Game of Stones
Problem Statement :
Two players called and are playing a game with a starting number of stones. Player always plays first, and the two players move in alternating turns. The game's rules are as follows: In a single move, a player can remove either , , or stones from the game board. If a player is unable to make a move, that player loses the game. Given the starting number of stones, find and print the name of the winner. is named First and is named Second. Each player plays optimally, meaning they will not make a move that causes them to lose the game if a winning move exists. For example, if , can make the following moves: removes stones leaving . will then remove stones and win. removes stones leaving . cannot move and loses. would make the second play and win the game. Function Description Complete the gameOfStones function in the editor below. It should return a string, either First or Second. gameOfStones has the following parameter(s): n: an integer that represents the starting number of stones Input Format The first line contains an integer , the number of test cases. Each of the next lines contains an integer , the number of stones in a test case. Output Format On a new line for each test case, print First if the first player is the winner. Otherwise print Second.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int dp[n+1];
dp[0]=dp[1]=1;
dp[4]=dp[2]=dp[3]=dp[5]=0;
int i;
for(i=6;i<=n;i++)
{
if(dp[i-5]||dp[i-2]||dp[i-3])
dp[i]=0;
else
dp[i]=1;
}
if(dp[n])
printf("Second\n");
else
printf("First\n");
}
return 0;
}
Solution in C++ :
In C++ :
#include <bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
long DP[101];
bool dp(long x)
{
if (DP[x]!=-1) return DP[x];
bool res=!dp(x-2);
if (x>3) res=res|(!dp(x-3));
if (x>5) res=res|(!dp(x-5));
DP[x]=res;
return res;
}
int main()
{
long nTest,n;
memset(DP,-1,sizeof(DP));
DP[1]=0;
DP[2]=DP[3]=DP[5]=1;
scanf("%ld",&nTest);
while (nTest--)
{
scanf("%ld",&n);
puts(dp(n)?"First":"Second");
}
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int no_of_testcases=Integer.parseInt(br.readLine());
while(no_of_testcases-->0){
int number=Integer.parseInt(br.readLine());
if(number%7==0 || number%7==1){
System.out.println("Second");
}
else{
System.out.println("First");
}
}
}
}
Solution in Python :
In Python3 :
def get_winner(n):
ans=[None]*(n+1)
ans[0]=False
ans[1]=False
for i in range(n+1):
if ans[i]==False:
for j in [i+2,i+3,i+5]:
if j<n+1:
ans[j]=True
elif ans[i] is None:
for j in [i-2, i-3, i-5]:
if j>0 and ans[j]!=True:
print('??',i,j,ans[j])
ans[i]=False
for j in [i+2,i+3,i+5]:
if j<n+1:
ans[j]=True
# print(i,ans)
return ans
w=get_winner(100)
for _ in range(int(input().strip())):
print('First' if w[int(input().strip())] else 'Second')
View More Similar Problems
Insert a node at a specific position in a linked list
Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e
View Solution →Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →Print in Reverse
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing
View Solution →Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →
