描述

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.

输入

You will be given at most 1000000 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

输出

For each test case, print the number of days to the next triple peak, in the form:

样例输入

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

样例输出

21252
21152
19575
16994
8910
10789

学信安的我表示学过中国剩余定理但是做题还是不会…..

还要看别人博客才知道要用到中国剩余定理,妈卖批!!!

中国剩余定理:

基本公式为:x+d=M1*y1*p+M2*y2*e+M3*y3*i

x+d=p(mod 23);

x+d=e(mod 23);

x+d=i(mod 23);

M=23*28*33=21252,M1=924,M2=759,M3=644;

924y1=1(mod 23);

644y3=1mod 33);

(这个是由Mk*yk=1(mod mk 可以知道 至于为 什么 我还是从一 篇别人的博客里知道的 至于原因 也没有写)

可以得出来:y1=6,y2=19,y3=2;

进而:x+d=(5544*p+14421*e+1288*i)%21252;

注意:x不能是负数,所以修改为x=(5544*p+14421*e+1288*i-d+21252)%21252;

#include<cstdio>
using namespace std;
int main()
{
int p, e, i, d, icase = 0, x;
while (scanf("%d%d%d%d", &p, &e, &i, &d), ~p)
{
x = (5544 * p + 14421 * e + 1288 * i - d + 21252) % 21252;
if (x == 0)
x = 21252;
printf("%d\n",x);
}
return 0;
}

nyoj 151 Biorhythms的更多相关文章

  1. NYOJ 8 一种排序(comparator排序)

    一种排序 时间限制: 3000 ms  |  内存限制: 65535 KB 难度: 3   描述 现在有很多长方形,每一个长方形都有一个编号,这个编号可以重复:还知道这个长方形的宽和长,编号.长.宽都 ...

  2. NYOJ 1007

    在博客NYOJ 998 中已经写过计算欧拉函数的三种方法,这里不再赘述. 本题也是对欧拉函数的应用的考查,不过考查了另外一个数论基本定理:如何用欧拉函数求小于n且与n互质所有的正整数的和. 记eule ...

  3. NYOJ 998

    这道题是欧拉函数的使用,这里简要介绍下欧拉函数. 欧拉函数定义为:对于正整数n,欧拉函数是指不超过n且与n互质的正整数的个数. 欧拉函数的性质:1.设n = p1a1p2a2p3a3p4a4...pk ...

  4. poj1006 / hdu1370 Biorhythms (中国剩余定理)

    Biorhythms 题意:读入p,e,i,d 4个整数,已知(n+d)%23=p;   (n+d)%28=e;   (n+d)%33=i ,求n .        (题在文末) 知识点:中国剩余定理 ...

  5. NYOJ 333

    http://www.cppblog.com/RyanWang/archive/2009/07/19/90512.aspx?opt=admin 欧拉函数 E(x)表示比x小的且与x互质的正整数的个数. ...

  6. NYOJ 99单词拼接(有向图的欧拉(回)路)

    /* NYOJ 99单词拼接: 思路:欧拉回路或者欧拉路的搜索! 注意:是有向图的!不要当成无向图,否则在在搜索之前的判断中因为判断有无导致不必要的搜索,以致TLE! 有向图的欧拉路:abs(In[i ...

  7. nyoj 10 skiing 搜索+动归

    整整两天了,都打不开网页,是不是我提交的次数太多了? nyoj 10: #include<stdio.h> #include<string.h> ][],b[][]; int ...

  8. NYOJ题目1080年龄排序

    aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAAtMAAAJVCAIAAACTf+6jAAAgAElEQVR4nO3dO1Lj3NbG8W8Szj0QYg ...

  9. poj 1006:Biorhythms(水题,经典题,中国剩余定理)

    Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Des ...

随机推荐

  1. 【hdu5381】维护区间内所有子区间的gcd之和-线段树

    题意:给定n个数,m个询问,每次询问一个区间内所有连续子区间的gcd的和.n,m<=10^5 题解: 这题和之前比赛的一题很像.我们从小到大枚举r,固定右端点枚举左端点,维护的区间最多只有log ...

  2. 【poj3621】最优比率环

    题意: 给定n个点,每个点有一个开心度F[i],每个点有m条单向边,每条边有一个长度d,要求一个环,使得它的 开心度的和/长度和 这个比值最大.n<=1000,m<=5000 题解: 最优 ...

  3. szoj657 【AHSDFZNOI 7.2 WuHongxun】Odd

    [题目大意] 给出$n$个数$a_1, a_2, ..., a_n$,求有多少个区间$[l, r]$,满足每个数都出现了奇数次. $1 \leq n \leq 2 * 10^5, 0 \leq a_i ...

  4. bootstrap框架的搭建

    bootstrap框架 Bootstrap,来自 Twitter,是目前最受欢迎的前端框架.Bootstrap 是基于 HTML.CSS.JAVASCRIPT 的,它简洁灵活,使得 Web 开发更加快 ...

  5. 超详细的Java面试题总结(三)之Java集合篇常见问题

    List,Set,Map三者的区别及总结 List:对付顺序的好帮手 List接口存储一组不唯一(可以有多个元素引用相同的对象),有序的对象 Set:注重独一无二的性质 不允许重复的集合.不会有多个元 ...

  6. [bzoj1005][HNOI2008]明明的烦恼-Prufer编码+高精度

    Brief Description 给出标号为1到N的点,以及某些点最终的度数,允许在 任意两点间连线,可产生多少棵度数满足要求的树? Algorithm Design 结论题. 首先可以参考这篇文章 ...

  7. bzoj 1912 tree_dp

    这道题我们加一条路可以减少的代价为这条路两端点到lca的路径的长度,相当于一条链,那么如果加了两条链的话,这两条链重复的部分还是要走两遍,反而对答案没有了贡献(其实这个可以由任意两条链都可以看成两条不 ...

  8. 一个简单插件this传值的跟踪

    <!DOCUTYPE html> <html> <head> <meta charset="UTF-8"> <script s ...

  9. Coursera课程《Machine Learning》吴恩达课堂笔记

    强烈安利吴恩达老师的<Machine Learning>课程,讲得非常好懂,基本上算是无基础就可以学习的课程. 课程地址 强烈建议在线学习,而不是把视频下载下来看.视频中间可能会有一些问题 ...

  10. Linux内核基础--事件通知链(notifier chain)good【转】

    转自:http://www.cnblogs.com/pengdonglin137/p/4075148.html 阅读目录(Content) 1.1. 概述 1.2.数据结构 1.3.  运行机理 1. ...