Biorhythms

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2481    Accepted Submission(s): 1091

Problem 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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 
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
1

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<stdio.h>
using namespace std;
int ext_gcd(int a,int b,int *x,int *y)
{
if(b==)
{
*x=,*y=;
return a;
}
int r = ext_gcd(b,a%b,x,y);
int t =*x;
*x=*y;
*y=t-a/b**y;
return r;
}
int chinese_remainder(int a[],int w[],int len)//a存放余数,w存放两两互质的数
{
int i,d,x,y,m,n,ret;
ret=;
n=;
for(i=; i<len; i++)
{
n*=w[i];
}
for(i=; i<len; i++)
{
m=n/w[i];
d=ext_gcd(w[i],m,&x,&y);
ret=(ret+y*m*a[i])%n;
}
return(ret%n+n)%n;
}
int main()
{
int a[];
int w[]= {,,};
int t;
scanf("%d",&t);
while(t--)
{
int cas=;
int d;
while(scanf("%d%d%d%d",&a[],&a[],&a[],&d))
{
if(a[]==-) break;
for(int i=; i<; i++)
a[i]%=w[i];
int ans=chinese_remainder(a,w,);
ans=ans-d;
if(ans<=) ans+=;
printf("Case %d: the next triple peak occurs in %d days.\n",cas++,ans); }
}
return ;
}

点这里看我整理中国剩余定理的分析

点这里看注释代码

hdu 1370 Biorthythms 中国剩余定理的更多相关文章

  1. hdu 5668 Circle 中国剩余定理

    Circle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem D ...

  2. hdu 3579 Hello Kiki 不互质的中国剩余定理

    Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Probl ...

  3. 《孙子算经》之"物不知数"题:中国剩余定理

    1.<孙子算经>之"物不知数"题 今有物不知其数,三三数之剩二,五五数之剩七,七七数之剩二,问物几何? 2.中国剩余定理 定义: 设 a,b,m 都是整数.  如果 m ...

  4. POJ 1006 中国剩余定理

    #include <cstdio> int main() { // freopen("in.txt","r",stdin); ; while(sca ...

  5. [TCO 2012 Round 3A Level3] CowsMooing (数论,中国剩余定理,同余方程)

    题目:http://community.topcoder.com/stat?c=problem_statement&pm=12083 这道题还是挺耐想的(至少对我来说是这样).开始时我只会60 ...

  6. poj1006中国剩余定理

    Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 103506   Accepted: 31995 Des ...

  7. (伪)再扩展中国剩余定理(洛谷P4774 [NOI2018]屠龙勇士)(中国剩余定理,扩展欧几里德,multiset)

    前言 我们熟知的中国剩余定理,在使用条件上其实是很苛刻的,要求模线性方程组\(x\equiv c(\mod m)\)的模数两两互质. 于是就有了扩展中国剩余定理,其实现方法大概是通过扩展欧几里德把两个 ...

  8. 洛谷P2480 [SDOI2010]古代猪文(费马小定理,卢卡斯定理,中国剩余定理,线性筛)

    洛谷题目传送门 蒟蒻惊叹于一道小小的数论题竟能涉及这么多知识点!不过,掌握了这些知识点,拿下这道题也并非难事. 题意一行就能写下来: 给定\(N,G\),求\(G^{\sum \limits _{d| ...

  9. 洛谷P3868 [TJOI2009]猜数字(中国剩余定理,扩展欧几里德)

    洛谷题目传送门 90分WA第二个点的看过来! 简要介绍一下中国剩余定理 中国剩余定理,就是用来求解这样的问题: 假定以下出现数都是自然数,对于一个线性同余方程组(其中\(\forall i,j\in[ ...

随机推荐

  1. pgbouncer介绍

    一.Pgbouncer 的介绍 Pgbouncer是一个针对PostgreSQL数据库的轻量级连接池,任何目标应用都可以把 pgbouncer 当作一个 PostgreSQL 服务器来连接,然后pgb ...

  2. 比较两个mysql数据库表结构的差异

    需求来源:一个线上系统,一个开发系统,现在要把开发系统更新到线上,但是开发系统的数据库结构与线上的略有差异,所以需要找出两个数据库的表结构差异. 数据库表结构的差异 注:操作均在Linux系统下完成 ...

  3. iOS基础框架的搭建/国际化操作

    1.基础框架的搭建 1.1 pod引入常用的第三方类库 1.2 创建基础文件夹结构/目录结构 Resource———存放声音/图片/xib/storyboard 等资源文件 Define——宏定义, ...

  4. 修改EsayUi 中 tree 的原有样式,变为according 之类的样式 ,且子菜单显示在右侧

    easyUi 中 tree 框架的属性有: 修改原有展开样式代码如下: onExpand:function(node,param){ $(this).children("li"). ...

  5. 【leetcode】Min Stack(easy)

    Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu ...

  6. HDU 5901 Count primes (1e11内的素数个数) -2016 ICPC沈阳赛区网络赛

    题目链接 题意:求[1,n]有多少个素数,1<=n<=10^11.时限为6000ms. 官方题解:一个模板题, 具体方法参考wiki或者Four Divisors. 题解:给出两种代码. ...

  7. UIView CALayer 的区别

    UIView与CALayer的区别,很详细 研究Core Animation已经有段时间了,关于Core Animation,网上没什么好的介绍.苹果网站上有篇专门的总结性介绍,但是似乎原理性的东西不 ...

  8. Linux(CentOS)系统下设置nginx开机自启动

    Nginx 是一个很强大的高性能Web和反向代理服务器.下面介绍在linux下安装后,如何设置开机自启动.首先,在linux系统的/etc/init.d/目录下创建nginx文件,使用如下命令:vi ...

  9. Java集合源码学习(五)几种常用集合类的比较

    这篇笔记对几个常用的集合实现,从效率,线程安全和应用场景进行综合比较. >>ArrayList.LinkedList与Vector的对比 (1)相同和不同都实现了List接口,使用类似.V ...

  10. Java 解析XML的几种方法

    XML现在已经成为一种通用的数据交换格式,它的平台无关性,语言无关性,系统无关性,给数据集成与交互带来了极大的方便. XML在不同的语言里解析方式都是一样的,只不过实现的语法不同而已. 基本的解析方式 ...