1+ class SB_250135 {
2+ private static boolean overlapHour (Time cur , Time nxt ) {
3+ if (cur .dh > cur .ds && nxt .dh <=nxt .ds ) return true ;
4+ return cur .ds == 354 && cur .dh > 354 ;
5+ }
6+ private static boolean overlapMin (Time cur , Time nxt ) {
7+ if (cur .dm > cur .ds && nxt .dm <=nxt .ds ) return true ;
8+ return cur .ds == 354 && cur .dm > 354 ; // 초침이 정각으로 갈때 0도 처리
9+ }
10+ public static int solution (int h1 , int m1 , int s1 , int h2 , int m2 , int s2 ) {
11+ int startTime = new Time (h1 , m1 , s1 ).getSecond ();
12+ int endTime = new Time (h2 , m2 , s2 ).getSecond ();
13+ 14+ int cnt = 0 ;
15+ for (int i = startTime ; i < endTime ; i ++) {
16+ Time cur = new Time (i );
17+ Time nxt = new Time (i + 1 );
18+ 19+ boolean isMinRing = overlapMin (cur , nxt );
20+ boolean isHourRing = overlapHour (cur , nxt );
21+ 22+ if (isMinRing && isHourRing ){ // 시침 분침 모두 겹쳤을때
23+ if (nxt .dh == nxt .dm ) cnt ++; // 시침분침의 위치가 같으면 한번만 카운트
24+ else cnt +=2 ;
25+ } else if (isMinRing || isHourRing ) cnt ++;
26+ }
27+ if (startTime ==0 || startTime ==43200 ) cnt ++; // 예외처리: 처음에 시분초 모두 같이 있을 경우(정오, 정각)
28+ return cnt ;
29+ }
30+ 31+ static class Time {
32+ int h , m , s ;
33+ double dh , dm , ds ;
34+ 35+ public Time (int h , int m , int s ) {
36+ this .h = h ;
37+ this .m = m ;
38+ this .s = s ;
39+ setDegrees ();
40+ }
41+ 42+ public Time (int s ) {
43+ this .h = s / 3600 ;
44+ this .m = (s % 3600 ) / 60 ;
45+ this .s = (s % 3600 ) % 60 ;
46+ setDegrees ();
47+ }
48+ 49+ 50+ private int getSecond () {
51+ return h * 3600 + m * 60 + s ;
52+ }
53+ 54+ private void setDegrees () {
55+ this .dh = (h % 12 ) * 30d + m * 0.5 + s * (0.5 / 60 );
56+ this .dm = m * 6 + s * 0.1 ;
57+ this .ds = s * 6 ;
58+ }
59+ }
60+ }
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