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. FFmpeg-20160422-snapshot-bin

    ESC 退出 0 进度条开关 1 屏幕原始大小 2 屏幕1/2大小 3 屏幕1/3大小 4 屏幕1/4大小 S 下一帧 [ -2秒 ] +2秒 ; -1秒 ' +1秒 下一个帧 -> -5秒 F ...

  2. 【剑指offer】题目20 顺时针打印矩阵

    输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字,例如,如果输入如下矩阵: 1   2   3  4 5   6   7  8 9  10 11 12 13 14 15 16 则依次打印出 ...

  3. rsync实现同步

    一.备份客户端: 1.创建/etc/rsyncd.secrets 权限配置600 (写服务器端的账户密码) 2.客户端配置文件: port=873log file=/var/log/rsync.log ...

  4. 数据库TSQL语句

    一.创建数据库create database test3;二.删除数据库drop database test3;三.如何创建表create(创建) table(表) test(表名)(此处写列 var ...

  5. c语言中的浮点数

    一.浮点数常量(小数) 0.11L, 0.0f ,0.0,1.88,2.5f ,0.188E1 E3表示103        比如 1.88E 3=1.88*1000=1880.0f E-3表示10- ...

  6. 数据存储--sqlite总结

    SQLite SQLite(轻量级的数据库,关系型数据库) 辅助工具:Navicat Premium 等 原理:ios针对存储问题封装了sqlite数据库(c语言数据库). 1 app获取沙盒地址命名 ...

  7. python异步爬虫

    本文主要包括以下内容 线程池实现并发爬虫 回调方法实现异步爬虫 协程技术的介绍 一个基于协程的异步编程模型 协程实现异步爬虫 线程池.回调.协程 我们希望通过并发执行来加快爬虫抓取页面的速度.一般的实 ...

  8. 磁盘空间占满inode结点没用完 并删除了文件但是释放不了

    lsof  |grep delete lsof(list system open file )可显示系统打开的文件,以root身份运行. 很多时候文件正在被占用,即使删除了,也无法释放空间,只有停 了 ...

  9. MVC部分视图(Partial View)

    分部视图,也就是整体视图的一部分.单个视图页面展示在整体页面之上,使用步骤如下 1.创建视图数据也就是viewmodel public class FooterViewModel { public s ...

  10. 算法系列:XXX

    转载自http://www.cnblogs.com/skynet/p/3372855.html 这次分享的宗旨是——让大家学会创建与使用静态库.动态库,知道静态库与动态库的区别,知道使用的时候如何选择 ...