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C++ code to count order collected when client calls
Suppose we have three numbers n, m and z. An office receives calls in every n minutes, and some deliveries come to office in every m minutes. Office is open for z minutes. We have to count the minimum number of orders are collected so there are no pending orders when client calls. Consider taking orders and talking with clients take exactly 1 minutes.
So, if the input is like n = 1; m = 2; z = 5, then the output will be 2, because we need to collect orders which comes in second and fourth minutes.
Steps
To solve this, we will follow these steps −
return z / ((n * m) / (gcd of n and m))
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h> using namespace std; int solve(int n, int m, int z){ return z / ((n * m) / __gcd(n, m)); } int main(){ int n = 1; int m = 2; int z = 5; cout << solve(n, m, z) << endl; }
Input
1, 2, 5
Output
2
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