Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
C++ Program to find out the distinct elements in a given sequence
Suppose we are given three integer numbers n, x, y, and z. We have to make a sequence from the given integers where the first item of the sequence is (x modulo 231). Other than the first element, other elements in the sequence ai = (a(i-1) * y + z) modulo 231 , where 1 ≤ i ≤ n - 1. We have to find out the number of distinct integers in the sequence we made.
So, if the input is like n = 5, x = 1, y = 2, z = 1, then the output will be 5.
The unique values are {1, 3, 7, 15, 31}. So, the answer is 5.
To solve this, we will follow these steps −
- MOD := 2^31
- Define an array temp
- resize the array temp to the value of MOD
- p := x mod MOD
- temp[p] := true
- ans := 1
- for initialize i := 1, when i < n, update (increase i by 1), do −
- p := ((p * y) + z) mod MOD
- if temp[p] is true, then −
- Come out from the loop
- increase ans by 1
- temp[p] := true
- return ans
Example
Let us see the following implementation to get better understanding −
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const long long MOD = 2147483648;
int solve(int n, long long x, long long y, long long z) {
vector<bool> temp;
temp.resize(MOD);
long long p = x % MOD;
temp[p] = true;
int ans = 1;
for (int i = 1; i < n; ++i) {
p = ((p * y) + z) % MOD;
if (temp[p])
break;
++ans;
temp[p] = true;
}
return ans;
}
int main() {
cout<< solve(5, 1, 2, 1) <<endl;
return 0;
}Input
5, 1, 2, 1
Output
5
Updated on: 20-Oct-2021
264 Views
Advertisements