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多校8 1008 Clock

Clock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1320 Accepted Submission(s): 601

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

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

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

  1 #include <stdio.h>
  2 
  3 int main()
  4 {
  5     int T;
  6     int i,j,k,t;
  7     char s[10];
  8     scanf("%d",&T);
  9     while(T--)
 10     {
 11         t=0;
 12         scanf("%s",s);
 13         t=t+(s[7]-'0')+(s[6]-'0')*10;
 14         t=t+((s[4]-'0')+(s[3]-'0')*10)*60;
 15         t=t+((s[1]-'0')+(s[0]-'0')*10)*3600;
 16         //printf("t:%d
",t);
 17         int a,A,b,B,c;
 18         c=(t*6)%360;
 19         b=t%3600,B=10;
 20         a=t%(120*360),A=120;
 21         //printf("a:%d b:%d c:%d
",a,b,c);
 22 
 23         int x,y;
 24         y=120;
 25         if(b*12-a>=0)
 26             x=b*12-a;
 27         else
 28             x=a-b*12;
 29         if(x>(120*180))
 30         {
 31             x=120*360-x;
 32         }
 33         if(x%120==0)
 34             printf("%d ",x/120);
 35         else
 36         {
 37             while(x%2==0 && y%2==0)
 38             {
 39                 x=x/2,y=y/2;
 40             }
 41             while(x%3==0 && y%3==0)
 42             {
 43                 x=x/3,y=y/3;
 44             }
 45             while(x%5==0 && y%5==0)
 46             {
 47                 x=x/5,y=y/5;
 48             }
 49             printf("%d/%d ",x,y);
 50         }
 51 
 52         y=120;
 53         if(c*120-a>=0)
 54             x=c*120-a;
 55         else
 56             x=a-c*120;
 57         if(x>(120*180))
 58         {
 59             x=120*360-x;
 60         }
 61         if(x%120==0)
 62             printf("%d ",x/120);
 63         else
 64         {
 65             while(x%2==0 && y%2==0)
 66             {
 67                 x=x/2,y=y/2;
 68             }
 69             while(x%3==0 && y%3==0)
 70             {
 71                 x=x/3,y=y/3;
 72             }
 73             while(x%5==0 && y%5==0)
 74             {
 75                 x=x/5,y=y/5;
 76             }
 77             printf("%d/%d ",x,y);
 78         }
 79 
 80         y=10;
 81         if(c*10-b>=0)
 82             x=c*10-b;
 83         else
 84             x=b-c*10;
 85         if(x>(10*180))
 86         {
 87             x=10*360-x;
 88         }
 89         //printf("x:%d
",x);
 90         if(x%10==0)
 91             printf("%d ",x/10);
 92         else
 93         {
 94             while(x%2==0 && y%2==0)
 95             {
 96                 x=x/2,y=y/2;
 97             }
 98             while(x%5==0 && y%5==0)
 99             {
100                 x=x/5,y=y/5;
101             }
102             printf("%d/%d ",x,y);
103         }
104         printf("
");
105     }
106     return 0;
107 }

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原文地址：https://www.cnblogs.com/cyd308/p/4771392.html