Misère Nim


Problem Statement :


Two people are playing game of Misère Nim. The basic rules for this game are as follows:

The game starts with  piles of stones indexed from  to . Each pile  (where ) has  stones.
The players move in alternating turns. During each move, the current player must remove one or more stones from a single pile.
The player who removes the last stone loses the game.
Given the value of  and the number of stones in each pile, determine whether the person who wins the game is the first or second person to move. If the first player to move wins, print First on a new line; otherwise, print Second. Assume both players move optimally.


There is no permutation of optimal moves where player 2 wins. For example, player 1 chooses the first pile. If player 2 chooses 1 from another pile, player 1 will choose the pile with 2 left. If player 2 chooses a pile of 2, player 1 chooses 1 from the remaining pile leaving the last stone for player 2. Return First.

Function Description

Complete the misereNim function in the editor below.

misereNim has the following parameters:

int s[n]: the number of stones in each pile

Returns

string: either First or Second

Input Format

The first line contains an integer, , the number of test cases.
Each test case is described over two lines:

An integer, , the number of piles.
 space-separated integers, , that describe the number of stones at pile .



Solution :



title-img


                            Solution in C :

In  C  :




#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {


    int num_tests = 0;
    int number_of_piles = 0;
    int *piles = NULL;
    int idx, idx1;
    int total = 0;
    int pile_more_than_one_true = 0;

    scanf("%d", &num_tests);
    
    for (idx = 0; idx < num_tests; idx++) {
        scanf("%d", &number_of_piles);
        piles =  malloc(number_of_piles*sizeof(int));
        
        for (idx1 = 0; idx1 < number_of_piles; idx1++){
            scanf("%d", &piles[idx1]);
            if (piles[idx1] > 1)
                pile_more_than_one_true = 1;
            total ^= piles[idx1];
        }

        if (pile_more_than_one_true && total == 0)
            printf("Second\n");
        else if (!pile_more_than_one_true && total == 1)
            printf("Second\n");
        else
            printf("First\n");
   
        total = 0;
        number_of_piles = 0;
        free(piles);
        pile_more_than_one_true = 0;
    }
    return 0;
}
                        


                        Solution in C++ :

In  C++  :





#include <bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

int main()
{
	long nTest,n;
	scanf("%ld",&nTest);
	while (nTest--)
	{
		long res=0,cnt=0,x;
		scanf("%ld",&n);
		for (long i=0; i<n; ++i)
		{
			scanf("%ld",&x);
			if (x==1) ++cnt;
			res^=x;
		}
		if (cnt==n) puts((cnt&1)?"Second":"First");
		else puts((res>0)?"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 {
//Day 2: Misere Nim
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        int[] s = new int[100];
        int i,j,n, max;
        
        int nimsum;
        for(int tt =0;tt<t;tt++){
            n = in.nextInt();
            max = 0;
            for(i = 0;i<n;i++){
                s[i]=in.nextInt();
                max = Math.max(s[i], max);
            }
            
            nimsum = s[0];
            for(i = 1;i<n;i++){
                nimsum^=s[i];
            }
            
            if (max==1 && nimsum == 1 || max>1 && nimsum==0) System.out.println("Second");
            else System.out.println("First");
        }
    }
}
                    


                        Solution in Python : 
                            
In  Python3 :






test = int (input ())
for _ in range (test):
    pile = int (input ())
    ar = list (map (int, input (). strip (). split ()))
    xor = 0
    for n in ar:
        xor = xor ^ n
    if ar == [1] * len (ar):
        if len (ar)% 2 == 0:
            print ('First')
        else:
            print ('Second')
    else:
        if xor == 0:
            print ('Second')
        else:
            print ('First')
                    


View More Similar Problems

Jesse and Cookies

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

View Solution →

Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

View Solution →

Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

View Solution →

Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

View Solution →

Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

View Solution →

Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

View Solution →