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~
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<vector> 5 #include<cmath> 6 #include<sstream> 7 #include<iostream> 8 using namespace std; 9 const double eps = 1e-8; 10 int hh, mm, ss; 11 int gcd(int a, int b) 12 { 13 return b == 0 ? a : gcd(b, a%b); 14 } 15 void sub(int a, int b, int c, int d) 16 { //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 17 if(c == 0) //c/d = 0，直接输出a/b 18 { 19 if(b == 1) printf("%d", abs(a)); 20 else printf("%d/%d", abs(a), abs(b)); 21 return; 22 } 23 int t1 = b*d, t2 = a*d - b*c; 24 if(t1 == 0 || t2 == 0) //分子或分子为0，直接输出0 25 { 26 printf("0"); 27 return; 28 } 29 int t12 = gcd(t1, t2); //化简 30 int tt1 = t1/t12; 31 int tt2 = t2/t12; 32 if(tt1 == 1) printf("%d", abs(tt2)); 33 else printf("%d/%d", abs(tt2), abs(tt1)); 34 } 35 36 void get_ang(double a1, double a2, int t11, int t12, int t21, int t22) 37 { 38 if(fabs(a1 - a2) < eps) //两角相等 39 { 40 printf("0"); 41 return ; 42 } 43 if(fabs(a1 - a2) > 180) //注意两角相差大于180度 44 { 45 if(a1 > a2) 46 { 47 t11 = t11 - 360*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 48 sub(t11, t12, t21, t22); 49 } 50 else 51 { 52 t21 = t21 - 360*t22; 53 sub(t21, t22, t11, t12); 54 } 55 } 56 else if(fabs(a1 - a2) <= 180) 57 { 58 if(a1 > a2) 59 sub(t11, t12, t21, t22); 60 else 61 sub(t21, t22, t11, t12); 62 } 63 } 64 void solve() 65 { 66 int h = hh*3600 + mm*60 + ss; //化成秒为单位 67 int m = mm*60 + ss; 68 int s = ss; 69 int ansh1, ansh2, ansm1, ansm2, anss1, anss2; //主要化简成分数计算 70 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 71 double ah, am, as; //用double比较三针走过的度数大小 72 if(h) 73 { 74 gh = gcd(h, 120); 75 ansh1 = h/gh, ansh2 = 120/gh; 76 ah = (double)ansh1/ansh2; 77 } 78 else 79 { 80 ansh1 = 0, ansh2 = 0; 81 ah = 0.0; 82 } 83 if(m) 84 { 85 gm = gcd(m, 10); 86 ansm1 = m/gm, ansm2 = 10/gm; 87 am = (double)ansm1/ansm2; 88 } 89 else 90 { 91 ansm1 = 0, ansm2 = 0; 92 am = 0.0; 93 } 94 if(s) 95 { 96 anss1 = s*6, anss2 = 1; 97 as = (double)anss1/anss2; 98 } 99 else 100 { 101 anss1 = 0, anss2 = 0; 102 as = 0.0; 103 } 104 get_ang(ah, am, ansh1, ansh2, ansm1, ansm2); printf(" "); 105 get_ang(ah, as, ansh1, ansh2, anss1, anss2); printf(" "); 106 get_ang(am, as, ansm1, ansm2, anss1, anss2); printf(" "); 107 } 108 int main() 109 { 110 int T; 111 scanf("%d", &T); 112 getchar(); 113 while(T--) 114 { 115 string str; 116 cin >> str; 117 for(int i = 0; i < str.length(); i++) 118 if(str[i] == ':') 119 str[i] = ' '; //方便用stringstream，用这个耗时会比较大，处理方法随意~ 120 121 stringstream SS(str); 122 SS >> hh; 123 SS >> mm; 124 SS >> ss; 125 if(hh == 12 && (mm || ss)) hh -= 12; 126 else if(hh > 12) hh -= 12; //24小时制转为12小时 127 solve(); 128 printf(" "); 129 } 130 return 0; 131 }