"巴卡斯杯" 中国大学生程序设计竞赛 - 2016年女生赛--D--Clock

Given a time HH:MM:SS and one parameter  aa, you need to calculate next time satisfying following conditions: 

1. The angle formed by the hour hand and the minute hand is  aa. 
2. The time may not be a integer(e.g. 12:34:56.78), rounded down(the previous example 12:34:56). 

InputThe input contains multiple test cases. 

Each test case contains two lines. 
The first line is the time HH:MM:SS (0≤HH<12,0≤MM<60,0≤SS<60)(0≤HH<12,0≤MM<60,0≤SS<60). 
The second line contains one integer  a(0≤a≤180)a(0≤a≤180). 
OutputFor each test case, output a single line contains test case number and the answer HH:MM:SS. 
Sample Input

0:59:59
30
01:00:00
30

Sample Output

Case #1: 01:00:00
Case #2: 01:10:54

精度是一个问题，另一个问题是思考的角度，总是想着想着就想偏了

表盘有12个格，所以一个格的角度为360/12=30°

3600s-----1h----360/12=30°

1s-----------------（1/120）°

300s-----5min---360/12=30°

1s-----------------（1/10）°

1s时针和分针的夹角变化为11/120°

一开始可以凭借这个算一下开始的夹角，然后算出需要改变的夹角的值

那么    △角*120/11 就是需要的秒数，将这个秒数加到当前时间上去就是答案

因为可能有精度问题所以把所有的数都乘一个120

#include <bits/stdc++.h>
#define ms(a) memset(a,0,sizeof(a))
typedef long long ll;
using namespace std;

int main()
{
    int hour,minute,second;
    int t=0;
    while(~scanf("%d:%d:%d",&hour,&minute,&second))
    {
        int a;
        scanf("%d",&a);

        int tangle = (60*minute+second)*12 - (3600*hour+60*minute+second);
        tangle=(tangle+120*360)%(120*360);
        int countSecond=0;

        if(tangle>120*180)
        {
            if(360*120-tangle>120*a)
                countSecond=(360*120-tangle-120*a)/11;
            else
                countSecond=(120*a+360*120-tangle)/11;
        }
        else
        {
            if(tangle<=120*a)
                countSecond=(120*a-tangle)/11;
            else
                countSecond=(360*120-tangle-120*a)/11;
        }

        second+=countSecond,minute+=second/60,hour+=minute/60;

        second%=60,minute%=60,hour%=12;

        printf("Case #%d: %02d:%02d:%02d\n",++t,hour,minute,second);
    }
    return 0;
}