Clock（数学题）

Clock
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1195    Accepted Submission(s): 417

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

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).

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

把时间都乘上120，进行整数之间的计算。

#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<map>
#include<algorithm>
#define maxn 10000
#define MAXN 10000
#define MAXM 10005
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
int main(){
    double a,b,c;int test=0;
    while(~scanf("%lf:%lf:%lf",&a,&b,&c)){
        int d;scanf("%d",&d);
        int aa=(int)a,bb=(int)b,cc=(int)c;
        int d1=1,d2=12,d3,d4,d5,ans,d6;
        int x=cc+bb*60+aa*60*60;
        for(int i=x+1;i<=x+46799;i++){
            i%=46799;
            d3=(d1*i)%(360*120);d4=(d2*i)%(360*120);
            d5=abs(d3-d4);
            if(d5>(180*120))d6=(360*120-d5);
            if(abs(d5-d*120)<=10||abs(d6-d*120)<=10){ans=i;break;}
        }
        //cout<<ans<<endl;
        int ans1,ans2,ans3;
        ans1=ans/3600;ans1%=12;
        ans2=(ans%3600)/60;ans3=(ans%3600)%60;
        printf("Case #%d: %02d:%02d:%02d
",++test,ans1,ans2,ans3);
    }
}