Codeforces Round #305 (Div. 1) A. Mike and Frog 暴力
A. Mike and Frog
Time Limit: 20 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/547/problem/A
Description
Mike has a frog and a flower. His frog is named Xaniar and his flower is named Abol. Initially(at time 0), height of Xaniar is h1 and height of Abol is h2. Each second, Mike waters Abol and Xaniar.
So, if height of Xaniar is h1 and height of Abol is h2, after one second height of Xaniar will become and height of Abol will become where x1, y1, x2 and y2 are some integer numbers and denotes the remainder of a modulo b.
Mike is a competitive programmer fan. He wants to know the minimum time it takes until height of Xania is a1 and height of Abol is a2.
Mike has asked you for your help. Calculate the minimum time or say it will never happen.
Input
The first line of input contains integer m (2 ≤ m ≤ 106).
The second line of input contains integers h1 and a1 (0 ≤ h1, a1 < m).
The third line of input contains integers x1 and y1 (0 ≤ x1, y1 < m).
The fourth line of input contains integers h2 and a2 (0 ≤ h2, a2 < m).
The fifth line of input contains integers x2 and y2 (0 ≤ x2, y2 < m).
It is guaranteed that h1 ≠ a1 and h2 ≠ a2.
Output
Print the minimum number of seconds until Xaniar reaches height a1 and Abol reaches height a2 or print -1 otherwise.
Sample Input
5
4 2
1 1
0 1
2 3
Sample Output
3
HINT
题意
h1=(h1*x1+y1)%m,h2=(h2*x2+y2)%m,求最短时间,h1到达a1的同时h2到达a2
题解:
暴力美学,首先先跑一法,打表出h1到达a1的时间,打表出h2到达a2的时间
首先,我们因为循环2*m次,根据抽屉原理,如果有一次达到了a1,那么必然会有第二次到达a1
然后ans1[0]储存的是h1到达a1的时间,ans2[0]储存的是h2到达a2的时间
然后周期显然就是ans1[1]-ans1[0]和ans2[1]-ans2[0]
然后暴力找就好了!
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** vector<int> ans1;
vector<int> ans2;
int main()
{
//test;
ll m,h1,h2,a1,a2,x1,x2,y1,y2;
m=read();
h1=read(),a1=read();
x1=read(),y1=read();
h2=read(),a2=read();
x2=read(),y2=read(); int tot=;
while(tot<=*m)
{
if(h1==a1)
ans1.push_back(tot);
if(h2==a2)
ans2.push_back(tot);
tot++;
h1=(x1*h1+y1)%m;
h2=(x2*h2+y2)%m;
}
if(ans1.empty()||ans2.empty())
{
cout<<"-1"<<endl;
return ;
}
ll t1=ans1[],t2=ans2[];
ll add1=ans1[]-ans1[],add2=ans2[]-ans2[];
for(int i=;i<;i++)
{
if(t1==t2)
{
printf("%lld\n",t1);
return ;
}
if(t1<t2)
t1+=add1;
else
t2+=add2;
}
cout<<"-1"<<endl;
return ;
}
Codeforces Round #305 (Div. 1) A. Mike and Frog 暴力的更多相关文章
- 数论/暴力 Codeforces Round #305 (Div. 2) C. Mike and Frog
题目传送门 /* 数论/暴力:找出第一次到a1,a2的次数,再找到完整周期p1,p2,然后以2*m为范围 t1,t2为各自起点开始“赛跑”,谁落后谁加一个周期,等到t1 == t2结束 详细解释:ht ...
- Codeforces Round #305 (Div. 2) B. Mike and Fun 暴力
B. Mike and Fun Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/548/pro ...
- Codeforces Round #305 (Div. 2) A. Mike and Fax 暴力回文串
A. Mike and Fax Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/548/pro ...
- set+线段树 Codeforces Round #305 (Div. 2) D. Mike and Feet
题目传送门 /* 题意:对于长度为x的子序列,每个序列存放为最小值,输出长度为x的子序列的最大值 set+线段树:线段树每个结点存放长度为rt的最大值,更新:先升序排序,逐个添加到set中 查找左右相 ...
- 暴力 Codeforces Round #305 (Div. 2) B. Mike and Fun
题目传送门 /* 暴力:每次更新该行的num[],然后暴力找出最优解就可以了:) */ #include <cstdio> #include <cstring> #includ ...
- 字符串处理 Codeforces Round #305 (Div. 2) A. Mike and Fax
题目传送门 /* 字符串处理:回文串是串联的,一个一个判断 */ #include <cstdio> #include <cstring> #include <iostr ...
- Codeforces Round #305 (Div. 1) B. Mike and Feet 单调栈
B. Mike and Feet Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/547/pro ...
- Codeforces Round #305 (Div. 2) D. Mike and Feet 单调栈
D. Mike and Feet time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Codeforces Round #305 (Div. 2) D. Mike and Feet
D. Mike and Feet time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
随机推荐
- 一个Servlet处理增删改查的方法
处理的思路是在servlet中定义不同的增删改查方法,页面请求 的时候携带请求的参数,根据参数判断调用不同的方法. package cn.xm.small.Servlet; import java.i ...
- 浅谈iOS多线程
浅谈iOS多线程 首先,先看看进程和线程的概念. 图1.1 这一块不难理解,重点点下他们的几个重要区别: 1,地址空间和资源:进程可以申请和拥有系统资源,线程不行.资源进程间相互独立,同一进程的各线程 ...
- Android的休眠与唤醒
Android 休眠(suspend),在一个打过android补丁的内核中,state_store()函数会走另外一条路,会进入到request_suspend_state()中,这个文件在earl ...
- Ural Sport Programming Championship 2015
Ural Sport Programming Championship 2015 A - The First Day at School 题目描述:给出课程安排,打印一个课程表. solution 暴 ...
- HOJ 1108
题目链接:HOJ-1108 题意为给定N和M,找出最小的K,使得K个N组成的数能被M整除.比如对于n=2,m=11,则k=2. 思路是抽屉原理,K个N组成的数modM的值最多只有M个. 具体看代码: ...
- C# 获取计算机cpu,硬盘,内存相关的信息
using System;using System.Management; namespace MmPS.Common.Helper{ /// <summary> /// 获取计算机相关的 ...
- csu 1329 一行盒子(链表操作)
1329: 一行盒子 Time Limit: 1 Sec Memory Limit: 128 MB Submit: 693 Solved: 134 [Submit][Status][Web Boa ...
- csu 1767(循环节)
1767: 想打架吗?算我一个!所有人,都过来!(2) Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 99 Solved: 18[Submit][St ...
- 为什么可以这么快! awk 与python的应用
这几天刚处理一个排序问题 源文件: 可以看到有11G大小,需要根据最后一列的热度来做一下排序.如果让你来做这样的排序,在linux环境下,你会如何处理呢? xch27@lanzhou:/asrdata ...
- BOM知识整理
1.窗口位置: 1-1.window,screenLeft获取窗口距离屏幕左边的距离 1-2.window.screenTop获取窗口距离屏幕顶端的距离 1-3.window.screenX和wind ...