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')
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