7
10 35 5 10 15 5 20
10 35 5 10 15 5 20

20

Nias is fond of tug-of-war. One day, he organized a tug-of-war game and invited a group of friends to take part in.

Nias will divide them into two groups. The strategy is simple, sorting them into a row according to their height from short to tall, then let them say one and two alternately (i.e. one, two, one, two...). The people who say one are the members of the red team while others are the members of the blue team.

We know that the team which has a larger sum of weight has advantages in the tug-of-war. Now give you the guys‘ heights and weights, please tell us which team has advantages.

The first line of input contains an integer T, indicating the number of test cases.

The first line of each test case contains an integer N (N is even and 6 ≤ N ≤ 100). 

Each of the next N lines contains two real numbers X and Y, representing the height and weight of a friend respectively.

One line for each test case. If the red team is more likely to win, output "red", if the blue team is more likely to win, output "blue". If both teams have the same weight, output "fair".

1
6 
170 55 
165.3 52.5 
180.2 60.3 
173.3 62.3 
175 57 
162.2 50

blue

One day, zbybr is playing a game with blankcqk, here are the rules of the game:

There is a circle of N stones, zbybr and blankcqk take turns taking the stones.

Each time, one player can choose to take one stone or take two adjacent stones.

You should notice that if there are 4 stones, and zbybr takes the 2nd, the 1st and 3rd stones are still not adjacent.

The winner is the one who takes the last stone.

Now, the game begins and zbybr moves first.

If both of them will play with the best strategy, can you tell me who will win the game?

The first line of input contains an integer T, indicating the number of test cases (T≈100000).

For each case, there is a positive integer N (N ≤ 10^18).

Output the name of the winner.

2
1
2

zbybr
zbybr

题意：zbybr先手，谁将环形石子最后一个拿走谁赢

对于环形的博弈，先手只可能在他能拿石子个数的范围内取胜，否则后手赢。因为当石子个数大于先手能拿的个数，若先手拿完后，剩下奇数个后手拿与先手对应位置的石子
即可，为偶则拿对应的两个，总可以让先手输

int main()
{
    IO;
    int T;
    cin>>T;
    while(T--)
    {
        long long  n;
        cin>>n;
        if(n==1||n==2)
            cout<<"zbybr"<<endl;
        else
            cout<<"blankcqk"<<endl;
    }
    return 0;
}

Association for Couples Math (ACM) is a non-profit organization which is engaged in helping single people to find his/her other half. As November 11th is “Single Day”, on this day, ACM invites a large group of singles to the party. People round together, chatting with others, and matching partners.

There are N gentlemen and M ladies in the party, each gentleman should only match with a lady and vice versa. To memorize the Singles Day, ACM decides to divides to divide people into 11 groups, each group should have the same amount of couples and no people are left without the groups.

Can ACM achieve the goal?

The first line of the input is a positive integer T. T is the number of test cases followed. Each test case contains two integer N and M (0 ≤ N, M ≤ 10^1000), which are the amount of gentlemen and ladies.

For each test case, output “YES” if it is possible to find a way, output “NO” if not.

3
1 1
11 11
22 11

NO
YES
NO

题意：

给定n个绅士和m个美女，要求能把他们分成11对，每对一男一女。
思路：大数模拟或者java或者每次对11取余
代码：

int main()
{
    IO;
    int T;
    cin>>T;
    while(T--)
    {
        char n[100000];
        char m[100000];
        cin>>n>>m;
        if(strcmp(n,m)!=0)
            cout<<"NO"<<endl;
        else
        {
            int sum=0;
            for(int i=0;i<strlen(n);i++)
            {
                sum=(sum*10+n[i]-48)%11;
            }
            if(sum==0)
                cout<<"YES"<<endl;
            else
                cout<<"NO"<<endl;
        }
    }
    return 0;
}