HDOJ 1006 Tick and Tick_演算法

Problem Description The three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate how much time in a day that all the hands are happy.
 
Input The input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1.
 
Output For each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places.
 
Sample Input

0 120 90 -1  
Sample Output

100.000 0.000 6.251

////  main.cpp//  1006////  Created by 張嘉韜 on 16/2/17.//  Copyright  2016年 張嘉韜. All rights reserved.//改種方法並為解決這個問題，以後有時間再填坑#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int d;double trf(double a){    if(a-180.0>0) return 360.0-a;    else return a;}int func(int h1,int m1,int s1){    double h,m,s,d1,d2,d3,r1,r2,r3;    s=s1;    m=m1+s1*(1.0/60.0);    h=h1+m1*(1.0/60.0)+s1*(1.0/3600.0);    d1=6.0*s;    d2=6.0*m;    d3=30.0*h;    r1=trf(abs(d1-d2));    r2=trf(abs(d1-d3));    r3=trf(abs(d2-d3));    if(r1-d>=0&&r2-d>=0&&r3-d>=0) return 1;    else return 0;}int main(int argc, const char * argv[]) {    while(scanf("%d",&d)==1&&d!=-1)    {        float counter=0;        double res=0;        int s,m,h;        s=m=h=0;        for(int i=1;i<=43200;i++)        {            s++;            if(s==60)            {                m++;                s=0;                if(m==60)                {                    h++;                    m=0;                    if(h==12)                        h=0;                }            }            //if(i==43200) cout<<h<<" "<<m<<" "<<s<<endl;            if(func(h,m,s)==1) {counter++;}        }        res=100*(counter/43200);        printf("%.3f",res);    }    return 0;}