POJ——T 1006 Biorhythms
http://poj.org/problem?id=1006
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 138219 | Accepted: 44274 |
Description
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
Output
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. CRT+细节处理
#include <algorithm>
#include <cstdio> using namespace std; int a[],str,m[],p[],cnt,tot,ans; void exgcd(int a,int b,int &x,int &y)
{
if(!b) { x=; y=; return ; }
exgcd(b,a%b,x,y);
int tmp=x; x=y;
y=tmp-a/b*y;
}
int CRT()
{
int ret=;
for(int i=;i<=;i++)
{
int t=tot/p[i],x,y;
exgcd(t,p[i],x,y);
ret=(ret+t*x*a[i])%tot;
}
return ret>=?ret:(ret+tot);
} int main()
{
p[]=; p[]=; p[]=;
tot=p[]*p[]*p[];
for(;scanf("%d%d%d%d",&a[],&a[],&a[],&str);)
{
if(a[]==-&&a[]==-&&a[]==-&&str==-) break;
ans=CRT()-str; cnt++;
for(;ans<=;) ans+=tot;
printf("Case %d: the next triple peak occurs in %d days.\n",cnt,ans);
}
return ;
}
POJ——T 1006 Biorhythms的更多相关文章
- POJ.1006 Biorhythms (拓展欧几里得+中国剩余定理)
POJ.1006 Biorhythms (拓展欧几里得+中国剩余定理) 题意分析 不妨设日期为x,根据题意可以列出日期上的方程: 化简可得: 根据中国剩余定理求解即可. 代码总览 #include & ...
- POJ 1006 - Biorhythms (中国剩余定理)
B - Biorhythms Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Subm ...
- Poj 1006 / OpenJudge 2977 1006 Biorhythms/生理周期
1.链接地址: http://poj.org/problem?id=1006 http://bailian.openjudge.cn/practice/2977 2.题目: Biorhythms Ti ...
- POJ 1006 Biorhythms(中国剩余定理)
题目地址:POJ 1006 学习了下中国剩余定理.參考的该博客.博客戳这里. 中国剩余定理的求解方法: 假如说x%c1=m1,x%c2=m2,x%c3=m3.那么能够设三个数R1,R2,R3.R1为c ...
- poj 1006 Biorhythms (中国剩余定理模板)
http://poj.org/problem?id=1006 题目大意: 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 ...
- [POJ 1006] Biorhythms C++解题
Biorhythms Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 107569 Accepted: 33365 ...
- 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 ...
- POJ 1006 Biorhythms (中国剩余定理)
在POJ上有译文(原文右上角),选择语言:简体中文 求解同余方程组:x=ai(mod mi) i=1~r, m1,m2,...,mr互质利用中国剩余定理令M=m1*m2*...*mr,Mi=M/mi因 ...
- [POJ] #1006# Biorhythms : 最小公倍数/同余问题
一. 题目 Biorhythms Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 127263 Accepted: 403 ...
随机推荐
- 烦人的Facebook分享授权
开发端授权app权限 facebook要求提交应用到他们平台, 并且还限制100mb, 坑爹死了, 果断使用google drive分享给他们, 最開始不确定分享给他们什么样的程序包, 结果审核没通过 ...
- git如何从远程非master分支更新到本地对应分支
git如何从远程非master分支更新到本地对应分支 自己实例 正确步骤 如果本地有分支,那就删除本地分支 删除本地分支::git branch -d 2018_4_18_second 切换分支: g ...
- 如何搭建Eclipse +Apache Tomcat配置Java开发环境
Linux平台下如何搭建Eclipse +Apache Tomcat配置的Java开发环境 本文出自 "李晨光原创技术博客" 博客,请务必保留此出处http://chenguang ...
- java关于File.separator
写好代码在模拟环境下测试,完全没问 题:但linux+tomcat下用的时候,却老是报告“No such file or diretory ”的异常,上传不了.后来发现是文件路径的问题.我的模拟测试环 ...
- 最近学习了一下DeepLearning,发现时NB。
持续关注,有空放个算法到线上的推荐上.
- Windows操作系统下将Redis安装为服务
安装服务: E:/Redis/Redis-x64-3.2.100/redis-server.exe --service-install E:/Redis/Redis-x64-3.2.100/redis ...
- zoj 2317 Nice Patterns Strike Back(矩阵乘法)
problemId=1317">http://acm.zju.edu.cn/onlinejudge/showProblem.do? problemId=1317 给出一个n*m的矩阵( ...
- JavaFX 一 出生新手村(阅读小规则)
我就不讲IDE怎么装的,网上有的是,我仅仅是说说我学习过程中遇到的,该注意的东西 1.JavaFX刚開始出是基于脚本script开发的语言,所以网上会有流传比較多关于script的JavaFX,对于被 ...
- JAVA多态学习2
好啦,这里紧接着上次没说的向下转型,请读者结合上次讨论的向上转型的样例一起探讨(事实上本次代码也有加上,放心). 我们这里希望从向下转型会出现的两种基本情况进行探讨 package downcasti ...
- Http1.1和http2.0
HTTP2.0 最近在读一本书叫<web性能权威指南>谷歌公司高性能团队核心成员的权威之作. 一直听说HTTP2.0,对此也仅仅是耳闻,没有具体研读过,这次正好有两个篇章,分别讲HTTP1 ...