hdu5387 六校 Clock（水题 模拟）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=5387

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 370    Accepted Submission(s): 255

Problem Description

Give a time.(hh:mm:ss)，you should answer the angle between any two of the minute.hour.second hand
Notice that the answer must be not more 180 and not less than 0

 

Input

There are T(1≤T≤104) test cases
for each case,one line include the time

0≤hh<24,0≤mm<60,0≤ss<60

 

Output

for each case,output there real number like A/B.(A and B are coprime).if it's an integer then just print it.describe the angle between hour and minute,hour and second hand,minute and second hand.

 

Sample Input

400:00:0006:00:0012:54:5504:40:00

 

Sample Output

0 0 0 180 180 0 1391/24 1379/24 1/2 100 140 120 
Hint
每行输出数据末尾均应带有空格
 

 

一道比较水的题=，=，稍微用到了一点计算几何的东西

AC代码

#include<stdio.h>#include<math.h>#define eps 1e-8int com(double m){    if(fabs(m)<eps) return 0;    if(m<0) return -1;    return 1;}void print2(double k){    for(int i=1;i<=360;i++)    {        if(com((i*k)-int(i*k+0.5))==0){            if(i==1)                printf("%d ",int(i*k));            else                printf("%d/%d ",int(i*k+0.5),i);                return;        }    }}void print(double x,double y){    double k;    k=fabs(x-y);    if(k>=180)        k=360-k;    print2(k);}int main(){    int hour,t;    double h,m,s,minn,sec;    scanf("%d",&t);    while(t--)    {        scanf("%d:%lf:%lf",&hour,&minn,&sec);        h=hour%12*30+minn*0.5+sec/120;        m=minn*6+sec*0.1;        s=sec*6;        print(h,m);        print(h,s);        print(m,s);        printf("\n");    }}