(int[] a, int n) {
for (int i = 0; i < n; i++) {
int j = rand.nextInt(i + 1);
int tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
Arrays.sort(a, 0, n);
}
public long gcd(long a,long b){
if(b==0){
return a;
}
return gcd(b,a%b);
}
// public void dfs(ArrayList<Integer> arrayList){
// for (int i = 0; i < arrayList.size(); i++) {
// if(v1[arrayList.get(i)]==false){
// v1[arrayList.get(i)]=true;
// dfs(arrayLists.get(i));
// }
// }
// }
public double fact(double h){
double sum=1;
while(h>=1){
sum=(sum%e9)*(h%e9);
h--;
}
return sum%e9;
}
public long primef(double r){
long c=0;
long ans=1;
long g=0;
while(r%2==0){
c++;
r=r/2;
}
if(c>0){
ans*=2;
}
c=0;
// System.out.println(ans+" "+r);
for (int i = 3; i <=Math.sqrt(r) ;i+=2) {
g=0;
while(r%i==0){
// System.out.println(i);
c++;
g++;
r=r/i;
}
if(c>0){
ans*=i;
}
c=0;
}
if(r>2){
ans*=r;
}
return ans;
}
public long divisor(double r){
long c=0;
// if(n==3)
// System.out.println(r+" "+n+" "+k+" "+"l");
for (int i =2; i <=Math.sqrt(r); i++) {
if(r%i==0){
if(r/i==i){
c++;
}
else{
if(r/i<n){
c+=2;
}else{
c++;
}
}
}
}
return c;
}
}
class Pair {
int x;
int y;
int z;
public Pair(int x,int y) {
this.x = x;
this.y = y;
// this.z=z;
}
// @Override
// public int hashCode() {
// int hash = 37;
// return this.x * hash + this.y;
// }
//
// @Override
// public boolean equals(Object o1) {
// if (o1 == null || o1.getClass() != this.getClass()) {
// return false;
// }
// Pair o = (Pair) o1;
// if (o.x == this.x && o.y == this.y) {
// return true;
// }
// return false;
// }
}
class Sorting implements Comparator<Pair> {
public int compare(Pair p1, Pair p2) {
if(p1.x==p2.x){
return Long.compare(p1.y,p2.y);
}else {
return Long.compare(p1.x, p2.x);
}
}
}
class Edge{
int s=0;
int d=0;
int c=0;
public Edge(int s,int d,int c){
this.s=s;
this.d=d;
this.c=c;
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 328edaab4929a2945ad2239256a1f585 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class ProblemB {
public static void solve(Scanner sc) {
int n = sc.nextInt();
sc.nextLine();
String s = sc.nextLine();
List<Integer> list = new ArrayList<>();
int cntOfContiStr = 0;
for (int i = 0; i < n - 1; i++) {
if (s.charAt(i) == s.charAt(i + 1)) {
cntOfContiStr++;
} else {
list.add(cntOfContiStr + 1);
cntOfContiStr = 0;
}
}
list.add(cntOfContiStr + 1);
int ans = 0;
int temp = 0;
boolean flag = false;
for (int i = 0; i < list.size(); i++) {
if (list.get(i) % 2 == 1) {
if (!flag) {
temp = i;
} else if (flag) {
ans += i - temp;
temp = 0;
}
flag = !flag;
}
}
System.out.println(ans);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t > 0) {
solve(sc);
t--;
}
sc.close();
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 44b15243abedadf758dd0cbcce89a6fb | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
/*
!!!!Hello World,Prakhar here!!!!
codechef handle prakhar_3011
codeforces handle prakhar_30
trying to get good at CP
PEACE OUT.........
*/
/*
11 10 01 10 00
*/
import java.io.*;
import java.util.*;
public class goodstring {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
String s=sc.next();int ans=0;
Stack<Character> c=new Stack<>();
for(int i=0;i<s.length();i++){
if(c.empty()==false&&c.peek()==s.charAt(i)){
c.pop();
}
else if(c.empty()==false&&c.peek()!=s.charAt(i)){
ans++;c.pop();
}
else{
c.push(s.charAt(i));
}
}
System.out.println(ans);
}
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
static void revsort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l, Collections.reverseOrder());
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
/* ......FAST SCANNER template taken from secondthread...... */
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());
}
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 nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 41bed42171a990fbb47b83c56f9e7f7b | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
char[] c = sc.next().toCharArray();
int ans = 0;
for (int i = 0; i < n; i+=2) {
if (c[i] != c[i+1])
ans++;
}
out.println(ans);
}
out.close();
}
static boolean inc(int[] a) {
for (int i = 0; i < a.length-1; i++)
if (a[i] >= a[i+1]) return false;
return true;
}
static boolean decOrEq(int[] a) {
for (int i = 0; i < a.length-1; i++)
if (a[i] < a[i+1]) return false;
return true;
}
static final Random random = new Random();
static void shuffleSort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int r = random.nextInt(n), temp = a[r];
a[r] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static class 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();}
int[] readArray(int n) throws IOException {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | eaba3c2907e820d9efb474fe76aa611c | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //package Div2.B;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
public class TokitsukazeAndGood01StringEasy {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t>0){
int n=Integer.parseInt(br.readLine());
String s=br.readLine();
List<Integer> list=new ArrayList<>();
for(int i=0;i<n;i++){
int cnt=1;
while(i+1<n && s.charAt(i)==s.charAt(i+1)){
cnt++;
i++;
}
list.add(cnt);
}
int operations = 0;
for(int i=0;i<list.size()-1;){
int cnt1=list.get(i);
int cnt2=list.get(i+1);
if(cnt1%2==0 && cnt2%2==0){
i+=2;
}else if(cnt1%2!=0 && cnt2%2!=0){
i+=2;
operations+=1;
}else if(cnt1%2==0 && cnt2%2!=0){
i+=1;
}else {
operations+=1;
list.set(i+1, cnt2-1);
i+=1;
}
}
System.out.println(operations);
t--;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 78d6bda0827386a464778fb2a2de0306 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) {
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int T = in.nextInt();
for(int ttt = 1; ttt <= T; ttt++) {
int n = in.nextInt();
String str = in.next();
char[] s = str.toCharArray();
List<Integer> list = new ArrayList<>();
int cur = 1, cnt = 1;
while(cur < s.length) {
if(s[cur] == s[cur-1]) cnt++;
else {
list.add(cnt);
cnt = 1;
}
cur++;
}
list.add(cnt);
int ans = 0;
for(int i = 0; i < list.size(); i++) {
if(list.get(i)%2==1) {
ans++;
list.set(i+1, list.get(i+1)+1);
}
}
out.println(ans);
// out.println(list);
}
out.close();
}
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;
}
int[] readInt(int size) {
int[] arr = new int[size];
for(int i = 0; i < size; i++)
arr[i] = Integer.parseInt(next());
return arr;
}
long[] readLong(int size) {
long[] arr = new long[size];
for(int i = 0; i < size; i++)
arr[i] = Long.parseLong(next());
return arr;
}
int[][] read2dArray(int rows, int cols) {
int[][] arr = new int[rows][cols];
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++)
arr[i][j] = Integer.parseInt(next());
}
return arr;
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 69aa220b27cfc3106f9506a4ff2b8b57 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.awt.datatransfer.StringSelection;
import java.util.*;
import javax.management.Query;
import java.io.*;
public class practice {
static String s;
static int n,k;
static int[] a;
static long[][] memo;
static HashMap<Long,Integer>hm;
public static void main(String[] args) throws Exception {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while (t-->0){
int n=sc.nextInt();
String s=sc.next();
int ones=0,zeros=0;
for (int i = 0; i <n ; i++) {
if(s.charAt(i)=='1')
ones++;
else
zeros++;
}
if(zeros==0||ones==0)
pw.println(0);
else{
int res=0;
Stack<Integer>st=new Stack<>();
for (int i = 0; i < n; i++) {
if(st.isEmpty())
st.push(s.charAt(i)-'0');
else if(st.peek().equals(s.charAt(i)-'0'))
st.pop();
else{
res++;
st.pop();
}
}
pw.println(res);
}
}
pw.close();
}
static long nCr(int n, int r)
{
return 1l*fact(n) / 1l*(fact(r) *
fact(n - r));
}
static long fact(int n)
{
int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
private static double customLog(double base, double logNumber) {
return Math.log(logNumber) / Math.log(base);
}
public static int next(long[] arr, int target,int days)
{
// pw.println("days="+days);
int start = 0, end = arr.length-1;
// Minimum size of the array should be 1
// If target lies beyond the max element, than the index of strictly smaller
// value than target should be (end - 1)
if (target >= 0l+arr[end]+1l*(end+1)*days) return end;
int ans = -1;
while (start <= end) {
int mid = (start + end) / 2;
// Move to the left side if the target is smaller
if (0l+arr[mid]+1l*(mid+1)*days > target) {
end = mid - 1;
}
// Move right side
else {
ans = mid;
start = mid + 1;
}
}
return ans;
}
public static long factorial(int n){
int y=1;
for (int i = 2; i <=n ; i++) {
y*=i;
}
return y;
}
public static void sort(int[] in) {
shuffle(in);
Arrays.sort(in);
}
public static void shuffle(int[] in) {
for (int i = 0; i < in.length; i++) {
int idx = (int) (Math.random() * in.length);
int tmp = in[i];
in[i] = in[idx];
in[idx] = tmp;
}
}
static LinkedList getfact(int n){
LinkedList<Integer>ll=new LinkedList<>();
LinkedList<Integer>ll2=new LinkedList<>();
for (int i = 1; i <= Math.sqrt(n); i++) {
if(n%i==0) {
ll.add(i);
if(i!=n/i)
ll2.addLast(n/i);
}
}
while (!ll2.isEmpty()){
ll.add(ll2.removeLast());
}
return ll;
}
static void rev(int n){
String s1=s.substring(0,n);
s=s.substring(n);
for (int i = 0; i <n ; i++) {
s=s1.charAt(i)+s;
}
}
static class SegmentTree { // 1-based DS, OOP
int N; //the number of elements in the array as a power of 2 (i.e. after padding)
long[] array, sTree;
Long[]lazy;
SegmentTree(long[] in)
{
array = in; N = in.length - 1;
sTree = new long[N<<1]; //no. of nodes = 2*N - 1, we add one to cross out index zero
lazy = new Long[N<<1];
build(1,1,N);
}
void build(int node, int b, int e) // O(n)
{
if(b == e)
sTree[node] = array[b];
else
{
int mid = b + e >> 1;
build(node<<1,b,mid);
build(node<<1|1,mid+1,e);
sTree[node] = sTree[node<<1]+sTree[node<<1|1];
}
}
void update_point(int index, int val) // O(log n)
{
index += N - 1;
sTree[index] += val;
while(index>1)
{
index >>= 1;
sTree[index] = sTree[index<<1] + sTree[index<<1|1];
}
}
void update_range(int i, int j, int val) // O(log n)
{
update_range(1,1,N,i,j,val);
}
void update_range(int node, int b, int e, int i, int j, int val)
{
if(i > e || j < b)
return;
if(b >= i && e <= j)
{
sTree[node] = (e-b+1)*val;
lazy[node] = val*1l;
}
else
{
int mid = b + e >> 1;
propagate(node, b, mid, e);
update_range(node<<1,b,mid,i,j,val);
update_range(node<<1|1,mid+1,e,i,j,val);
sTree[node] = sTree[node<<1] + sTree[node<<1|1];
}
}
void propagate(int node, int b, int mid, int e)
{
if(lazy[node]!=null) {
lazy[node << 1] = lazy[node];
lazy[node << 1 | 1] = lazy[node];
sTree[node << 1] = (mid - b + 1) * lazy[node];
sTree[node << 1 | 1] = (e - mid) * lazy[node];
}
lazy[node] = null;
}
long query(int i, int j)
{
return query(1,1,N,i,j);
}
long query(int node, int b, int e, int i, int j) // O(log n)
{
if(i>e || j <b)
return 0;
if(b>= i && e <= j)
return sTree[node];
int mid = b + e >> 1;
propagate(node, b, mid, e);
long q1 = query(node<<1,b,mid,i,j);
long q2 = query(node<<1|1,mid+1,e,i,j);
return q1 + q2;
}
}
// public static long dp(int idx) {
// if (idx >= n)
// return Long.MAX_VALUE/2;
// return Math.min(dp(idx+1),memo[idx]+dp(idx+k));
// }
// if(num==k)
// return dp(0,idx+1);
// if(memo[num][idx]!=-1)
// return memo[num][idx];
// long ret=0;
// if(num==0) {
// if(s.charAt(idx)=='a')
// ret= dp(1,idx+1);
// else if(s.charAt(idx)=='?') {
// ret=Math.max(1+dp(1,idx+1),dp(0,idx+1) );
// }
// }
// else {
// if(num%2==0) {
// if(s.charAt(idx)=='a')
// ret=dp(num+1,idx+1);
// else if(s.charAt(idx)=='?')
// ret=Math.max(1+dp(num+1,idx+1),dp(0,idx+1));
// }
// else {
// if(s.charAt(idx)=='b')
// ret=dp(num+1,idx+1);
// else if(s.charAt(idx)=='?')
// ret=Math.max(1+dp(num+1,idx+1),dp(0,idx+1));
// }
// }
// }
static void putorrem(long x){
if(hm.getOrDefault(x,0)==1){
hm.remove(x);
}
else
hm.put(x,hm.getOrDefault(x,0)-1);
}
public static int max4(int a,int b, int c,int d) {
int [] s= {a,b,c,d};
Arrays.sort(s);
return s[3];
}
public static double logbase2(int k) {
return( (Math.log(k)+0.0)/Math.log(2));
}
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();
}
}
static class pair implements Comparable<pair> {
long x;
long y;
public pair(long x, long y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return Long.compare(this.y, other.y);
}
return Long.compare(this.x, other.x);
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
public tuble add(tuble t){
return new tuble(this.x+t.x,this.y+t.y,this.z+t.z);
}
}
static long mod = 1000000007;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 9d8da834da3786c9194c45aabec92d36 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B1_Tokitsukaze_and_Good_01_String_easy_version {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int testCases, n, iterations = 0;
static char a[];
/*
10
1110011000
-> 11 00 11 11 00
*/
static void solve(int index, ArrayList<Integer> list, int len) {
if (index >= n) {
return;
}
len = 1;
while (index + 1 < n && a[index] == a[index + 1]) {
++len;
++index;
}
list.add(len);
++iterations;
solve(index + 1, list, len);
}
static void solve(int t) {
iterations = 0;
ArrayList<Integer> list = new ArrayList<>();
solve(0, list, 1);
long arr[] = new long[list.size()];
int index = 0;
while (!list.isEmpty()) {
arr[index++] = list.get(0);
list.popFront();
}
int ans1 = 0, m = arr.length;
iterations = 0;
for (int i = 0; i < m - 1; i++) {
if (arr[i] % 2 == arr[i + 1] % 2 && arr[i] % 2 == 1) {
arr[i]--;
arr[i + 1]++;
ans1++;
//++iterations;
} else if (arr[i] % 2 != arr[i + 1] % 2 && arr[i + 1] % 2 == 0) {
arr[i]--;
arr[i + 1]++;
ans1++;
}
}
ans.append(ans1);
if (t != testCases) {
ans.append("\n");
}
}
public static void main(String[] amit) throws IOException {
testCases = in.nextInt();
for (int t = 0; t < testCases; ++t) {
n = in.nextInt();
a = in.next().toCharArray();
solve(t + 1);
}
out.print(ans.toString());
out.flush();
in.close();
}
static boolean isSmaller(String str1, String str2) {
int n1 = str1.length(), n2 = str2.length();
if (n1 < n2) {
return true;
}
if (n2 < n1) {
return false;
}
for (int i = 0; i < n1; i++) {
if (str1.charAt(i) < str2.charAt(i)) {
return true;
} else if (str1.charAt(i) > str2.charAt(i)) {
return false;
}
}
return false;
}
static String sub(String str1, String str2) {
if (isSmaller(str1, str2)) {
String t = str1;
str1 = str2;
str2 = t;
}
String str = "";
int n1 = str1.length(), n2 = str2.length();
int diff = n1 - n2;
int carry = 0;
for (int i = n2 - 1; i >= 0; i--) {
int sub
= (((int) str1.charAt(i + diff) - (int) '0')
- ((int) str2.charAt(i) - (int) '0')
- carry);
if (sub < 0) {
sub = sub + 10;
carry = 1;
} else {
carry = 0;
}
str += String.valueOf(sub);
}
for (int i = n1 - n2 - 1; i >= 0; i--) {
if (str1.charAt(i) == '0' && carry > 0) {
str += "9";
continue;
}
int sub = (((int) str1.charAt(i) - (int) '0')
- carry);
if (i > 0 || sub > 0) {
str += String.valueOf(sub);
}
carry = 0;
}
return new StringBuilder(str).reverse().toString();
}
static String sum(String str1, String str2) {
if (str1.length() > str2.length()) {
String t = str1;
str1 = str2;
str2 = t;
}
String str = "";
int n1 = str1.length(), n2 = str2.length();
int diff = n2 - n1;
int carry = 0;
for (int i = n1 - 1; i >= 0; i--) {
int sum = ((int) (str1.charAt(i) - '0')
+ (int) (str2.charAt(i + diff) - '0') + carry);
str += (char) (sum % 10 + '0');
carry = sum / 10;
}
for (int i = n2 - n1 - 1; i >= 0; i--) {
int sum = ((int) (str2.charAt(i) - '0') + carry);
str += (char) (sum % 10 + '0');
carry = sum / 10;
}
if (carry > 0) {
str += (char) (carry + '0');
}
return new StringBuilder(str).reverse().toString();
}
static long detect_sum(int i, long a[], long sum) {
if (i >= a.length) {
return sum;
}
return detect_sum(i + 1, a, sum + a[i]);
}
static String mul(String num1, String num2) {
int len1 = num1.length();
int len2 = num2.length();
if (len1 == 0 || len2 == 0) {
return "0";
}
int result[] = new int[len1 + len2];
int i_n1 = 0;
int i_n2 = 0;
for (int i = len1 - 1; i >= 0; i--) {
int carry = 0;
int n1 = num1.charAt(i) - '0';
i_n2 = 0;
for (int j = len2 - 1; j >= 0; j--) {
int n2 = num2.charAt(j) - '0';
int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
carry = sum / 10;
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
if (carry > 0) {
result[i_n1 + i_n2] += carry;
}
i_n1++;
}
int i = result.length - 1;
while (i >= 0 && result[i] == 0) {
i--;
}
if (i == -1) {
return "0";
}
String s = "";
while (i >= 0) {
s += (result[i--]);
}
return s;
}
static class Node<T> {
T data;
Node<T> next;
public Node() {
this.next = null;
}
public Node(T data) {
this.data = data;
this.next = null;
}
public T getData() {
return data;
}
public void setData(T data) {
this.data = data;
}
public Node<T> getNext() {
return next;
}
public void setNext(Node<T> next) {
this.next = next;
}
@Override
public String toString() {
return this.getData().toString() + " ";
}
}
static class ArrayList<T> {
Node<T> head, tail;
int len;
public ArrayList() {
this.head = null;
this.tail = null;
this.len = 0;
}
int size() {
return len;
}
boolean isEmpty() {
return len == 0;
}
int indexOf(T data) {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
int index = -1, i = 0;
while (temp != null) {
if (temp.getData() == data) {
index = i;
}
i++;
temp = temp.getNext();
}
return index;
}
void add(T data) {
Node<T> newNode = new Node<>(data);
if (isEmpty()) {
head = newNode;
tail = newNode;
len++;
} else {
tail.setNext(newNode);
tail = newNode;
len++;
}
}
void see() {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
while (temp != null) {
out.print(temp.getData().toString() + " ");
out.flush();
temp = temp.getNext();
}
out.println();
out.flush();
}
void inserFirst(T data) {
Node<T> newNode = new Node<>(data);
Node<T> temp = head;
if (isEmpty()) {
head = newNode;
tail = newNode;
len++;
} else {
newNode.setNext(temp);
head = newNode;
len++;
}
}
T get(int index) {
if (isEmpty() || index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
if (index == 0) {
return head.getData();
}
Node<T> temp = head;
int i = 0;
T data = null;
while (temp != null) {
if (i == index) {
data = temp.getData();
}
i++;
temp = temp.getNext();
}
return data;
}
void addAt(T data, int index) {
if (index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> newNode = new Node<>(data);
int i = 0;
Node<T> temp = head;
while (temp.next != null) {
if (i == index) {
newNode.setNext(temp.next);
temp.next = newNode;
}
i++;
temp = temp.getNext();
}
// temp.setNext(temp);
len++;
}
void popFront() {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
if (head == tail) {
head = null;
tail = null;
} else {
head = head.getNext();
}
len--;
}
void removeAt(int index) {
if (index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
if (index == 0) {
this.popFront();
return;
}
Node<T> temp = head;
int i = 0;
Node<T> n = new Node<>();
while (temp != null) {
if (i == index) {
n.next = temp.next;
temp.next = n;
break;
}
i++;
n = temp;
temp = temp.getNext();
}
tail = n;
--len;
}
void clearAll() {
this.head = null;
this.tail = null;
}
}
static void merge(long a[], int left, int right, int mid) {
int n1 = mid - left + 1, n2 = right - mid;
long L[] = new long[n1];
long R[] = new long[n2];
for (int i = 0; i < n1; i++) {
L[i] = a[left + i];
}
for (int i = 0; i < n2; i++) {
R[i] = a[mid + 1 + i];
}
int i = 0, j = 0, k1 = left;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
a[k1] = L[i];
i++;
} else {
a[k1] = R[j];
j++;
}
k1++;
}
while (i < n1) {
a[k1] = L[i];
i++;
k1++;
}
while (j < n2) {
a[k1] = R[j];
j++;
k1++;
}
}
static void sort(long a[], int left, int right) {
if (left >= right) {
return;
}
int mid = (left + right) / 2;
sort(a, left, mid);
sort(a, mid + 1, right);
merge(a, left, right, mid);
}
static class Scanner {
BufferedReader in;
StringTokenizer st;
public Scanner() {
in = new BufferedReader(new InputStreamReader(System.in));
}
String next() throws IOException {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
String nextLine() throws IOException {
return in.readLine();
}
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 {
in.close();
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 47bba199bff717ca57539fb93b647805 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class CF_1678B1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
char s[] = sc.next().toCharArray();
int ans =0;
for(int i = 0;i<n;i+=2){
if(s[i]!=s[i+1]){
ans+=1;
}
}
System.out.println(ans);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 24a20a6510138e3213f533656642c52a | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.IOException;
import java.io.InputStream;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;
import java.util.*;
import static java.util.Arrays.*;
public class CodeforcesTemp {
public static void main(String[] args) throws IOException {
Reader scan = new Reader();
FastPrinter out = new FastPrinter();
int t = scan.nextInt();
for (int tc = 1; tc <= t; tc++) {
int n=scan.nextInt();
String s=scan.next();
int[] cnt=new int[1];
cnt[0]=0;
int[] segment=new int[cnt[0]];
doWork(s,cnt,false,segment);
segment=new int[cnt[0]];
doWork(s,cnt,true,segment);
int op=0;
for (int i = 0; i < cnt[0]/2; i++) {
if(segment[i]%2==1){
segment[i]++;
segment[i+1]--;
op++;
}
else{
continue;
}
}
out.println(op);
out.flush();
}
out.close();
}
private static void doWork(String s,int[] cnt,boolean flag,int[] segment) {
int curr_cnt=1;
int p=0;
int n=s.length();
for (int i = 0; i < s.length(); i++) {
if((i+1)<n && s.charAt(i)==s.charAt(i+1)){
curr_cnt++;
continue;
}else{
cnt[0]++;
if(flag){
segment[p++]=curr_cnt;
}
curr_cnt=1;
}
}
}
static class Reader {
private final InputStream in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private static final long LONG_MAX_TENTHS = 922337203685477580L;
private static final int LONG_MAX_LAST_DIGIT = 7;
private static final int LONG_MIN_LAST_DIGIT = 8;
public Reader(InputStream in) {
this.in = in;
}
public Reader() {
this(System.in);
}
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
} else {
ptr = 0;
try {
buflen = in.read(buffer);
} catch (IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte()) return buffer[ptr++];
else return -1;
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
public boolean hasNext() {
while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
return hasNextByte();
}
public String next() {
if (!hasNext())
throw new NoSuchElementException();
StringBuilder sb = new StringBuilder();
int b = readByte();
while (isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext())
throw new NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
int digit = b - '0';
if (n >= LONG_MAX_TENTHS) {
if (n == LONG_MAX_TENTHS) {
if (minus) {
if (digit <= LONG_MIN_LAST_DIGIT) {
n = -n * 10 - digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("not number"));
}
}
} else {
if (digit <= LONG_MAX_LAST_DIGIT) {
n = n * 10 + digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("not number"));
}
}
}
}
throw new ArithmeticException(
String.format(" overflows long."));
}
n = n * 10 + digit;
} else if (b == -1 || !isPrintableChar(b)) {
return minus ? -n : n;
} else {
throw new NumberFormatException();
}
b = readByte();
}
}
public int nextInt() {
long nl = nextLong();
if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
throw new NumberFormatException();
return (int) nl;
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long[] nextLongArray(int length) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = this.nextLong();
return array;
}
public long[] nextLongArray(int length, java.util.function.LongUnaryOperator map) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsLong(this.nextLong());
return array;
}
public int[] nextIntArray(int length) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = this.nextInt();
return array;
}
public int[][] nextIntArrayMulti(int length, int width) {
int[][] arrays = new int[width][length];
for (int i = 0; i < length; i++) {
for (int j = 0; j < width; j++)
arrays[j][i] = this.nextInt();
}
return arrays;
}
public int[] nextIntArray(int length, java.util.function.IntUnaryOperator map) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsInt(this.nextInt());
return array;
}
public double[] nextDoubleArray(int length) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = this.nextDouble();
return array;
}
public double[] nextDoubleArray(int length, java.util.function.DoubleUnaryOperator map) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsDouble(this.nextDouble());
return array;
}
public long[][] nextLongMatrix(int height, int width) {
long[][] mat = new long[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextLong();
}
return mat;
}
public int[][] nextIntMatrix(int height, int width) {
int[][] mat = new int[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextInt();
}
return mat;
}
public double[][] nextDoubleMatrix(int height, int width) {
double[][] mat = new double[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextDouble();
}
return mat;
}
public char[][] nextCharMatrix(int height, int width) {
char[][] mat = new char[height][width];
for (int h = 0; h < height; h++) {
String s = this.next();
for (int w = 0; w < width; w++) {
mat[h][w] = s.charAt(w);
}
}
return mat;
}
}
static class FastPrinter extends PrintWriter {
public FastPrinter(PrintStream stream) {
super(stream);
}
public FastPrinter() {
super(System.out);
}
private static String dtos(double x, int n) {
StringBuilder sb = new StringBuilder();
if (x < 0) {
sb.append('-');
x = -x;
}
x += Math.pow(10, -n) / 2;
sb.append((long) x);
sb.append(".");
x -= (long) x;
for (int i = 0; i < n; i++) {
x *= 10;
sb.append((int) x);
x -= (int) x;
}
return sb.toString();
}
@Override
public void print(float f) {
super.print(dtos(f, 20));
}
@Override
public void println(float f) {
super.println(dtos(f, 20));
}
@Override
public void print(double d) {
super.print(dtos(d, 20));
}
@Override
public void println(double d) {
super.println(dtos(d, 20));
}
public void printArray(int[] array, String separator) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(array[i]);
super.print(separator);
}
super.println(array[n - 1]);
}
public void printArray(int[] array) {
this.printArray(array, " ");
}
public void printArray(int[] array, String separator, java.util.function.IntUnaryOperator map) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(map.applyAsInt(array[i]));
super.print(separator);
}
super.println(map.applyAsInt(array[n - 1]));
}
public void printArray(int[] array, java.util.function.IntUnaryOperator map) {
this.printArray(array, " ", map);
}
public void printArray(long[] array, String separator) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(array[i]);
super.print(separator);
}
super.println(array[n - 1]);
}
public void printArray(long[] array) {
this.printArray(array, " ");
}
public void printArray(long[] array, String separator, java.util.function.LongUnaryOperator map) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(map.applyAsLong(array[i]));
super.print(separator);
}
super.println(map.applyAsLong(array[n - 1]));
}
public void printArray(long[] array, java.util.function.LongUnaryOperator map) {
this.printArray(array, " ", map);
}
public void printMatrix(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
this.printArray(arr[i]);
}
}
}
static Random __r = new Random();
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 3a3c0c0f236533064bd2b53e5d39f083 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 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) {
int n = sc.nextInt();
String s = sc.next();
List<Integer> list = new ArrayList<>();
int count = 0;
for (int i = 0; i < n - 1; i++) {
count++;
if (s.charAt(i) != s.charAt(i + 1)) {
list.add(count);
count = 0;
}
}
list.add(count + 1);
int res = 0;
for (int i = 0; i < list.size() - 1; i++) {
if (list.get(i) % 2 != 0) {
res++;
list.set(i + 1, list.get(i + 1) - 1);
}
}
System.out.println(res);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 5d3e6d947f2b573ed3d5be24a5ae9c37 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
public static int inf = (int)1e9+7;
public static int mod = (int)1e9+7;
public static void main(String[] args) throws IOException, InterruptedException {
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
MyScanner sc = new MyScanner();
//code part
int t = sc.nextInt();
while (t-- > 0){
int n = sc.nextInt();
String s = sc.next();
int cnt = 0;
int ans = 0;
for (int i = 0; i < s.length(); i++){
if(i == 0 || s.charAt(i) == s.charAt(i - 1)){
cnt++;
}else{
if(cnt % 2 == 0){
cnt = 1;
}else{
cnt = 2;
ans++;
}
}
}
writer.write(ans + "\n");
}
//code part
writer.flush();
writer.close();
}
//快读模板
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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | f4bb70b3443abb1cc1c08f1e8b117786 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.*;
import java.util.*;
public class codeMaster {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int tc = sc.nextInt();
while(tc-->0){
int n = sc.nextInt();
char[] arr = sc.next().toCharArray();
ArrayList<Integer> seq = new ArrayList<>();
int cnt = 1;
for(int i = 0; i<n-1; i++){
if(arr[i] == arr[i+1]){
cnt++;
}else{
seq.add(cnt);
cnt = 1;
}
}
seq.add(cnt);
int carry = 0;
int ans = 0;
for(int x:seq){
x+=carry;
carry = 0;
if(x%2 != 0){
ans++;
carry++;
}
}
pw.println(ans);
}
pw.flush();
}
static class SegmentTree{
long[] tree;
int N;
public SegmentTree(long[] arr){
N = arr.length;
tree = new long[2*N - 1];
build(tree, arr);
}
public void build(long[] tree, long[] arr){
for(int i = N-1, j = 0; i<tree.length; i++, j++)tree[i] = arr[j];
for(int i = tree.length - 1, j = i - 1, k = N-2; k>=0; i -= 2, j-= 2, k--){
tree[k] = tree[i] + tree[j];
}
}
public void update(int idx, int val){
tree[idx + N - 2] = val;
boolean f = true;
int i = idx + N - 2;
int j = i - 1;
if(i % 2 != 0){
i++;
j++;
}
for(int k = (tree.length - N - 1) - ((tree.length - 1 - i)/2); k>=0; ){
tree[k] = tree[i] + tree[j];
f = !f;
i = k;
j = k - 1;
if(k % 2 != 0){
i++;
j++;
}
k = (tree.length - N - 1) - ((tree.length - 1 - i)/2);
}
}
}
public static boolean isSorted(Long[] arr){
boolean f = true;
for(int i = 1; i<arr.length; i++){
if(arr[i] < arr[i - 1]){
f = false;
break;
}
}
return f;
}
public static int binary_Search(long key, long[] arr){
int low = 0;
int high = arr.length;
int mid = (low + high) / 2;
while(low < high){
mid = (low + high) / 2;
if(arr[mid] == key)break;
else if(arr[mid] > key){
high = mid - 1;
}else{
low = mid + 1;
}
}
return mid;
}
public static int differences(int n, int test){
int changes = 0;
while(test > 0){
if(test % 10 != n % 10)changes++;
test/=10;
n/=10;
}
return changes;
}
static int maxSubArraySum(int a[], int size)
{
int max_so_far = Integer.MIN_VALUE,
max_ending_here = 0,start = 0,
end = 0, s = 0;
for (int i = 0; i < size; i++)
{
max_ending_here += a[i];
if (max_so_far < max_ending_here)
{
max_so_far = max_ending_here;
start = s;
end = i;
}
if (max_ending_here < 0)
{
max_ending_here = 0;
s = i + 1;
}
}
return start;
}
static int maxSubArraySum2(int a[], int size)
{
int max_so_far = Integer.MIN_VALUE,
max_ending_here = 0,start = 0,
end = 0, s = 0;
for (int i = 0; i < size; i++)
{
max_ending_here += a[i];
if (max_so_far < max_ending_here)
{
max_so_far = max_ending_here;
start = s;
end = i;
}
if (max_ending_here < 0)
{
max_ending_here = 0;
s = i + 1;
}
}
return end;
}
public static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static class Pair implements Comparable<Pair>{
int x;
int y;
public Pair(int x, int y){
this.x = x;
this.y = y;
}
public int compareTo(Pair x){
if(this.y - x.y != 0) return this.y - x.y;
else return this.x - x.x;
}
public String toString(){
return "("+this.x + ", " + this.y + ")";
}
}
public static boolean isSubsequence(char[] arr, String s){
boolean ans = false;
for(int i = 0, j = 0; i<arr.length; i++){
if(arr[i] == s.charAt(j)){
j++;
}
if(j == s.length()){
ans = true;
break;
}
}
return ans;
}
public static void sortIdx(long[]a,long[]idx) {
mergesortidx(a, idx, 0, a.length-1);
}
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 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 | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 5d86be94dc404bddf240a20b6be77a58 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main{
static int mod = (int)1e9+7;
static PrintWriter out;
static String[] memo;
static int n;
static String line;
public static void main(String[] args) throws IOException {
//BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
Scanner sc=new Scanner(System.in);
int t = sc.nextInt();
while(t-- !=0)
{
int n = sc.nextInt();
char [] m = sc.next().toCharArray();
//System.out.println(Arrays.toString(m));
int ans = 0;
for(int i=0;i<n;)
{
int tmp=i;
while(i<n && m[i]==m[tmp])i++;
// out.println(tmp+" "+n+" "+i);
if((i-tmp)%2!=0)
{
i++;
ans++;
}
}
out.println(ans);
}
out.flush();
}
public static int log2(int N)
{
// calculate log2 N indirectly
// using log() method
int result = (int)(Math.log(N) / Math.log(2));
if((int)(Math.log(N) % Math.log(2))!=0)
result++;
return result;
}
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();
}
}
static class pair implements Comparable<pair> {
int x;
int y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return Long.compare(this.y, other.y);
}
return Long.compare(this.x, other.x);
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | a25b6dcb79cebc4bcc7700883d8e2779 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | /**
* @Jai_Bajrang_Bali
* @Har_Har_Mahadev
*/
/* Created by IntelliJ IDEA.
* Author: Sanat Kumar Dubey (sanat04)
* Date: 02-05-2022
* Time: 19:43
* File: B.java
*/
import java.util.Scanner;
public class B {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n=sc.nextInt();
String str=sc.next();
char[] ch=str.toCharArray();
int count=0;
for (int i = 0; i < n; i+=2) {
if(ch[i]!=ch[i+1]) count++;
}
System.out.println(count);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 39e759b193fb54984e1228d7f0dab186 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | // package faltu;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.Map.Entry;
public class Main {
public static int upperBound(long[] arr, long m, int l, int r) {
while(l<=r) {
int mid=(l+r)/2;
if(arr[mid]<=m) l=mid+1;
else r=mid-1;
}
return l;
}
public static int lowerBound(long[] a, long m, int l, int r) {
while(l<=r) {
int mid=(l+r)/2;
if(a[mid]<m) l=mid+1;
else r=mid-1;
}
return l;
}
public static long getClosest(long val1, long val2,long target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
static void ruffleSort(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 ruffleSort(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);
}
int ceilIndex(int input[], int T[], int end, int s){
int start = 0;
int middle;
int len = end;
while(start <= end){
middle = (start + end)/2;
if(middle < len && input[T[middle]] < s && s <= input[T[middle+1]]){
return middle+1;
}else if(input[T[middle]] < s){
start = middle+1;
}else{
end = middle-1;
}
}
return -1;
}
public static int findIndex(long arr[], long t)
{
if (arr == null) {
return -1;
}
int len = arr.length;
int i = 0;
while (i < len) {
if (arr[i] == t) {
return i;
}
else {
i = i + 1;
}
}
return -1;
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a,long b)
{
return (a / gcd(a, b)) * b;
}
public static int[] swap(int a[], int left, int right)
{
int temp = a[left];
a[left] = a[right];
a[right] = temp;
return a;
}
public static void swap(long x,long max1)
{
long temp=x;
x=max1;
max1=temp;
}
public static int[] reverse(int a[], int left, int right)
{
// Reverse the sub-array
while (left < right) {
int temp = a[left];
a[left++] = a[right];
a[right--] = temp;
}
return a;
}
static int lowerLimitBinarySearch(ArrayList<Integer> A,int B) {
int n =A.size();
int first = 0,second = n;
while(first <second) {
int mid = first + (second-first)/2;
if(A.get(mid) > B) {
second = mid;
}else {
first = mid+1;
}
}
if(first < n && A.get(first) < B) {
first++;
}
return first; //1 index
}
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;
}
}
// *******----segement tree implement---*****
// -------------START--------------------------
void buildTree (int[] arr,int[] tree,int start,int end,int treeNode)
{
if(start==end)
{
tree[treeNode]=arr[start];
return;
}
buildTree(arr,tree,start,end,2*treeNode);
buildTree(arr,tree,start,end,2*treeNode+1);
tree[treeNode]=tree[treeNode*2]+tree[2*treeNode+1];
}
void updateTree(int[] arr,int[] tree,int start,int end,int treeNode,int idx,int value)
{
if(start==end)
{
arr[idx]=value;
tree[treeNode]=value;
return;
}
int mid=(start+end)/2;
if(idx>mid)
{
updateTree(arr,tree,mid+1,end,2*treeNode+1,idx,value);
}
else
{
updateTree(arr,tree,start,mid,2*treeNode,idx,value);
}
tree[treeNode]=tree[2*treeNode]+tree[2*treeNode+1];
}
// disjoint set implementation --start
static void makeSet(int n)
{
parent=new int[n];
rank=new int[n];
for(int i=0;i<n;i++)
{
parent[i]=i;
rank[i]=0;
}
}
static void union(int u,int v)
{
u=findpar(u);
v=findpar(v);
if(rank[u]<rank[v])parent[u]=v;
else if(rank[v]<rank[u])parent[v]=u;
else
{
parent[v]=u;
rank[u]++;
}
}
private static int findpar(int node)
{
if(node==parent[node])return node;
return parent[node]=findpar(parent[node]);
}
static int parent[];
static int rank[];
// *************end
static void presumbit(int[][]prebitsum) {
for(int i=1;i<=200000;i++) {
int z=i;
int j=0;
while(z>0) {
if((z&1)==1) {
prebitsum[i][j]+=(prebitsum[i-1][j]+1);
}else {
prebitsum[i][j]=prebitsum[i-1][j];
}
z=z>>1;
j++;
}
}
}
public static int[] sort(int[] arr) {
ArrayList<Integer> al = new ArrayList<>();
for(int i=0;i<arr.length;i++) al.add(arr[i]);
Collections.sort(al);
for(int i=0;i<arr.length;i++) arr[i]=al.get(i);
return arr;
}
static ArrayList<String>powof2s;
static void powof2S() {
long i=1;
while(i<(long)2e18) {
powof2s.add(String.valueOf(i));
i*=2;
}
}
static boolean coprime(int a, long l){
return (gcd(a, l) == 1);
}
static int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
static Long MOD=(long) (1e9+7);
static int prebitsum[][];
static ArrayList<Integer>arr;
static boolean[] vis;
static ArrayList<ArrayList<Integer>>adj;
public static int[] readIntArray(int tokens) {
int[] ret = new int[tokens];
for (int i = 0; i < tokens; i++) {
ret[i] =s.nextInt();
}
return ret;
}
public static long[] readLongArray(int tokens) {
long[] ret = new long[tokens];
for (int i = 0; i < tokens; i++) {
ret[i] =s.nextLong();
}
return ret;
}
static long[]a;
static FastReader s = new FastReader();
public static void main(String[] args) throws IOException
{
// sieve();
// prebitsum=new int[200001][18];
// presumbit(prebitsum);
// powof2S();
int tt = s.nextInt();
while(tt-->0) {
// int n=s.nextInt();
// int[]a=new int[n];
// for(int i=0;i<n;i++)a[i]=s.nextInt();
// Arrays.sort(a);
// solver(a,n);
int n=s.nextInt();
String str=s.next();
char[]ch=str.toCharArray();
ArrayList<Integer>a=new ArrayList<Integer>();
for(int i=0;i<ch.length;i++) {
if(ch[i]=='0') {
int cnt1=0;
while(i<n&&ch[i]=='0') {
i++;
cnt1++;
}
i--;
a.add(cnt1);
}
else {
int cnt1=0;
while(i<n&&ch[i]=='1') {
i++;
cnt1++;
}
i--;
a.add(cnt1);
}
}
int max=-1;
int ans=0;
for(int i=0;i<a.size();i++) {
if(a.get(i)%2!=0) {
++i;
int len=1;
while(i<a.size()&&a.get(i)%2==0) {
len++;
i++;
}
ans+=len;
}
}
max=ans;
ans=0;
for(int i=a.size()-1;i>=0;i--) {
if(a.get(i)%2!=0) {
--i;
int len=1;
while(i>=0&&a.get(i)%2==0) {
len++;
i--;
}
ans+=len;
}
}
max=Math.min(max,ans);
System.out.println(max);
}
}
static void solver(int[]a,int n) {
boolean f=false;
int cnt=0;
for(int i=0;i<n-1;i++) {
if(a[i]==a[i+1]) {a[i]=0;cnt++;f=true;}
}
int cntt=0;
for(int i:a) {
if(i!=0)cntt++;
else f=true;
}
if(f) {
int ans=cnt+cntt;
System.out.println(ans);
}
else System.out.println(n+1);
}
static void pc2d(char[][]a) {
int n=a.length;
int m=a[0].length;
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
static void pi2d(int[][]a) {
int n=a.length;
int m=a[0].length;
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
static void DFSUtil(int v, boolean[] vis)
{
vis[v] = true;
Iterator<Integer> it = adj.get(v).iterator();
while (it.hasNext()) {
int n = it.next();
if (!vis[n])
DFSUtil(n, vis);
}
}
static long DFS(int n)
{
vis = new boolean[n+1];
long cnt=0;
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
DFSUtil(i, vis);
cnt++;
}
}
return cnt;
}
public static String revStr(String str){
String input = str;
StringBuilder input1 = new StringBuilder();
input1.append(input);
input1.reverse();
return input1.toString();
}
public static String sortString(String inputString){
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
static long myPow(long n, long i){
if(i==0) return 1;
if(i%2==0) return (myPow(n,i/2)%MOD * myPow(n,i/2)%MOD)%MOD;
return (n%MOD* myPow(n,i-i)%MOD)%MOD;
}
static void palindromeSubStrs(String str) {
HashSet<String>set=new HashSet<>();
char[]a =str.toCharArray();
int n=str.length();
int[][]dp=new int[n][n];
for(int g=0;g<n;g++){
for(int i=0,j=g;j<n;j++,i++){
if(!set.contains(str.substring(i,i+1))&&g==0) {
dp[i][j]=1;
set.add(str.substring(i,i+1));
}
else {
if(!set.contains(str.substring(i,j+1))&&isPalindrome(str,i,j)) {
dp[i][j]=1;
set.add(str.substring(i,j+1));
}
}
}
}
int ans=0;
for(int i=0;i<n;i++) {
for(int j=0;j<n;j++) {
System.out.print(dp[i][j]+" ");
if(dp[i][j]==1)ans++;
}
System.out.println();
}
System.out.println(ans);
}
static boolean isPalindrome(String str,int i,int j)
{
while (i < j) {
if (str.charAt(i) != str.charAt(j))
return false;
i++;
j--;
}
return true;
}
static boolean sign(long num) {
return num>0;
}
static boolean isSquare(long x){
if(x==1)return true;
long y=(long) Math.sqrt(x);
return y*y==x;
}
static long Power(long a,long b) {
if(b == 0){
return 1;
}
long ans = Power(a,b/2);
ans *= ans%MOD;
if(b % 2!=0){
ans *= a%MOD;
}
return ans%MOD;
}
static void swap(StringBuilder sb,int l,int r)
{
char temp = sb.charAt(l);
sb.setCharAt(l,sb.charAt(r));
sb.setCharAt(r,temp);
}
// function to reverse the string between index l and r
static void reverse(StringBuilder sb,int l,int r)
{
while(l < r)
{
swap(sb,l,r);
l++;
r--;
}
}
// function to search a character lying between index l and r
// which is closest greater (just greater) than val
// and return it's index
static int binarySearch(StringBuilder sb,int l,int r,char val)
{
int index = -1;
while (l <= r)
{
int mid = (l+r)/2;
if (sb.charAt(mid) <= val)
{
r = mid - 1;
}
else
{
l = mid + 1;
if (index == -1 || sb.charAt(index) >= sb.charAt(mid))
index = mid;
}
}
return index;
}
// this function generates next permutation (if there exists any such permutation) from the given string
// and returns True
// Else returns false
static boolean nextPermutation(StringBuilder sb)
{
int len = sb.length();
int i = len-2;
while (i >= 0 && sb.charAt(i) >= sb.charAt(i+1))
i--;
if (i < 0)
return false;
else
{
int index = binarySearch(sb,i+1,len-1,sb.charAt(i));
swap(sb,i,index);
reverse(sb,i+1,len-1);
return true;
}
}
private static int lps(int m ,int n,String s1,String s2,int[][]mat)
{
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(s1.charAt(i-1)==s2.charAt(j-1))mat[i][j]=1+mat[i-1][j-1];
else mat[i][j]=Math.max(mat[i-1][j],mat[i][j-1]);
}
}
return mat[m][n];
}
static int lcs(String X, String Y, int m, int n)
{
int[][] L = new int[m+1][n+1];
// Following steps build L[m+1][n+1] in bottom up fashion. Note
// that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i-1) == Y.charAt(j-1))
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
}
}
return L[m][n];
// Following code is used to print LCS
// int index = L[m][n];
// int temp = index;
//
// // Create a character array to store the lcs string
// char[] lcs = new char[index+1];
// lcs[index] = '\u0000'; // Set the terminating character
//
// // Start from the right-most-bottom-most corner and
// // one by one store characters in lcs[]
// int i = m;
// int j = n;
// while (i > 0 && j > 0)
// {
// // If current character in X[] and Y are same, then
// // current character is part of LCS
// if (X.charAt(i-1) == Y.charAt(j-1))
// {
// // Put current character in result
// lcs[index-1] = X.charAt(i-1);
//
// // reduce values of i, j and index
// i--;
// j--;
// index--;
// }
//
// // If not same, then find the larger of two and
// // go in the direction of larger value
// else if (L[i-1][j] > L[i][j-1])
// i--;
// else
// j--;
// }
// return String.valueOf(lcs);
// Print the lcs
// System.out.print("LCS of "+X+" and "+Y+" is ");
// for(int k=0;k<=temp;k++)
// System.out.print(lcs[k]);
}
static long lis(long[] aa2, int n)
{
long lis[] = new long[n];
int i, j;
long max = 0;
for (i = 0; i < n; i++)
lis[i] = 1;
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (aa2[i] >= aa2[j] && lis[i] <= lis[j] + 1)
lis[i] = lis[j] + 1;
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
static boolean isPalindrome(String str)
{
int i = 0, j = str.length() - 1;
while (i < j) {
if (str.charAt(i) != str.charAt(j))
return false;
i++;
j--;
}
return true;
}
static boolean issafe(int i, int j, int r,int c, char ch)
{
if (i < 0 || j < 0 || i >= r || j >= c|| ch!= '1')return false;
else return true;
}
static long power(long a, long b)
{
a %=MOD;
long out = 1;
while (b > 0) {
if((b&1)!=0)out = out * a % MOD;
a = a * a % MOD;
b >>= 1;
a*=a;
}
return out;
}
static long[] sieve;
public static void sieve()
{
int nnn=(int) 1e6+1;
long nn=(int) 1e6;
sieve=new long[(int) nnn];
int[] freq=new int[(int) nnn];
sieve[0]=0;
sieve[1]=1;
for(int i=2;i<=nn;i++)
{
sieve[i]=1;
freq[i]=1;
}
for(int i=2;i*i<=nn;i++)
{
if(sieve[i]==1)
{
for(int j=i*i;j<=nn;j+=i)
{
if(sieve[j]==1)
{
sieve[j]=0;
}
}
}
}
}
}
class decrease implements Comparator<Long> {
// Used for sorting in ascending order of
// roll number
public int compare(long a, long b)
{
return (int) (b - a);
}
@Override
public int compare(Long o1, Long o2) {
// TODO Auto-generated method stub
return (int) (o2-o1);
}
}
class pair{
long x;
long y;
long c;
char ch;
public pair(long x,long y) {
this.x=x;
this.y=y;
}
public pair(long x,char ch) {
this.x=x;
this.ch=ch;
}
public pair(long x,long y,long c)
{
this.x=x;
this.y=y;
this.c=c;
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 1d5c05f53563e2db20ec52808dc1bce7 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.util.*;
public class B1{
public static int solve(String str){
ArrayList<Integer> arr=new ArrayList<>();
int count=1;
for(int i=0;i<str.length()-1;i++){
char ch1=str.charAt(i);
char ch2=str.charAt(i+1);
if(ch1!=ch2){
arr.add(count);
count=1;
}else{
count++;
}
}
char ch1=str.charAt(str.length()-1);
char ch2=str.charAt(str.length()-2);
if(ch1!=ch2){
arr.add(count);
arr.add(1);
}else{
count++;
arr.add(count);
}
count=0;
for(int i=0;i<arr.size();i++){
if(arr.get(i)%2==1&& i+1!=arr.size()){
int no=arr.get(i);
arr.set(i,no-1);
arr.set(i+1,arr.get(i+1)+1);
count++;
}else{
continue;
}
}
return count;
}
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=0;i<n;i++){
int len=sc.nextInt();
String str=sc.next();
int op=solve(str);
System.out.println(op);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 54a95b0614423cab91315a29c3522ca4 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
public class test {
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
int n=scn.nextInt();
while(n>0){
int len=scn.nextInt();
String str=scn.next();
char[] arr=new char[len];
for(int i=0;i<len;i++){
arr[i]=str.charAt(i);
}
int count=0;
int odd=0;
int even=0;
if(arr[0]=='0')even++;
else{
odd++;
}
for(int i=1;i<len;i++){
if(arr[i]=='0'){
if(arr[i-1]=='1' && odd%2==1){
count+=1;
arr[i]='1';
odd=0;
}else{
even+=1;
}
}else{
if(arr[i-1]=='0' && even%2==1){
count+=1;
arr[i]='0';
even=0;
}else{
odd+=1;
}
}
}
System.out.println(count);
n--;
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | cf9fb66f7322cd437136c5a377cfeea4 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// write your code here
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
int n = in.nextInt();
for (int i = 0; i < n; i++) {
int m = in.nextInt();
String a = in.next();
String b[] = a.split("");
int s = 1;
int sum = 0;
for (int j = 1; j < m; j++) {
if (b[j - 1].equals(b[j]))
s++;
else if (!b[j - 1].equals(b[j]) && s % 2 != 0) {
b[j] = b[j - 1];
sum++;
s ++;
}else
s = 1;
}
System.out.println(sum);
}
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 01e3730305113d170b08e8c7c3c92996 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
public class Solution{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String str;
int t, row, ops;
char last;
for(int i = 0; i < n; i++){
t = sc.nextInt();
sc.nextLine();
str = sc.nextLine();
row = 1;
ops = 0;
last = str.charAt(0);
for(int j = 1; j < str.length(); j++){
if(row != 0){
if(str.charAt(j) != last){
if(row % 2 != 0){
ops++;
row = 0;
}else{
row = 1;
last = str.charAt(j);
}
}else{
row++;
}
}else{
last = str.charAt(j);
row = 1;
}
}
System.out.println(ops);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 45c0fe0f5e6b677dd6d4b2a56637a51d | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
public class forces {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int flag = 1;
// int n = sc.nextInt();
int n = sc.nextInt();
String str = sc.next();
char a[] = new char[n + 1];
for (int i = 0; i < n; i++) {
a[i] = str.charAt(i);
}
ArrayList<Integer> list = new ArrayList<>();
int c = 1;
for (int i = 1; i < n + 1; i++) {
if (a[i - 1] == a[i]) {
c++;
} else {
list.add(c);
c = 1;
}
}
// list.add(c);
int l = list.size();
int b[] = new int[l];
int m = 0;
for (int x : list) {
b[m++] = x;
}
int ans = 0;
for (int i = 0; i < l; i++) {
if (b[i] % 2 == 1) {
ans++;
if (i + 1 < n) {
b[i + 1]++;
}
}
}
// System.out.println(list);
System.out.println(ans);
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | fb1993d73a822f06715b87a162e72d7d | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | /*
* Click nbfs://nbhost/SystemFileSystem/Templates/Licenses/license-default.txt to change this license
* Click nbfs://nbhost/SystemFileSystem/Templates/Classes/Main.java to edit this template
*/
import java.lang.invoke.MethodHandles;
import java.util.Scanner;
import javax.swing.text.PasswordView;
import java.util.*;
/**
*
* @author HauPC
*/
public class JavaApplication4 {
/**
* @param args the command line arguments
*
*/
public static void main(String[] args) {
// TODO code application logic here
// 6 9 15
Scanner input = new Scanner(System.in);
int t=Integer.parseInt(input.nextLine());
while(t-->0)
{
int n;
int res=0;
n=Integer.parseInt(input.nextLine());
String binaryString=input.nextLine();
int count =1;
for(int i=1;i<n;i++)
{
if(binaryString.charAt(i-1)!=binaryString.charAt(i)&&(count%2)!=0)
{
res++;
i++;
count=1;
}
else{
count++;
}
}
System.out.println(res);
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 81cea9a08fe16dfcb0a7466075b21547 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long startTime = System.currentTimeMillis();
// for global initializations and methods starts here
// global initialisations and methods end here
static void run() {
boolean tc = true;
//AdityaFastIO r = new AdityaFastIO();
FastReader r = new FastReader();
try (OutputStream out = new BufferedOutputStream(System.out)) {
//long startTime = System.currentTimeMillis();
int testcases = tc ? r.ni() : 1;
int tcCounter = 1;
// Hold Here Sparky------------------->>>
// Solution Starts Here
start:
while (testcases-- > 0) {
int n = r.ni();
char[] s = r.word().toCharArray();
int count = 0;
int res = 0;
char charc = '1';
{
int i = 0;
while (i < n) {
if (s[i] == s[i + 1]) {
charc = s[i + 1];
break;
}
i += 2;
}
}
{
int i = 0;
while (i < s.length) {
if (s[i] == s[i + 1]) {
charc = s[i + 1];
} else {
count++;
s[i] = charc;
s[i + 1] = charc;
}
i += 2;
}
}
charc = 'b';
int i = 0;
while (i < n) {
if (s[i] != charc) {
res++;
charc = s[i];
}
i++;
}
out.write((count + " ").getBytes());
//out.write((res + " ").getBytes());
out.write(("\n").getBytes());
}
// Solution Ends Here
} catch (IOException e) {
e.printStackTrace();
}
}
static class AdityaFastIO {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public BufferedReader br;
public StringTokenizer st;
public AdityaFastIO() {
br = new BufferedReader(new InputStreamReader(System.in));
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public AdityaFastIO(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public String readLine() throws IOException {
byte[] buf = new byte[100000001]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int ni() 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 nl() 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 nd() 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 Exception {
run();
}
static int[] readIntArr(int n, AdityaFastIO r) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = r.ni();
return arr;
}
static long[] readLongArr(int n, AdityaFastIO r) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = r.nl();
return arr;
}
static List<Integer> readIntList(int n, AdityaFastIO r) throws IOException {
List<Integer> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.ni());
return al;
}
static List<Long> readLongList(int n, AdityaFastIO r) throws IOException {
List<Long> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.nl());
return al;
}
static long mod = 998244353;
static long modInv(long base, long e) {
long result = 1;
base %= mod;
while (e > 0) {
if ((e & 1) > 0) result = result * base % mod;
base = base * base % mod;
e >>= 1;
}
return result;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int ni() {
return Integer.parseInt(word());
}
long nl() {
return Long.parseLong(word());
}
double nd() {
return Double.parseDouble(word());
}
}
static int MOD = (int) (1e9 + 7);
static long powerLL(long x, long n) {
long result = 1;
while (n > 0) {
if (n % 2 == 1) result = result * x % MOD;
n = n / 2;
x = x * x % MOD;
}
return result;
}
static long powerStrings(int i1, int i2) {
String sa = String.valueOf(i1);
String sb = String.valueOf(i2);
long a = 0, b = 0;
for (int i = 0; i < sa.length(); i++) a = (a * 10 + (sa.charAt(i) - '0')) % MOD;
for (int i = 0; i < sb.length(); i++) b = (b * 10 + (sb.charAt(i) - '0')) % (MOD - 1);
return powerLL(a, b);
}
static long gcd(long a, long b) {
if (a == 0) return b;
else return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long lower_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] >= x) r = m;
else l = m;
}
return r;
}
static int upper_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] <= x) l = m;
else r = m;
}
return l + 1;
}
static void addEdge(ArrayList<ArrayList<Integer>> graph, int edge1, int edge2) {
graph.get(edge1).add(edge2);
graph.get(edge2).add(edge1);
}
public static class Pair implements Comparable<Pair> {
int first;
int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(Pair o) {
// TODO Auto-generated method stub
if (this.first != o.first)
return (int) (this.first - o.first);
else return (int) (this.second - o.second);
}
}
public static class PairC<X, Y> implements Comparable<PairC> {
X first;
Y second;
public PairC(X first, Y second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(PairC o) {
// TODO Auto-generated method stub
return o.compareTo((PairC) first);
}
}
static boolean isCollectionsSorted(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) <= list.get(i - 1)) return false;
return true;
}
static boolean isCollectionsSortedReverseOrder(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) >= list.get(i - 1)) return false;
return true;
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | c096b1b8c6d96c6397703288035aabe4 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.io.*;
import static java.lang.Math.*;
import static java.lang.System.out;
public class Main {
static class Reader{
BufferedReader br;
StringTokenizer st;
public Reader(boolean f) throws IOException{
if(f) {
br = new BufferedReader(new FileReader("input.txt"));
}else{
br=new BufferedReader(new InputStreamReader(System.in));
}
}
String next(){
while(st==null || !st.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class Writer {
private final BufferedWriter bw;
public Writer(boolean f) throws IOException {
if(f) {
this.bw = new BufferedWriter(new FileWriter("output.txt"));
}else{
this.bw = new BufferedWriter(new OutputStreamWriter(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();
}
}
public static void main(String[] args){
try {
Reader in=new Reader(false);
Writer out =new Writer(false);
int t = 1;
t=in.nextInt();
while(t-- > 0){
// write code here
int n = in.nextInt();
String s = in.nextLine();
int count = 0;
for (int i = 0; i <=n-1; i++) {
if((s.charAt(i)=='1'&&i%2==0&&s.charAt(i)!=s.charAt(i+1))||(s.charAt(i)=='0'&&i%2==0&&s.charAt(i)!=s.charAt(i+1))){
count++;
}
}
out.println(count);
}
out.close();
} catch (Exception e) {
return;
}
}
} | Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | f08ae876ec05b0553b42631cea1a1039 | train_108.jsonl | 1652020500 | This is the easy version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1'($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? | 256 megabytes | //CP-MASS
import java.util.*;
import java.io.*;
public class Codechef{
static FastReader in=new FastReader();
static final Random random=new Random();
static long mod=1000000007L;
static HashMap<String,Integer>map=new HashMap<>();
public static void main(String args[]) throws IOException {
try{
int t=in.nextInt();
int cse=1;
loop:
while(t-->0)
{
StringBuilder res=new StringBuilder();
//res.append("Hello"+"\n");
int n=in.nextInt();
ArrayList<Integer> al=new ArrayList<>();
String s=in.next();
int c=1;
int f=0;
for(int i=1;i<n;i++)
{
if(s.charAt(i)==s.charAt(i-1))
{
c++;
}
else{
al.add(c);
c=1;
}
}
if(s.charAt(n-1)==s.charAt(n-2))
{
al.add(c);
}
else{
al.add(1);
}
for(int i=0;i<al.size();i++)
{
if(((int)al.get(i))%2!=0)
{
f=1;
break;
}
}
if(f==0)
{
println("0");
}
else{
int op=0;
for(int i=0;i<al.size()-1;i++)
{
if((int)al.get(i)%2==1 && (int)al.get(i+1)%2==1)
{
op++;
al.set(i,(int)al.get(i)+1);
al.set(i+1,(int)al.get(i+1)-1);
}
else if((int)al.get(i)%2==1 && (int)al.get(i+1)%2==0)
{
op++;
al.set(i,(int)al.get(i)+1);
al.set(i+1,(int)al.get(i+1)-1);
}
}
println(op);
}
//println(op);
}
}catch(Exception e)
{
return;
}
}
static int max(int a, int b)
{
if(a<b)
return b;
return a;
}
static void ruffleSort(int[] a) {
int n=a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static < E > void print(E res)
{
System.out.print(res);
}
static < E > void println(E res)
{
System.out.println(res);
}
static int gcd(int a,int b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static int abs(int a)
{
if(a<0)
return -1*a;
return a;
}
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
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;
}
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;
}
int [] readintarray(int n) {
int res [] = new int [n];
for(int i = 0; i<n; i++)res[i] = nextInt();
return res;
}
long [] readlongarray(int n) {
long res [] = new long [n];
for(int i = 0; i<n; i++)res[i] = nextLong();
return res;
}
}
}
| Java | ["5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110"] | 1 second | ["3\n0\n0\n0\n3"] | NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100".In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. | Java 8 | standard input | [
"implementation"
] | aded139ff4d3f313f8e4d81559760f12 | The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, print a single line with one integer — the minimum number of operations to make $$$s$$$ good. | standard output | |
PASSED | 71fd48abacae88905aa6194c5a091ea4 | train_108.jsonl | 1652020500 | Tokitsukaze has a sequence $$$a$$$ of length $$$n$$$. For each operation, she selects two numbers $$$a_i$$$ and $$$a_j$$$ ($$$i \ne j$$$; $$$1 \leq i,j \leq n$$$). If $$$a_i = a_j$$$, change one of them to $$$0$$$. Otherwise change both of them to $$$\min(a_i, a_j)$$$. Tokitsukaze wants to know the minimum number of operations to change all numbers in the sequence to $$$0$$$. It can be proved that the answer always exists. | 256 megabytes | import java.util.*;
import java.io.*;
public class G {
static FastScanner fs = new FastScanner();
static PrintWriter pw = new PrintWriter(System.out);
static StringBuilder sb = new StringBuilder("");
static Scanner scn = new Scanner(System.in);
public static void main(String[] args) {
int t = fs.nextInt();
//int t = 1;
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = fs.nextInt();
sort(a);
int cnt = 0;
int z = 0;
for (int i = 0; i < n-1; i++) {
if (a[i] == 0) z++;
if (a[i] == a[i+1] && a[i] != 0) cnt++;
}
if (a[n-1] == 0) z++;
if (z > 0) sb.append(n-z + "\n");
else if (cnt > 0) sb.append(n + "\n");
else sb.append(n+1 + "\n");
}
pw.print(sb.toString());
pw.close();
}
static void sort(int[] a) {
ArrayList<Integer> al = new ArrayList<>();
for (int i : a)
al.add(i);
Collections.sort(al);
for (int i = 0; i < a.length; i++)
a[i] = al.get(i);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {while (!st.hasMoreTokens())try {st = new StringTokenizer(br.readLine());} catch (IOException e) {}return st.nextToken();}
int nextInt() {return Integer.parseInt(next());}
long nextLong() {return Long.parseLong(next());}
double nextDouble() {return Double.parseDouble(next());}
int[] readArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) {a[i] = nextInt();} return a;}
}
} | Java | ["3\n3\n1 2 3\n3\n1 2 2\n3\n1 2 0"] | 1 second | ["4\n3\n2"] | NoteIn the first test case, one of the possible ways to change all numbers in the sequence to $$$0$$$:In the $$$1$$$-st operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = a_1 = 1$$$. Now the sequence $$$a$$$ is $$$[1,1,3]$$$.In the $$$2$$$-nd operation, $$$a_1 = a_2 = 1$$$, after the operation, $$$a_1 = 0$$$. Now the sequence $$$a$$$ is $$$[0,1,3]$$$.In the $$$3$$$-rd operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,3]$$$.In the $$$4$$$-th operation, $$$a_2 < a_3$$$, after the operation, $$$a_3 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,0]$$$.So the minimum number of operations is $$$4$$$. | Java 11 | standard input | [
"implementation"
] | ad242f98f1c8eb8d30789ec672fc95a0 | The first line contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the length of the sequence $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the sequence $$$a$$$. | 800 | For each test case, print a single integer — the minimum number of operations to change all numbers in the sequence to $$$0$$$. | standard output | |
PASSED | 51b46caa66a0398d07c4e44f3a2f2ae2 | train_108.jsonl | 1652020500 | Tokitsukaze has a sequence $$$a$$$ of length $$$n$$$. For each operation, she selects two numbers $$$a_i$$$ and $$$a_j$$$ ($$$i \ne j$$$; $$$1 \leq i,j \leq n$$$). If $$$a_i = a_j$$$, change one of them to $$$0$$$. Otherwise change both of them to $$$\min(a_i, a_j)$$$. Tokitsukaze wants to know the minimum number of operations to change all numbers in the sequence to $$$0$$$. It can be proved that the answer always exists. | 256 megabytes | import java.util.*;
import java.lang.*;
public class Main{
public static String s;
public static int n;
public static void main(String args[]){
Scanner inp=new Scanner(System.in);
int t=inp.nextInt();
for(int i=0;i<t;i++){
int count=0,a=0,b=0,coun=0;
/*n=inp.nextInt();
comeback2 obj[] = new comeback2[n];
int x=inp.nextInt();
inp.nextLine();
for(int j=0;j<n;j++){
obj[j].s=inp.nextLine();
System.out.println(obj[j].s);*/
n=inp.nextInt();
int arr[]=new int[n];
for(int j=0;j<n;j++){
arr[j]=inp.nextInt();
if(arr[j]==0)
count++;
}
Arrays.sort(arr);
if(arr[0]==0 && arr[n-1]==0){
System.out.println(0);
continue;
}
for(int j=0;j<n;j++){
if(arr[j]==0){
System.out.println(n-count);
a++;
break;
}
}
if(a!=0)
continue;
for(int j=0;j<n-1;j++){
if(arr[j]==arr[j+1]){
coun=1;
coun=coun+(n-1);
System.out.println(coun);
b++;
break;
}
}
if(b!=0)
continue;
int run=2+(n-1);
System.out.println(run);
}
/* for(int j=0;j<n;j++){
for(int k=0;k<x;k++){
char first = 'e';
char last = 'o';
int diff = last-first;
}
}*/
//System.out.println(diff);
}
}
| Java | ["3\n3\n1 2 3\n3\n1 2 2\n3\n1 2 0"] | 1 second | ["4\n3\n2"] | NoteIn the first test case, one of the possible ways to change all numbers in the sequence to $$$0$$$:In the $$$1$$$-st operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = a_1 = 1$$$. Now the sequence $$$a$$$ is $$$[1,1,3]$$$.In the $$$2$$$-nd operation, $$$a_1 = a_2 = 1$$$, after the operation, $$$a_1 = 0$$$. Now the sequence $$$a$$$ is $$$[0,1,3]$$$.In the $$$3$$$-rd operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,3]$$$.In the $$$4$$$-th operation, $$$a_2 < a_3$$$, after the operation, $$$a_3 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,0]$$$.So the minimum number of operations is $$$4$$$. | Java 11 | standard input | [
"implementation"
] | ad242f98f1c8eb8d30789ec672fc95a0 | The first line contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the length of the sequence $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the sequence $$$a$$$. | 800 | For each test case, print a single integer — the minimum number of operations to change all numbers in the sequence to $$$0$$$. | standard output | |
PASSED | 713995cf2b5cef522cb67ed03492a860 | train_108.jsonl | 1652020500 | Tokitsukaze has a sequence $$$a$$$ of length $$$n$$$. For each operation, she selects two numbers $$$a_i$$$ and $$$a_j$$$ ($$$i \ne j$$$; $$$1 \leq i,j \leq n$$$). If $$$a_i = a_j$$$, change one of them to $$$0$$$. Otherwise change both of them to $$$\min(a_i, a_j)$$$. Tokitsukaze wants to know the minimum number of operations to change all numbers in the sequence to $$$0$$$. It can be proved that the answer always exists. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class ContestAns{
public static void main(String[] args){
Scanner scin = new Scanner(System.in);
int t = scin.nextInt();
while (t-->0){
int times = scin.nextInt();
int[] a = new int[times];
for (int i = 0 ; i < times ; i++){
a[i] = scin.nextInt();
}
Arrays.sort(a);
int count = 0;
for (int i = 0 ; i < times ; i++){
if (a[i] == 0){
count++;
}
}
if (count > 0){
System.out.println(times - count);
continue;
}
boolean x = false;
for (int i = 0 ; i < times - 1 ; i++){
if (a[i] == a[i+1]){
x = true;
break;
}
}
if (x){
System.out.println(times);
}
else{
System.out.println(times + 1);
}
}
}
} | Java | ["3\n3\n1 2 3\n3\n1 2 2\n3\n1 2 0"] | 1 second | ["4\n3\n2"] | NoteIn the first test case, one of the possible ways to change all numbers in the sequence to $$$0$$$:In the $$$1$$$-st operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = a_1 = 1$$$. Now the sequence $$$a$$$ is $$$[1,1,3]$$$.In the $$$2$$$-nd operation, $$$a_1 = a_2 = 1$$$, after the operation, $$$a_1 = 0$$$. Now the sequence $$$a$$$ is $$$[0,1,3]$$$.In the $$$3$$$-rd operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,3]$$$.In the $$$4$$$-th operation, $$$a_2 < a_3$$$, after the operation, $$$a_3 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,0]$$$.So the minimum number of operations is $$$4$$$. | Java 11 | standard input | [
"implementation"
] | ad242f98f1c8eb8d30789ec672fc95a0 | The first line contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the length of the sequence $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the sequence $$$a$$$. | 800 | For each test case, print a single integer — the minimum number of operations to change all numbers in the sequence to $$$0$$$. | standard output | |
PASSED | 9727a0c10d5a0672a4fa455939c4bcfd | train_108.jsonl | 1652020500 | Tokitsukaze has a sequence $$$a$$$ of length $$$n$$$. For each operation, she selects two numbers $$$a_i$$$ and $$$a_j$$$ ($$$i \ne j$$$; $$$1 \leq i,j \leq n$$$). If $$$a_i = a_j$$$, change one of them to $$$0$$$. Otherwise change both of them to $$$\min(a_i, a_j)$$$. Tokitsukaze wants to know the minimum number of operations to change all numbers in the sequence to $$$0$$$. It can be proved that the answer always exists. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class ContestAns {
public static void main(String[] koolProgrammer){
Scanner scin = new Scanner(System.in);
int t = scin.nextInt();
while (t-->0){
int times = scin.nextInt();
int[] a = new int[times];
for (int i = 0 ; i < times ; i++){
a[i] = scin.nextInt();
}
Arrays.sort(a);
int count = 0;
for (int i = 0 ; i < times ; i++){
if (a[i] == 0){
count++;
}
}
if (count > 0){
System.out.println(times - count);
continue;
}
boolean x = false;
for (int i = 1 ; i < times ; i++){
if (a[i] == a[i - 1]){
x = true;
}
}
if (x){
System.out.println(times);
}
else{
System.out.println(times + 1);
}
}
}
} | Java | ["3\n3\n1 2 3\n3\n1 2 2\n3\n1 2 0"] | 1 second | ["4\n3\n2"] | NoteIn the first test case, one of the possible ways to change all numbers in the sequence to $$$0$$$:In the $$$1$$$-st operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = a_1 = 1$$$. Now the sequence $$$a$$$ is $$$[1,1,3]$$$.In the $$$2$$$-nd operation, $$$a_1 = a_2 = 1$$$, after the operation, $$$a_1 = 0$$$. Now the sequence $$$a$$$ is $$$[0,1,3]$$$.In the $$$3$$$-rd operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,3]$$$.In the $$$4$$$-th operation, $$$a_2 < a_3$$$, after the operation, $$$a_3 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,0]$$$.So the minimum number of operations is $$$4$$$. | Java 11 | standard input | [
"implementation"
] | ad242f98f1c8eb8d30789ec672fc95a0 | The first line contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the length of the sequence $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the sequence $$$a$$$. | 800 | For each test case, print a single integer — the minimum number of operations to change all numbers in the sequence to $$$0$$$. | standard output | |
PASSED | 9d5ae94fe93ac1f1e636be1b49377851 | train_108.jsonl | 1652020500 | Tokitsukaze has a sequence $$$a$$$ of length $$$n$$$. For each operation, she selects two numbers $$$a_i$$$ and $$$a_j$$$ ($$$i \ne j$$$; $$$1 \leq i,j \leq n$$$). If $$$a_i = a_j$$$, change one of them to $$$0$$$. Otherwise change both of them to $$$\min(a_i, a_j)$$$. Tokitsukaze wants to know the minimum number of operations to change all numbers in the sequence to $$$0$$$. It can be proved that the answer always exists. | 256 megabytes | //package javaapplication1;
import java.util.*;
import java.io.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc.
ArrayList<Integer> A = new ArrayList<Integer>();
for (int i = 1; i <= t; i++) {
int a=in.nextInt();
int ans=0;
for(int j=0;j<a;j++){
A.add(in.nextInt());
}
for(int j=0;j<A.size();j++){
int b=Collections.frequency(A, A.get(j));
if(b>=2){
ans=A.size()-Collections.frequency(A, 0);
j=A.size();
}
}
if(ans==0){
if(A.contains(0)==true){
ans=(A.size())-Collections.frequency(A, 0);
}else{
ans=(A.size()+1);
}
}
System.out.println(ans);
A = new ArrayList<Integer>();
}
}
} | Java | ["3\n3\n1 2 3\n3\n1 2 2\n3\n1 2 0"] | 1 second | ["4\n3\n2"] | NoteIn the first test case, one of the possible ways to change all numbers in the sequence to $$$0$$$:In the $$$1$$$-st operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = a_1 = 1$$$. Now the sequence $$$a$$$ is $$$[1,1,3]$$$.In the $$$2$$$-nd operation, $$$a_1 = a_2 = 1$$$, after the operation, $$$a_1 = 0$$$. Now the sequence $$$a$$$ is $$$[0,1,3]$$$.In the $$$3$$$-rd operation, $$$a_1 < a_2$$$, after the operation, $$$a_2 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,3]$$$.In the $$$4$$$-th operation, $$$a_2 < a_3$$$, after the operation, $$$a_3 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,0]$$$.So the minimum number of operations is $$$4$$$. | Java 11 | standard input | [
"implementation"
] | ad242f98f1c8eb8d30789ec672fc95a0 | The first line contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the length of the sequence $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the sequence $$$a$$$. | 800 | For each test case, print a single integer — the minimum number of operations to change all numbers in the sequence to $$$0$$$. | standard output |
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