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;
}