1835.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include"stdio.h"
struct point
{
	int x, y, z;
};
inline point cross( point &a, point &b )
{
	point c;
	c.x = a.y*b.z - b.y*a.z;
	c.y = a.z*b.x - b.z*a.x;
	c.z = a.x*b.y - b.x*a.y;
	return c;
}
inline void back( point &p )
{
	p.x = -p.x;
	p.y = -p.y;
	p.z = -p.z;
}
void charge( point &up, point &towards, char c )
{
	point t;
	switch( c )
	{
	case 'f': break;
	case 'l': towards = cross( up, towards); break;
	case 'r': towards = cross( towards, up); break;
	case 'b': back( towards ); break;
	case 'u': t = towards; towards = up; up = t; back( up ); break;
	case 'd': t = up; up = towards; towards = t; back( towards ); break;
	}
	return;
}
point normal[] = { {1,0,0}, {0,-1,0}, {0,0,1}, {-1,0,0}, {0,1,0}, {0,0,-1} };
int main()
{
	int cas, n, x, i;
	char w[10];	
	point up, towards;
	__int64 xx,yy,zz;
	scanf( "%d", &cas );	
	while( cas-- )
	{
		xx = yy = zz = 0;
		up.z = 1, up.x = up.y = 0;
		towards.x = 1, towards.y = towards.z = 0;
		
		scanf( "%d", &n );
		while(n--)
		{
			scanf( "%s %d", w, &x );
			charge( up, towards, w[0] );
			
			xx += x * towards.x;
			yy += x * towards.y;
			zz += x * towards.z;
		}
		for( i=0; i<6; i++ )
			if( towards.x == normal[i].x && towards.y == normal[i].y && towards.z == normal[i].z )
				break;

		printf( "%I64d %I64d %I64d %d\n", xx, -yy, zz, i );
	}
	return 0;
}