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[20];

//const int inf=0x7fffffff;   //нчоч╢С

const int inf=0x3f3f3f3f;

inline ll read()

{

    ll x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    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=0;

    while(tot<=2*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 0;

    }

    ll t1=ans1[0],t2=ans2[0];

    ll add1=ans1[1]-ans1[0],add2=ans2[1]-ans2[0];

    for(int i=0;i<5000000;i++)

    {

        if(t1==t2)

        {

            printf("%lld\n",t1);

            return 0;

        }

        if(t1<t2)

            t1+=add1;

        else

            t2+=add2;

    }

    cout<<"-1"<<endl;

    return 0;

}