zoukankan      html  css  js  c++  java
  • Biorhythms(中国剩余定理)

    Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical, emotional, and intellectual cycles, and they have periods of lengths 23, 28, and 33 days, respectively. There is one peak in each period of a cycle. At the peak of a cycle, a person performs at his or her best in the corresponding field (physical, emotional or mental). For example, if it is the mental curve, thought processes will be sharper and concentration will be easier.
    Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak.

    Input

    You will be given a number of cases. The input for each case consists of one line of four integers p, e, i, and d. The values p, e, and i are the number of days from the beginning of the current year at which the physical, emotional, and intellectual cycles peak, respectively. The value d is the given date and may be smaller than any of p, e, or i. All values are non-negative and at most 365, and you may assume that a triple peak will occur within 21252 days of the given date. The end of input is indicated by a line in which p = e = i = d = -1.

    Output

    For each test case, print the case number followed by a message indicating the number of days to the next triple peak, in the form:

    Case 1: the next triple peak occurs in 1234 days.

    Use the plural form ``days'' even if the answer is 1.

    Sample Input

    0 0 0 0
    0 0 0 100
    5 20 34 325
    4 5 6 7
    283 102 23 320
    203 301 203 40
    -1 -1 -1 -1

    Sample Output

    Case 1: the next triple peak occurs in 21252 days.
    Case 2: the next triple peak occurs in 21152 days.
    Case 3: the next triple peak occurs in 19575 days.
    Case 4: the next triple peak occurs in 16994 days.
    Case 5: the next triple peak occurs in 8910 days.
    Case 6: the next triple peak occurs in 10789 days.
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    #include <cstring>
    using namespace std;
    typedef long long ll;
    ll exgcd(ll a,ll b,ll &x,ll &y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            return a;
        }
        ll d=exgcd(b,a%b,x,y);
        ll t=x;
        x=y;
        y=t-a/b*y;
        return d;
    }
    
    ll inv(ll a,ll n)
    {
        ll d,x,y;
        d=exgcd(a,n,x,y);
        return d==1?(x+n)%n:-1;
    }
    ll china(ll n, ll *a, ll *m)
    {
        ll M = 1, ret = 0;
        for(ll i = 0; i < n; i ++)
            M *= m[i];
        for(ll i = 0; i < n; i ++)
        {
            ll w = M / m[i];
            ret = (ret + w * inv(w, m[i]) * a[i]) % M;
        }
        return (ret + M) % M;
    }
    int main()
    {
        ll t=1,d;
        ll a[10],m[10];
        m[0]=23;
        m[1]=28;
        m[2]=33;
        while(1)
        {
            cin>>a[0]>>a[1]>>a[2]>>d;
            if(a[0]==-1&&a[1]==-1&&a[2]==-1&&d==-1)
                break;
            ll ans=china(3,a,m);
            if(ans<=d)
                ans+=21252;
            ans-=d;
            cout<<"Case "<<t<<": the next triple peak occurs in "<<ans<<" days."<<endl;
            t++;
        }
        return 0;
    }

    关于本题:

    使33×28被23除余1,用33×28×6=5544;

    使23×33被28除余1,用23×33×19=14421;

    使23×28被33除余1,用23×28×2=1288。

    (5544×p+14421×e+1288×i)%(23×28×33)=n+d

    n=(5544×p+14421×e+1288×i-d)%(23×28×33)

        #include <cstdio>
        int main ()
        {
            int p,e,i,d;
            int lcm=21252,Cas=1;  // lcm(23,28,33)
            while (scanf("%d%d%d%d",&p,&e,&i,&d),~p || ~e || ~i || ~d)
            {
                int n=(5544*p+14421*e+1288*i-d+lcm)%lcm;
                if (n==0) n=lcm;
                printf("Case %d: the next triple peak occurs in %d days.
    ",Cas++,n);
            }
        }
  • 相关阅读:
    微信小程序:模板消息推送提示{“errcode”:41030,”errmsg”:”invalid page hint: [gP1eXXXXXX]”}
    Linux 定时执行shell脚本命令之crontab
    Ubuntu16.04系统下 解决“无法获得锁 /var/lib/dpkg/lock -open (11:资源暂时不可用)、无法锁定管理目录(/var/lib/dpkg/),是否有其他进程正占用它?”的方法
    微信小程序:wx.request之post请求后端无法获取数据的问题
    jQuery获取浏览器参数
    Chrome等浏览器下出现net::ERR_BLOCKED_BY_CLIENT的解决办法
    Thinkphp3.2.3框架下封装公共的函数,例如封装CURL函数来获取接口数据
    Select下拉列表选择自动提交form表单数据
    一步一步从原理跟我学邮件收取及发送 5.C语言的socket示例
    一步一步从原理跟我学邮件收取及发送 4.不同平台下的socket
  • 原文地址:https://www.cnblogs.com/zcy19990813/p/9702693.html
Copyright © 2011-2022 走看看