zoukankan      html  css  js  c++  java
  • Biorhythms(poj1006+中国剩余定理)

    Biorhythms

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 117973   Accepted: 37026

    Description

    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.

    Source

     

    题意:

    生理周期

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 117973   Accepted: 37026

    Description

    人生来就有三个生理周期,分别为体力、感情和智力周期,它们的周期长度为23天、28天和33天。每一个周期中有一天是高峰。在高峰这天,人会在相应的方面表现出色。例如,智力周期的高峰,人会思维敏捷,精力容易高度集中。因为三个周期的周长不同,所以通常三个周期的高峰不会落在同一天。对于每个人,我们想知道何时三个高峰落在同一天。对于每个周期,我们会给出从当前年份的第一天开始,到出现高峰的天数(不一定是第一次高峰出现的时间)。你的任务是给定一个从当年第一天开始数的天数,输出从给定时间开始(不包括给定时间)下一次三个高峰落在同一天的时间(距给定时间的天数)。例如:给定时间为10,下次出现三个高峰同天的时间是12,则输出2(注意这里不是3)。

    Input

    输入四个整数:p, e, i和d。 p, e, i分别表示体力、情感和智力高峰出现的时间(时间从当年的第一天开始计算)。d 是给定的时间,可能小于p, e, 或 i。 所有给定时间是非负的并且小于365, 所求的时间小于21252。 

    当p = e = i = d = -1时,输入数据结束。

    Output

    从给定时间起,下一次三个高峰同天的时间(距离给定时间的天数)。 

    采用以下格式: 
    Case 1: the next triple peak occurs in 1234 days. 

    注意:即使结果是1天,也使用复数形式“days”。

    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.

    Source

    Translator

    北京大学程序设计实习2007, Xie Di
     
     

    解法:中国剩余定理,看了很多的博客终于总结了自己的模板,也是我入门题了。

    关于中国剩余定理的解释: http://www.cnblogs.com/yuyixingkong/p/4358300.html    与 http://baike.sogou.com/v1853008.htm  与  http://www.cnblogs.com/tom987690183/p/3260625.html  与  http://blog.csdn.net/u010468553/article/details/38346195  都可以作为参考的博客;

    关于欧几里得和逆元的解释:http://blog.csdn.net/cqlf__/article/details/7953039 与  http://blog.csdn.net/acdreamers/article/details/8220787  都可以作为参考的博客;  

    转载请注明出处:寻找&星空の孩子

    题目链接:poj1006

    #include<stdio.h>
    #define LL __int64
    LL m[5]={23,28,33};
    LL at[5];
    LL s;
    void exgcd(LL a,LL b,LL &d,LL &x,LL &y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            d=a;
            return ;
        }
        else
        {
            exgcd(b,a%b,d,y,x);
            y-=x*(a/b);
        }
    }
    LL China(int r)
    {
        LL Mc=1;
        LL i,Mi,x,y,d,as=0;
        for(i=0; i<r; i++)
            Mc*=m[i];
        for(i=0; i<r; i++)
        {
            Mi=Mc/m[i];
            exgcd(Mi,m[i],d,x,y);
            as=(as+Mi*x*at[i])%Mc;
        }
        as-=s;
        if(as<=0)
            as+=Mc;
        return as;
    }
    LL work()
    {
        if(!(at[0]%m[0]||at[1]%m[1]||at[2]%m[2]))
            return 23*28*33-s;
        else
            China(3);
    }
    int main()
    {
        int test=1;
        while(scanf("%I64d%I64d%I64d%I64d",&at[0],&at[1],&at[2],&s)!=EOF)
        {
            if(at[0]==-1&&at[1]==-1&&at[2]==-1&&s==-1) break;
            printf("Case %d: the next triple peak occurs in %I64d days.
    ",test++,work());
        }
        return 0;
    }
  • 相关阅读:
    PAT Advanced 1044 Shopping in Mars (25) [⼆分查找]
    PAT Advanced 1029 Median (25) [two pointers]
    PAT Advanced 1010 Radix(25) [⼆分法]
    PAT Basic 1070 结绳(25) [排序,贪⼼]
    PAT Basic 1023 组个最⼩数 (20) [贪⼼算法]
    PAT Basic 1020 ⽉饼 (25) [贪⼼算法]
    PAT Advanced 1070 Mooncake (25) [贪⼼算法]
    PAT Advanced 1067 Sort with Swap(0,*) (25) [贪⼼算法]
    PAT Advanced 1038 Recover the Smallest Number (30) [贪⼼算法]
    PAT Advanced 1037 Magic Coupon (25) [贪⼼算法]
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/4367068.html
Copyright © 2011-2022 走看看