hdu1209 Clock

/*m点n分k秒时时针从0点转过的角度a = [m + n/60 + k/3600]*30 = 30m + n/2 + k/120(度) 分针转过的角度b = [n/60 + k/3600]*360 = 6n + k/10(度) */#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#define MAXN 10using namespace std;struct node {int h,m;double angle;}times[MAXN];int cmp(node x,node y){if(fabs(x.angle-y.angle)<1e-6)//注意浮点数大小比较{if(x.h==y.h)return x.m<y.m;elsereturn x.h<y.h;}elsereturn x.angle<y.angle;}int main(){int test,i;scanf("%d",&test);while(test--){for (i=0;i<5;++i){scanf("%d:%d",×[i].h,×[i].m);if(times[i].h>12)times[i].angle=fabs(30.0*(times[i].h-12)+times[i].m/2.0-6.0*times[i].m);elsetimes[i].angle=fabs(30.0*times[i].h+times[i].m/2.0-6.0*times[i].m);if(times[i].angle>180)times[i].angle=360-times[i].angle;}sort(times,times+5,cmp);printf("%02d:%02d\n",times[2].h,times[2].m);}return 0;}