題意:
(from luckycat)
解法: 先排時間做排序,找最大差值。
TAG: ad hoc
注意: 他給的時間可能沒依照順序喔QQ
程式碼:
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/** | |
* Tittle: 10191 - Longest Nap | |
* Author: Cheng-Shih, Wong | |
* Date: 2015/05/09 | |
*/ | |
// include files | |
#include <iostream> | |
#include <cstdio> | |
#include <cstring> | |
#include <algorithm> | |
using namespace std; | |
// definitions | |
#define FOR(i,a,b) for( int i=(a),_n=(b); i<=_n; ++i ) | |
#define clr(x,v) memset( x, v, sizeof(x) ) | |
#define N 105 | |
class Time { | |
public: | |
int from, to; | |
Time( int _f = 0, int _t = 0 ): from(_f), to(_t) {} | |
const bool operator<( const Time &op ) const { | |
return from<op.from; | |
} | |
}; | |
// declarations | |
int n; | |
Time t[N]; | |
char buf[3*N]; | |
int maxi, maxv; | |
int day = 1; | |
// functions | |
// main function | |
int main( void ) | |
{ | |
int a, b, c, d; | |
// input | |
while( scanf( "%d", &n )==1 ) { | |
n += 2; | |
t[1] = Time( 0, 10*60 ); | |
FOR( i, 2, n-1 ) { | |
scanf( "%d:%d %d:%d", &a, &b, &c, &d ); | |
t[i].from = a*60+b; | |
t[i].to = c*60+d; | |
gets(buf); | |
} | |
t[n] = Time( 18*60, 18*60 ); | |
// solve | |
sort( t+2, t+n ); | |
// FOR( i, 1, n ) printf( "(%d,%d)\n", t[i].from, t[i].to ); | |
maxi = 1; | |
maxv = t[2].from-t[1].to; | |
FOR( i, 2, n-1 ) | |
if( maxv < (t[i+1].from-t[i].to) ) { | |
maxv = t[i+1].from-t[i].to; | |
maxi = i; | |
} | |
// output | |
printf( "Day #%d: the longest nap starts at ", day++ ); | |
printf( "%02d:%02d", t[maxi].to/60, t[maxi].to%60 ); | |
printf( " and will last for " ); | |
if( maxv >= 60 ) | |
printf( "%d hours and ", maxv/60 ); | |
printf( "%d minutes.\n", maxv%60 ); | |
} | |
return 0; | |
} |
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