hdoj-1006-c++

思路一：按秒数1s1s遍历43200秒，对满足条件的秒数进行统计

结果：答案错误，问题在于原题是连续时间（角度连续变化）的，用离散的方法肯定不行。

思路二：参考网上的解法，使用角速度相减算周期的方法

具体参考 https://blog.youkuaiyun.com/sinat_18897273/article/details/71191038的解法。

结果：顺利AC。提醒：需要将所有数据设置为double类型，设置为float类型在最后一位数上会有错误，心累。

#include <iostream>

#include <algorithm>

#include <iomanip>

using namespace std;

double max(double a,double b,double c){

    double result = a>b?a:b;

    result = result>c?result:c;

    return result;

}

double min(double a,double b,double c){

    double result = a<b?a:b;

    result = result<c?result:c;

    return result;

}

int main(){

    double angle;

    while(cin>>angle&&angle!=-1){

    double vsh,vsm,vmh,tsh,tsm,tmh,start,end,total=0;

    vsm = 6.-1./10.;vsh = 6.0-1./120.;vmh = 1./10.-1./120.;

    tsm = 360.0/vsm;tsh = 360.0/vsh;tmh = 360.0/vmh;

    double bsh,esh,bsm,esm,bmh,emh;

    bsh = angle/vsh;esh = (360.-angle)/vsh;

    bsm = angle/vsm;esm = (360.-angle)/vsm;

    bmh = angle/vmh;emh = (360.-angle)/vmh;

    while(bmh<43200){

        while(bsm<emh){

            while(bsh<emh&&bsh<esm){

                start = max(bsh,bsm,bmh);

                end = min(esh,esm,emh);

                total += (end-start)>0?(end-start):0;

                bsh = bsh+tsh;esh = esh+tsh;

            }

            bsh = bsh-tsh;esh = esh-tsh;

            bsm = bsm+tsm;esm = esm+tsm;

        }

        bsm = bsm-tsm;esm = esm-tsm;

        bmh = bmh+tmh;emh = emh+tmh;

    }

        cout<<fixed<<setprecision(3)<<total/432<<endl;}

    return 0;

}