HDU 多校 VIII 1008 clock

来源:互联网 发布:淘宝花草茶 那么便宜 编辑:程序博客网 时间:2024/05/22 06:23

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 377    Accepted Submission(s): 200


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(1T104) test cases
for each case,one line include the time

0hh<24,0mm<60,0ss<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

/* * In the name of god (^_^) */#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cmath>#include<stdlib.h>#include<map>#include<set>#include<time.h>#include<vector>#include<queue>#include<string>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define eps 1e-8#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LL long long#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))typedef pair<int , int> pii;int h, m, s;int t[4];int ans[3];int gcd(int a, int b){    return b == 0 ? a : gcd(b, a % b);}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%d:%d:%d", &h, &m, &s);        t[3] = s * 720;        t[2] = m * 720 + s * 12;        t[1] = (h % 12) * 3600 + m * 60 + s;        ans[0] = abs(t[1] - t[2]);        ans[1] = abs(t[1] - t[3]);        ans[2] = abs(t[2] - t[3]);        for(int i = 0; i < 3; i++)        {            if(ans[i] > 180 * 120) ans[i] = 360 * 120 - ans[i];            if(ans[i]%120 == 0)                printf("%d ", ans[i]/120);            else            {                int tmp = gcd(ans[i], 120);                printf("%d/%d ", ans[i]/tmp, 120/tmp);            }        }        printf("\n");    }    return 0;}/*400:00:0006:00:0012:54:5504:40:00*/


1 0