HDU 5387 Clock（计算时针与分针时针与秒针分针与秒针的夹角）

最新推荐文章于 2022-01-23 17:46:04 发布

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最新推荐文章于 2022-01-23 17:46:04 发布

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本文介绍了一道关于计算时钟中时针、分针及秒针之间角度的问题。通过具体的样例输入输出展示了如何计算任意两针之间的夹角，并确保角度范围在0到180度之间。使用了简单的数学和编程技巧来解决这个问题。

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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

There are

(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

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

Source

2015 Multi-University Training Contest 8

求得夹角必须都是小于180度的

点击打开链接

#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;
}