Suppose we have an array A containing K elements. Consider a game with N players and a game master. This game has K rounds. In round i of the game, the game master announces that A[i] children will be formed into groups. The remaining children then form as many groups of A[i] children as possible. A child cannot participate in more than one group. No one in the group leaves the game. Others advance to the next round. There may be no player losses in a round. Finally, after round K, only two children were left and they were declared the winners. We must either find the minimum and maximum number of children that may exist in the game before starting, or determine that N does not have a valid value.
So if the input is something like A = [3, 4 , 3, 2] then the output will be [6, 8] because if the game starts with 6 children then it will continue
In the first round, 6 of them formed two groups of 3 people each
They formed two groups of 4 and 2 respectively Children
Then a group of 1 child and 3 children, 1 will leave the game
The three of them form a group of 1 and 2 Group. 1 will leave.
The last 2 kids were declared the winners.
To solve this problem we will follow the following steps-
n := size of A Define a large array a, l, r, a of size: 100010. l := 2, r = 2 for initialize i := 1, when i <= n, update (increase i by 1), do: a[i] := A[i - 1] for initialize i := n, when i >= 1, update (decrease i by 1), do: x := a[i], L := (l + x - 1) if L > R, then: return -1, 0 l := L, r = R + x - 1 return l, r
Let us see the following implementation for better understanding -
#include <bits/stdc++.h> using namespace std; void solve(vector<int> A){ int n = A.size(); int l, r, a[100010]; l = 2, r = 2; for (int i = 1; i <= n; i++) a[i] = A[i - 1]; for (int i = n; i >= 1; i--){ int x = a[i], L = (l + x - 1) / x * x, R = r / x * x; if (L > R){ cout << "-1, 0"; } l = L, r = R + x - 1; } cout << l << ", " << r << endl; return; } int main(){ vector<int> A = { 3, 4, 3, 2 }; solve(A); }
{ 3, 4, 3, 2 }
6, 8
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