模拟 HDOJ 5387 Clock

题目传送门

 /*
     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)
         模拟题伤不起！计算公式在代码内(格式：hh/120, mm/120, ss/120)
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-13 13:04:31
 * File Name     :H.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 char str[];

 int GCD(int a, int b)   {
     return (b == ) ? a : GCD (b, a % b);
 }

 int main(void)    {     //HDOJ 5387 Clock
     int T;  scanf ("%d", &T);
     while (T--) {
         int h = , m = , s = ;
         int hh[], mm[], ss[];
         scanf ("%s", str + );
         h = (h + (str[] - '')) *  + str[] - '';
         m = (m + (str[] - '')) *  + str[] - '';
         s = (s + (str[] - '')) *  + str[] - '';

         if (h >= )    h -= ;        //WA点

         hh[] =  * h +  * m + s;      //角度
         mm[] =  * m +  * s;
         ss[] =  * s;

         int tmp = hh[];                    //计算角度差，要在0~180度内
         hh[] = abs (hh[] - mm[]);
         hh[] = min (hh[], * - hh[]);
         int tmp2 = mm[];
         mm[] = abs (tmp - ss[]);
         mm[] = min (mm[], * - mm[]);
         ss[] = abs (tmp2 - ss[]);
         ss[] = min (ss[], * - ss[]);

         hh[] = GCD (, hh[]);           //求GCD，化简
         mm[] = GCD (, mm[]);
         ss[] = GCD (, ss[]);

         if (hh[] %  == ) {             //输出
             printf ("%d ", hh[] / );
         }
         else   {
             printf ("%d/%d ", hh[] / hh[],  / hh[]);
         }
         if (mm[] %  == ) {
             printf ("%d ", mm[] / );
         }
         else   {
             printf ("%d/%d ", mm[] / mm[],  / mm[]);
         }
         if (ss[] %  == ) {
             printf ("%d \n", ss[] / );
         }
         else   {
             printf ("%d/%d \n", ss[] / ss[],  / ss[]);
         }
     }

     return ;
 }