Clock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 182 Accepted Submission(s): 129
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
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
4 00:00:00 06:00:00 12:54:55 04:40:00
Sample Output
Hint
每行输出数据末尾均应带有空格
Source
2015 Multi-University Training Contest 8
求得夹角必须都是小于180度的
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
using namespace std;
int x,y,z;
int gcd(int x, int y)
{
if (x%y) return gcd(y, x%y);
return y;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d:%d:%d",&x,&y,&z);
int zz = z*6*120;
int yy = (y*60+z)*12;///角度*10
int xx = ((x%12)*3600 + y*60 + z);///角度*120
int xy = abs(xx - yy);
int xz = abs(xx - zz);
int yz = abs(yy - zz);
if(xy>120*180)
{
xy = 360*120 - xy;
}
if(xz > 120*180)
{
xz = 360*120 - xz;
}
if(yz > 120 * 180)
{
yz = 360*120 - yz;
}
if(xy%120 == 0)
{
printf("%d ",xy/120);
}
else
{
int pp = gcd(xy,120);
printf("%d/%d ",xy/pp,120/pp);
}
if(xz%120 == 0)
{
printf("%d ",xz/120);
}
else
{
int pp = gcd(xz,120);
printf("%d/%d ",xz/pp,120/pp);
}
if(yz%120 == 0)
{
printf("%d ",yz/120);
}
else
{
int pp = gcd(yz,120);
printf("%d/%d ",yz/pp,120/pp);
}
printf("\n");
}
return 0;
}
