Mike and Frog（CF547A）

A. Mike and Frog

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 amodulo 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 test(s)

input

54 21 10 12 3

output

3

input

10231 21 01 21 1

output

-1

Note

In the first sample, heights sequences are following:

Xaniar: 

Abol: 

 

题意：就是说h1=(h1*x1+y1)%m;h2=(h2*x2+y2)%m;经过t次变化，问你能否使得h1==a1;h2==a2;如果可以就输出t；否则输出-1

思路：考虑循环节；情况一：不管变化几次都无法到达a1或a2；情况二：能到，但是a1或a2有一个不在，或者两个都不在循环节内；情况三：在循环节内，利用同余方程，扩展欧几里德求解；<此题有个大坑，那就是用欧几里德求解时候，x的值在运算过程中可能会溢出，要做特殊处理>

详见代码：

#include<stdio.h>#include<string.h>const int N = 1000006;#define LL __int64void exgcd(LL a,LL b,LL& d,LL& x,LL& y){    if(!b){d=a;x=1;y=0;}    else    {        exgcd(b,a%b,d,y,x);        y-=x*(a/b);    }}bool vis1[N],vis2[N];int main(){    LL m,h1,h2,a1,a2,x1,x2,y1,y2;    scanf("%I64d",&m);    scanf("%I64d%I64d",&h1,&a1);    scanf("%I64d%I64d",&x1,&y1);    scanf("%I64d%I64d",&h2,&a2);    scanf("%I64d%I64d",&x2,&y2);    LL h=h1;    memset(vis1,false,sizeof(vis1));    memset(vis2,false,sizeof(vis2));    bool bo=false;    vis1[h1]=true;    vis2[h2]=true;    LL cnt1=0,tp1=0;    LL cnt2=0,tp2=0;    while(1)    {        h=(x1*h+y1)%m;        if(!vis1[a1]) tp1++;        if(vis1[h]) break;        vis1[h]=true;    }    if(vis1[a1])    {        memset(vis1,0,sizeof(vis1));        vis1[a1]=true;        h=a1;        cnt1=0;        while(1)        {            h=(x1*h+y1)%m;            cnt1++;            if(vis1[h]) break;            vis1[h]=true;        }        if(h!=a1) cnt1=0;        h=h2;        while(1)        {            h=(x2*h+y2)%m;            cnt2++;            if(!vis2[a2]) tp2++;            if(vis2[h]) break;            vis2[h]=true;        }        if(vis2[a2])        {            memset(vis2,false,sizeof(vis2));            vis2[a2]=true;            h=a2;            cnt2=0;            while(1)            {                h=(x2*h+y2)%m;                cnt2++;                if(vis2[h]) break;                vis2[h]=true;            }            if(h!=a2) cnt2=0;            if(cnt1==0||cnt2==0)            {                if(cnt1==0&&cnt2!=0)                {                    LL tmp=tp1-tp2;                    if(tmp>=0&&tmp%cnt2==0) printf("%I64d\n",tp1);                    else printf("-1");                }                else if(cnt2==0&&cnt1!=0)                {                    LL tmp=tp2-tp1;                    if(tmp>=0&&tmp%cnt1==0) printf("%I64d\n",tp2);                    else printf("-1");                }                else                {                    if(tp1==tp2) printf("%I64d\n",tp1);                    else printf("-1\n");                }            }            else            {                if(tp1==tp2) printf("%I64d\n",tp1);                else                {                    LL a=cnt2;                    LL b=cnt1;                    LL c1=tp2;                    LL c2=tp1;                    LL c=c2-c1;                    LL d,x,y;                    exgcd(a,b,d,x,y);                    if(c%d) printf("-1\n");                    else if(d==1&&x*y>0)                    {                        LL ax=1,by=1;                        while((ax*a%m+c1)!=(by*b%m+c2))                        {                            if((ax*a%m+c1)-(by*b%m+c2)<0) ax++;                            else by++;                        }                        printf("%I64d\n",ax*a+c1);                    }                    else                    {                        x=x*(c/d);                        LL mm=b/d;                        x=(x%mm+mm)%mm;                        printf("%I64d\n",c1+a*x);                    }                }            }        }        else printf("-1\n");    }    else printf("-1\n");    return 0;}

 

 

 

大神代码：

 

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;#define out(x) {printf("%I64d\n",(long long)(x));return 0;}int m,h1,a1,x1,y1,h2,a2,x2,y2,p,q,c,dx,dy;int main(){    scanf("%d",&m);    scanf("%d%d%d%d",&h1,&a1,&x1,&y1);    scanf("%d%d%d%d",&h2,&a2,&x2,&y2);    for(p=1;p<=m;p++){        h1=(1LL*h1*x1+y1)%m;        h2=(1LL*h2*x2+y2)%m;        if(h1==a1)break;    }    if(p>m)out(-1);    if(h1==a1&&h2==a2)out(p);    for(c=1;c<=m;c++){        h1=(1LL*h1*x1+y1)%m;        if(h1==a1)break;    }    if(c>m)out(-1);    dx=1;dy=0;    for(int i(1);i<=c;i++){        dx=(1LL*dx*x2)%m;        dy=(1LL*dy*x2+y2)%m;    }    for(int i(1);i<=m;i++){        h2=(1LL*dx*h2+dy)%m;        if(h2==a2)out(p+1LL*c*i);    }    out(-1);    return 0;}