344A-magnet(A.Magnet)

-

#include<iostream>
using namespace std;
int main()
{
int n,z=1;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)cin>>arr[i];
int in=arr[0];
for(int i=1;i<n;i++)
{

if(arr[i-1]!=arr[i]){
z++;
}

}
cout<<z<<endl;
}

Comments

Post a Comment

Popular posts from this blog

Codeforce Problem 1703A. YES or YES?

-
A. YES or YES?   There is a string  s �  of length  3 3 , consisting of uppercase and lowercase English letters. Check if it is equal to " YES " (without quotes), where each letter can be in any case. For example, " yES ", " Yes ", " yes " are all allowable. Input The first line of the input contains an integer  t �  ( 1 ≤ t ≤ 10 3 1 ≤ � ≤ 10 3 ) — the number of testcases. The description of each test consists of one line containing one string  s �  consisting of three characters. Each character of  s �  is either an uppercase or lowercase English letter. Output For each test case, output " YES " (without quotes) if  s �  satisfies the condition, and " NO " (without quotes) otherwise. You can output " YES " and " NO " in any case (for example, strings " yES ", " yes " and " Yes " will be recognized as a positive response). Example input Copy 10 YES yES yes Yes YeS Noo orZ yEz Y...
Read more

Aptitude test assistant programmer 2018

-
 #include <stdio.h> #include <stdlib.h> int main() { char str[100]; int i; int space=0;     printf("Enter a string\n");     fgets(str,sizeof(str),stdin);     i=0;     while(i<=str[i]){  if(str[i]==' '){         space++;     }     i++; } printf("Total word count :%d" ,space+1);       return 0; }
Read more

1041A. Heist solution

-
#include <iostream> #include <algorithm> using namespace std ; int main () { long int n ; cin >> n ;   long long arr [ n ]; for ( int j = 0 ; j < n ; j ++){ cin >> arr [ j ];}   sort ( arr , arr + n );   int b = arr [ 0 ];   int l = arr [ n - 1 ];   int x = 0 ;   for ( int i = b ; i <= l ; i ++) {   x ++; }   cout << x - n ;   return 0 ; }
Read more