题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5705
题目：
Problem Description
Given a time HH:MM:SS and one parameter a , you need to calculate next time satisfying following conditions:
1. The angle formed by the hour hand and the minute hand is a
2. The time may not be a integer(e.g. 12:34:56.78), rounded down(the previous example 12:34:56).
Input
1. The angle formed by the hour hand and the minute hand is a
2. The time may not be a integer(e.g. 12:34:56.78), rounded down(the previous example 12:34:56).
Input
The 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)
The second line contains one integer a(0≤a≤180)
Each test case contains two lines.
The first line is the time HH:MM:SS(0≤HH<12,0≤MM<60,0≤SS<60)
The second line contains one integer a(0≤a≤180)
Output
For 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
题意：给出时间 HH:MM:SS 角度a 问下一个H和M的角度为a的时刻?
思路：
时针每小时30度,每分钟30/60度,每秒1/120 (3600,60,1)
分针每小时360,每分钟6度,每秒6/60,1/10度; (,720,12)避免精度问题全部乘以120
算出h,m初始角度 h,m之间角度为deg=abs(h-m),暴力枚举几秒后deg==a即可,O(T)
参考：http://www.bubuko.com/infodetail-2051662.html
代码：
#include <bits/stdc++.h> using namespace std; typedef pair<int,int> ii; const double eps=1e-3; const int N=2e5+20; int a; int main() { int cas=1; int h,m,s; while(~scanf("%d:%d:%d",&h,&m,&s)) { cin>>a; a*=120; int sum=h*3600+m*60+s;//总的时间(秒) //h,m初始角度 int hh=sum%(360*120); int mm=(sum*12)%(360*120); int ans=0;//暴力枚举几秒后 while(true) { hh=(hh+1)%(360*120);//时针每秒1/120度，但由于乘了120，所以每次+1度 mm=(mm+12)%(360*120);//分针每秒1/10度，但由于乘了120，所以分针1秒离散成12度 ans++; int deg=abs(hh-mm);//h,m角度 if(abs(deg-a)<=10)//误差在1度内 break; } int anss=(ans+s)%60; int ansm=((ans+s)/60+m)%60; int ansh=(((ans+s)/60+m)/60+h)%12; printf("Case #%d: %02d:%02d:%02d\n",cas++,ansh,ansm,anss); } return 0; }