在这里写题意
时针每过1分钟转0.5°.
(360/(12*60))
分钟每过1分钟转6°
(360/60);
根据这个就能算出时针和分针的角度之差了.
0
在这了写反思
#include
#include
#include
#include
#include
#include
typedef pair
typedef pair
//mt19937 myrand(time(0));
//int get_rand(int n){return myrand()%n + 1;}
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;
int h,m,s,t;
int main(){
//Open();
//Close();
int T;
ri(T);
while (T--){
ri(h),ri(m),ri(s);
ri(t);
t = t % 86400;
s = s + t;
if (s >= 60) {
m += s / 60;
s %= 60;
if (m >= 60) {
h += m / 60;
m %= 60;
if (h >= 24) {
h %= 24;
}
}
}
if (h >= 12) h -= 12;
double s1 = (s/60.0)\*0.5 + 1.0\*m\*0.5 + h\*30;
double s2 = (s/60.0)\*6.0 + m\*6.0;
double temp = abs(s1-s2);
printf("%.4f\\n",min(360-temp,temp));
}
return 0;
}
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