1018 hammer scissors cloth

Everyone should be able to play “ Hammer scissors ” The game of ： Both of them make gestures at the same time , The winning and losing rules are as shown in the figure ：

Now give the record of the two men's confrontation , Please count the wins of both sides 、 flat 、 Negative times , And give the two sides respectively out what gesture has the greatest chance of winning .

Input format ：

Enter the first 1 Line gives a positive integer  N（≤105）, That is, the number of confrontation between the two sides . And then  N  That's ok , Each line gives the message of a confrontation , Namely a 、 Gestures given by both parties at the same time .C  representative “ The hammer ”、J  representative “ scissors ”、B  representative “ cloth ”, The first 1 Two letters represent Party A , The first 2 On behalf of Party B , There is 1 A space .

Output format ：

Output No 1、2 Lines give a respectively 、 B's victory 、 flat 、 Negative times , Between the numbers 1 Space separation . The first 3 The line gives two letters , Each represents a 、 B the gesture that wins the most times , There is 1 A space . If the solution is not unique , The solution with the smallest alphabetical order is output .

sample input ：

10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J

sample output ：

5 3 2
2 3 5
B B

  The code is as follows ：

#include<stdio.h>
int main()
{
	int n, j, i, k;
	char a, b, f;
	char ar[3] = { 0 };
	char br[3] = { 0 };
	scanf("%d", &n);
	getchar();
	int arr[3] = { 0 };
	int shena[3] = { 0 };
	int shenb[3] = { 0 };
	for (i = 0; i < n; i++)
	{
		scanf("%c %c", &a, &b);
		getchar();
		if ((a == 'C' && b == 'J') || (a == 'J' && b == 'B') || (a == 'B' && b == 'C'))
		{
			arr[0]++;// - 
			if (a == 'C')
			{
				shena[0]++;
				ar[0] = 'C';
			}
			else if (a == 'J')
			{
				shena[1]++;
				ar[1] = 'J';
			}
			else
			{
				shena[2]++;
				ar[2] = 'B';
			}
		}
		else if ((a == 'C' && b == 'B') || (a == 'J' && b == 'C') || (a == 'B' && b == 'J'))
		{
			arr[1]++;// negative 
			if (b == 'C')
			{
				shenb[0]++;
				br[0] = 'C';
			}
			else if (b == 'J')
			{
				shenb[1]++;
				br[1] = 'J';
			}
			else
			{
				shenb[2]++;
				br[2] = 'B';
			}
		}
		else
			arr[2]++;
	}
	printf("%d %d %d\n", arr[0], arr[2], arr[1]);
	printf("%d %d %d\n", arr[1], arr[2], arr[0]);


	//shena 0-C 1-J 2-B
	for (i = 0; i < 3 - 1; i++)
	{
		for (j = 0; j < 3 - i - 1; j++)
		{
			if (shena[j] > shena[j + 1])
			{
				k = shena[j];
				shena[j] = shena[j + 1];
				shena[j + 1] = k;
				f = ar[j];
				ar[j] = ar[j + 1];
				ar[j + 1] = f;
			}
		}
	}
	if (shena[0] == shena[1] && shena[1] == shena[2])
		printf("B");
	else if (shena[1] == shena[2])
	{
		if (ar[1] < ar[2])
			printf("%c", ar[1]);
		else
			printf("%c", ar[2]);
	}
	else
		printf("%c", ar[2]);
	for (i = 0; i < 3 - 1; i++)
	{
		for (j = 0; j < 3 - i - 1; j++)
		{
			if (shenb[j] > shenb[j + 1])
			{
				k = shenb[j];
				shenb[j] = shenb[j + 1];
				shenb[j + 1] = k;
				f = br[j];
				br[j] = br[j + 1];
				br[j + 1] = f;
			}
		}
	}
	if (shenb[0] == shenb[1] && shenb[1] == shenb[2])
		printf(" B");
	else if (shenb[1] == shenb[2])
	{
		if (br[1] < br[2])
			printf(" %c", br[1]);
		else
			printf(" %c", br[2]);
	}
	else
		printf(" %c", br[2]);
    printf("\n");
	return 0;

}