【模拟，时针分针秒针两两夹角】【没有跳坑好兴奋】hdu - 5387 （多校#8 1008）

算是最好写的一道题了吧，最近模拟没手感，一次过也是很鸡冻o(*￣▽￣*)o

注意事项都在代码里，没有跳坑也不清楚坑点在哪~

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<vector>
 #include<cmath>
 #include<sstream>
 #include<iostream>
 using namespace std;
 const double eps = 1e-;
 int hh, mm, ss;
 int gcd(int a, int b)
 {
     return b ==  ? a : gcd(b, a%b);
 }
 void sub(int a, int b, int c, int d)
 { //分数相减（a/b - c/d = (ad-bc)/bd），各种凌乱的特判，注意取绝对值输出
     if(c == ) //c/d = 0，直接输出a/b
     {
         if(b == )  printf("%d", abs(a));
         else        printf("%d/%d", abs(a), abs(b));
         return;
     }
     int t1 = b*d, t2 = a*d - b*c;
     if(t1 ==  || t2 == ) //分子或分子为0，直接输出0
     {
         printf("");
         return;
     }
     int t12 = gcd(t1, t2); //化简
     int tt1 = t1/t12;
     int tt2 = t2/t12;
     if(tt1 == )    printf("%d", abs(tt2));
     else            printf("%d/%d", abs(tt2), abs(tt1));
 }

 void get_ang(double a1, double a2, int t11, int t12, int t21, int t22)
 {
     if(fabs(a1 - a2) < eps) //两角相等
     {
         printf("");
         return ;
     }
     if(fabs(a1 - a2) > ) //注意两角相差大于180度
     {
         if(a1 > a2)
         {
             t11 = t11 - *t12; //大角减360度，然后二者相减去绝对值，这里动手画一下就清楚了
             sub(t11, t12, t21, t22);
         }
         else
         {
             t21 = t21 - *t22;
             sub(t21, t22, t11, t12);
         }
     }
     else if(fabs(a1 - a2) <= )
     {
         if(a1 > a2)
             sub(t11, t12, t21, t22);
         else
             sub(t21, t22, t11, t12);
     }
 }
 void solve()
 {
     int h = hh* + mm* + ss; //化成秒为单位
     int m = mm* + ss;
     int s = ss;
     int ansh1, ansh2, ansm1, ansm2, anss1, anss2; //主要化简成分数计算
     int gh, gm;        //时针化为秒数与120的最大公约数，分针化为秒数与10的最大公约数
     double ah, am, as; //用double比较三针走过的度数大小
     if(h)
     {
         gh = gcd(h, );
         ansh1 = h/gh, ansh2 = /gh;
         ah = (double)ansh1/ansh2;
     }
     else
     {
         ansh1 = , ansh2 = ;
         ah = 0.0;
     }
     if(m)
     {
         gm = gcd(m, );
         ansm1 = m/gm, ansm2 = /gm;
         am = (double)ansm1/ansm2;
     }
     else
     {
         ansm1 = , ansm2 = ;
         am = 0.0;
     }
     if(s)
     {
         anss1 = s*, anss2 = ;
         as = (double)anss1/anss2;
     }
     else
     {
         anss1 = , anss2 = ;
         as = 0.0;
     }
     get_ang(ah, am, ansh1, ansh2, ansm1, ansm2); printf(" ");
     get_ang(ah, as, ansh1, ansh2, anss1, anss2); printf(" ");
     get_ang(am, as, ansm1, ansm2, anss1, anss2); printf(" ");
 }
 int main()
 {
     int T;
     scanf("%d", &T);
     getchar();
     while(T--)
     {
         string str;
         cin >> str;
         for(int i = ; i < str.length(); i++)
             if(str[i] == ':')
                 str[i] = ' '; //方便用stringstream，用这个耗时会比较大，处理方法随意~

         stringstream SS(str);
         SS >> hh;
         SS >> mm;
         SS >> ss;
         if(hh ==  && (mm || ss)) hh -= ;
         else if(hh > ) hh -= ; //24小时制转为12小时
         solve();
         printf("\n");
     }
     return ;
 }

hdu-5387