You are my brother

You are my brother

时间: 1ms        内存:128M

描述:

Little A gets to know a new friend, Little B, recently. One day, they realize that they are family 500 years ago. Now, Little A wants to know whether Little B is his elder, younger or brother.

输入:

There are multiple test cases.

For each test case, the first line has a single integer, n (n<=1000). The next n lines have two integers a and b (1<=a,b<=2000) each, indicating b is the father of a. One person has exactly one father, of course. Little A is numbered 1 and Little B is numbered 2.

Proceed to the end of file.

输出:

For each test case, if Little B is Little A’s younger, print “You are my younger”. Otherwise, if Little B is Little A’s elder, print “You are my elder”. Otherwise, print “You are my brother”. The output for each test case occupied exactly one line.

示例输入:

5
1 3
2 4
3 5
4 6
5 6
6
1 3
2 4
3 5
4 6
5 7
6 7

示例输出:

You are my elder
You are my brother

提示:

参考答案(内存最优[800]):

#include <stdio.h>



int t[10][3][3]=

{

	//0

	{

		{' ', '_', ' '},

		{'|', ' ', '|'},

		{'|', '_', '|'}

	},

	//1

	{

		{' ', ' ', ' '},

		{' ', ' ', '|'},

		{' ', ' ', '|'},

	},

	//2

	{

		{' ','_',' '},

		{' ','_','|'},

		{'|','_',' '},

	},

	//3

	{

		{' ','_',' '},

		{' ','_','|'},

		{' ','_','|'},

	},

	//4

	{

		{' ',' ',' '},

		{'|','_','|'},

		{' ',' ','|'},

	},

	//5

	{

		{' ','_',' '},

		{'|','_',' '},

		{' ','_','|'},

	},

	//6

	{

		{' ','_',' '},

		{'|','_',' '},

		{'|','_','|'},

	},

	//7

	{

		{' ','_',' '},

		{' ',' ','|'},

		{' ',' ','|'},

	},

	//8

	{

		{' ','_',' '},

		{'|','_','|'},

		{'|','_','|'},

	},

	//9

	{

		{' ','_',' '},

		{'|','_','|'},

		{' ','_','|'},

	},

};





int main()

{

	//freopen("in.in","r",stdin);

	//freopen("out.out","w",stdout);

	int a[4];

	int i,j,k;

	while(scanf("%d %d %d %d",&a[0], &a[1],&a[2],&a[3]) != EOF)

	{

		for(i=0; i<3; ++i)

		{

			for(j=0; j<4; ++j)

			{

				for(k=0;k<3; ++k)

				{

					printf("%c", t[a[j]][i][k]);

				}

			}

			printf("\n");

		}

	}

}

参考答案(时间最优[0]):

#include <stdio.h>



int t[10][3][3]=

{

	//0

	{

		{' ', '_', ' '},

		{'|', ' ', '|'},

		{'|', '_', '|'}

	},

	//1

	{

		{' ', ' ', ' '},

		{' ', ' ', '|'},

		{' ', ' ', '|'},

	},

	//2

	{

		{' ','_',' '},

		{' ','_','|'},

		{'|','_',' '},

	},

	//3

	{

		{' ','_',' '},

		{' ','_','|'},

		{' ','_','|'},

	},

	//4

	{

		{' ',' ',' '},

		{'|','_','|'},

		{' ',' ','|'},

	},

	//5

	{

		{' ','_',' '},

		{'|','_',' '},

		{' ','_','|'},

	},

	//6

	{

		{' ','_',' '},

		{'|','_',' '},

		{'|','_','|'},

	},

	//7

	{

		{' ','_',' '},

		{' ',' ','|'},

		{' ',' ','|'},

	},

	//8

	{

		{' ','_',' '},

		{'|','_','|'},

		{'|','_','|'},

	},

	//9

	{

		{' ','_',' '},

		{'|','_','|'},

		{' ','_','|'},

	},

};





int main()

{

	//freopen("in.in","r",stdin);

	//freopen("out.out","w",stdout);

	int a[4];

	int i,j,k;

	while(scanf("%d %d %d %d",&a[0], &a[1],&a[2],&a[3]) != EOF)

	{

		for(i=0; i<3; ++i)

		{

			for(j=0; j<4; ++j)

			{

				for(k=0;k<3; ++k)

				{

					printf("%c", t[a[j]][i][k]);

				}

			}

			printf("\n");

		}

	}

}

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