You are given a sorted sequence of numbers, ending with a -1.

Question

You are given a sorted (either in the increasing or in the decreasing order) sequence of numbers, ending with a -1. You can assume that are at least two numbers before the ending -1.

Let us call the sequence x0  x1 ... xn -1.

You have to output the number of distinct elements in the sorted sequence.

Kindly do not use arrays in the code.

Answer 1

#include<iostream>
using namespace std;
int main()
{
    int a=0,b,c,count=0;
    while(a!=-1)
    {
        cin>>a;
        if(c!=a)
            count++;
        c=a;
    }
    cout<<"Number Of Distinct Number is:"<<count-1;
}

Answer 2

#include<stdio.h>
void main()
{
    int a,count=0,x,y;
    scanf("%d",&a);
    while(a!=-1)
    {
        x=a;
        scanf("%d",&a);
        y=a;
        if(x!=y&&a!=-1)
                count++;   
      }
     if(count>=2)
            printf("1");
      else
            printf("0");
}