1 /*
2 模拟：这题没啥好说的，把指针转成角度处理就行了，有两个注意点：结果化简且在0~180内；小时13点以后和1以后是一样的(24小时)
3 模拟题伤不起！计算公式在代码内(格式：hh/120, mm/120, ss/120)
4 */
5 /************************************************
6 * Author :Running_Time
7 * Created Time :2015-8-13 13:04:31
8 * File Name :H.cpp
9 ************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 1e5 + 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36 char str[11];
37
38 int GCD(int a, int b) {
39 return (b == 0) ? a : GCD (b, a % b);
40 }
41
42 int main(void) { //HDOJ 5387 Clock
43 int T; scanf ("%d", &T);
44 while (T--) {
45 int h = 0, m = 0, s = 0;
46 int hh[2], mm[2], ss[2];
47 scanf ("%s", str + 1);
48 h = (h + (str[1] - '0')) * 10 + str[2] - '0';
49 m = (m + (str[4] - '0')) * 10 + str[5] - '0';
50 s = (s + (str[7] - '0')) * 10 + str[8] - '0';
51
52 if (h >= 12) h -= 12; //WA点
53
54 hh[0] = 3600 * h + 60 * m + s; //角度
55 mm[0] = 720 * m + 12 * s;
56 ss[0] = 720 * s;
57
58 int tmp = hh[0]; //计算角度差，要在0~180度内
59 hh[0] = abs (hh[0] - mm[0]);
60 hh[0] = min (hh[0], 360*120 - hh[0]);
61 int tmp2 = mm[0];
62 mm[0] = abs (tmp - ss[0]);
63 mm[0] = min (mm[0], 360*120 - mm[0]);
64 ss[0] = abs (tmp2 - ss[0]);
65 ss[0] = min (ss[0], 360*120 - ss[0]);
66
67 hh[1] = GCD (120, hh[0]); //求GCD，化简
68 mm[1] = GCD (120, mm[0]);
69 ss[1] = GCD (120, ss[0]);
70
71 if (hh[0] % 120 == 0) { //输出
72 printf ("%d ", hh[0] / 120);
73 }
74 else {
75 printf ("%d/%d ", hh[0] / hh[1], 120 / hh[1]);
76 }
77 if (mm[0] % 120 == 0) {
78 printf ("%d ", mm[0] / 120);
79 }
80 else {
81 printf ("%d/%d ", mm[0] / mm[1], 120 / mm[1]);
82 }
83 if (ss[0] % 120 == 0) {
84 printf ("%d
", ss[0] / 120);
85 }
86 else {
87 printf ("%d/%d
", ss[0] / ss[1], 120 / ss[1]);
88 }
89 }
90
91 return 0;
92 }
模拟 HDOJ 5387 Clock
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