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
- linux网络配置完全解析
概述:熟悉了windows下面的网络配置,对linux下的网络配置缺未必了解透彻.熟练掌握linux下的网络配置原理,能帮助我们更容易掌握网络传输原理:同时具备一些网络连接不通对应问题的排查能力.文本 ...
- 关于SQLite3 编译及交叉编译的一些问题
from : http://blog.sina.com.cn/s/blog_5f2e119b0101ibwn.html SQLite3 (http://www.sqlite.org)是一个非常强大的小 ...
- 修改vs17中的cordova模板
因为visual studio 2017创建的默认cordova-ios的版本自动编译带有swift语言的插件会出现异常,cordova-ios升级到4.3.1,并且配置build.json能解决问题 ...
- apache 各种配置
//apache 的网站配置文件 /usr/local/apache2/conf/extra/httpd-vhosts.conf -->在编辑这个文件前需要去httpd.conf把这个文件的注释 ...
- Tutorial 2: Requests and Responses
转载自:http://www.django-rest-framework.org/tutorial/2-requests-and-responses/ Tutorial 2: Requests and ...
- HTTPS握手过程
HTTPS在HTTP的基础上加入了SSL协议,SSL依靠证书来验证服务器的身份,并为浏览器和服务器之间的通信加密.具体是如何进行加密,解密,验证的,且看下图,下面的称为一次握手. 1. 客户端发起HT ...
- HDU 1043 Eight(反向BFS+打表+康托展开)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1043 题目大意:传统八数码问题 解题思路:就是从“12345678x”这个终点状态开始反向BFS,将各 ...
- C语言俄罗斯方块
#include <windows.h> #include <stdio.h> #include <time.h> #include <conio.h> ...
- K&R《C语言》书中的一个Bug
最近在重温K&R的C语言圣经,第二章中的练习题2-2引起了我的注意. 原题是: Write a loop equivalent to the for loop above without us ...
- CSS Sprites的原理(图片整合技术)(CSS精灵)/雪碧图
CSS Sprites的原理(图片整合技术)(CSS精灵)/雪碧图 一.将导航背景图片,按钮背景图片等有规则的合并成一张背景图,即将多张图片合为一张整图,然后用background-positio ...