hdu 1907 John&& hdu 2509 Be the Winner（基础nim博弈）

Problem Description

Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.

Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.

 

Input

The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.

Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747

 

Output

Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.

 

Sample Input

2
3
3 5 1
1
1

 

Sample Output

John
Brother

 

//第一道nim博弈题

// T异或为0 S异或为1

//必胜态：T0态，S1态，S2态必胜

//必败态：S0态，T2态必输

 

 

#include <iostream>

using namespace std;

int main()
{
    int t,n;
    int sum1,sum2,ans;
    int a[];
    cin>>t;
    while(t--)
    {
        cin>>n;
        sum1=;
        sum2=;
        ans=;
        for(int i=;i<n;i++)
        {
            cin>>a[i];
            ans=ans^a[i];
            if(a[i]>=)
            sum2++;//富裕堆
            else
            sum1++;//孤单堆
        }
        if((ans&&sum2)||(!ans&&!sum2))
        cout<<"John"<<endl;
        if((ans&&sum1%!=&&sum2==)||(!ans&&sum2>=))
        cout<<"Brother"<<endl;
    }
    return ;
}

Problem Description

Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is 
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).

 

Input

You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.

 

Output

If a winning strategies can be found, print a single line with "Yes", otherwise print "No".

 

Sample Input

2
2 2
1
3

 

Sample Output

No
Yes

#include <iostream>

using namespace std;

int main()
{
    int n,ans,sum1,sum2;
    int a[];
    while(cin>>n)
    {
        sum1=sum2=ans=;
        for(int i=;i<n;i++)
        {
            cin>>a[i];
            ans=ans^a[i];
            if(a[i]>=)
            sum2++;
            else
            sum1++;
        }
        if((ans&&sum2)||(!ans&&!sum2))
        cout<<"Yes"<<endl;
        if((ans&&sum1%!=&&sum2==)||(!ans&&sum2>=))
        cout<<"No"<<endl;
    }
    return ;
}