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  • HDU 1006 Tick and Tick(时钟,分钟,秒钟角度问题)

    传送门:

    http://acm.hdu.edu.cn/showproblem.php?pid=1006

    Tick and Tick

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 22203    Accepted Submission(s): 5877


    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
     
    Author
    PAN, Minghao
     
    Source
     
    题意:时针,分针和秒针都厌倦了其余两针,只有在与其余两针保持n度以上的距离才能感到高兴。问一天中,三针都高兴的时间占总时间的百分比。注意时间是连续的。
     
    分析:
    时间是连续的,不能枚举。。我一开始就去枚举。。。
    推荐一个大神的题解:(我的太垃圾)
    不要脸的贴一下代码:
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define max_v 10000
    int main()
    {
        int t;
        double n,sum,ft1,ft2,ft3,et1,et2,et3,max,min;
        double sm,sh,mh,tsm,tsh,tmh,fsm,fsh,fmh,esm,esh,emh;
        sm=10./59.;
        sh=120./719.;
        mh=120./11.;
        tsm=360*sm;
        tsh=360*sh;
        tmh=360*mh;
        while(~scanf("%lf",&n))
        {
            if(n<0)
                break;
            sum=0;
            fsm=sm*n;
            fsh=sh*n;
            fmh=mh*n;
            esm=tsm-fsm;
            esh=tsh-fsh;
            emh=tmh-fmh;
            for(ft3=fmh,et3=emh;et3<=43200;et3+=tmh,ft3+=tmh)
            {
                for(ft2=fsh,et2=esh;et2<=43200;et2+=tsh,ft2+=tsh)
                {
                    if(et2<ft3)
                        continue;
                    if(ft2>et3)
                        break;
                    for(t=0,ft1=fsm,et1=esm;et1<=43200;t=t+1,et1=esm+t*tsm,ft1=fsm+t*tsm)
                    {
                        if(et1<ft3 || et1<ft2)
                            continue;
                        if(ft1>et3 || ft1>et2)
                            break;
                        max=ft1;
                        if(ft2>max)
                            max=ft2;
                        if(ft3>max)
                            max=ft3;
                        min=et1;
                        if(et2<min)
                            min=et2;
                        if(et3<min)
                            min=et3;
                        sum+=min-max;
                    }
                }
            }
            sum/=432.;
            printf("%.3lf
    ",sum);
        }
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/yinbiao/p/9311896.html
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