2815.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

Problem Id:2815  User Id:fzk 
#include <stdio.h>
#include <memory.h>
#include <string.h>
#include <math.h>
double get_angle( double s, double f, double m ) {

	if( s >= 12 )
		s -= 12;
	s = ( s+f/60+m/7200 ) / 12 * 360;
	f = ( f+m/120 )/60 * 360;
	f -= s;
	if( f < 0 )
		f += 360;
	return f;
}
double ans[24][60][120];
int main( ) {
	int i, j, k, a, ii, jj, kk;
	double t;
	for( i=0; i<24; i++ )
	for( j=0; j<60; j++ )
	for( k=0; k<120; k++ ) 
		ans[i][j][k] = get_angle( i, j, k );
//	printf( "%lf %lf %lf\n", ans[9][16][21*2], ans[9][16][21*2+1], ans[9][16][22*2] );
	while( 1 ) {
		scanf( "%d %d:%d:%d", &a, &i, &j, &k );
		t = ans[i][j][k*2];		
		if( a < 0 )
			break;			
		if( fabs( t - a ) < 1e-7 ) {
			printf( "%02d:%02d:%02d\n", i, j, k );
			continue;
		}
		ii = i, jj = j, kk = k;
		for(  ; ; i=(i+1)%24 )
		for( j%=60; j<60; j++ )
		for( k%=60; k<60; k++ )
		if( k != kk )
		{
			if( ans[i][j][2*k] - t > 0 ) {
				if(	ans[i][j][k*2]-a > 0 && 0 <= a-t )
						goto over;
			}
			else if( a-t >= 0 || 0 < ans[i][j][2*k]-a )
				goto over;
			t = ans[i][j][2*k];
			ii=i, jj=j, kk=k;
//			printf( "%lf\n", t );
		}		
over:
		printf( "%02d:%02d:%02d\n", ii, jj, kk );
	}
	return 0;
}