Mike and Frog (CF547A), frogcf547a

Mike and Frog (CF547A), frogcf547a
A. Mike and Frogtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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 isH_{1 and height of Abol isH2. Each second, Mike waters Abol and Xaniar.}

So, if height of Xaniar isH_{1 and height of Abol isH2, after one second height of Xaniar will become and height of Abol will become whereX1, bytes,Y1, bytes,X2 andY2 are some integer numbers and denotes the remainderAModuloB.}

Mike is a competitive programmer fan. He wants to know the minimum time it takes until height of Xania isA_{1 and height of Abol isA2.}

Mike has asked you for your help. Calculate the minimum time or say it will never happen.

Input

The first line of input contains integerM(2 cores ≤ CoresMLimit ≤ limit 10^{6 ).}

The second line of input contains integersH_{1 andA1 (0 bytes ≤ bytesH1, bytes,A1 worker <workerM).}

The third line of input contains integersX_{1 andY1 (0 bytes ≤ bytesX1, bytes,Y1 worker <workerM).}

The fourth line of input contains integersH_{2 andA2 (0 bytes ≤ bytesH2, bytes,A2 bytes <averageM).}

The maximum th line of input contains integersX_{2 andY2 (0 bytes ≤ bytesX2, bytes,Y2 bytes <averageM).}

It is guaranteed thatH_{1 bytes = bytesA1 andH2 bytes = bytesA2.}

Output

Print the minimum number of seconds until Xaniar reaches heightA_{1 and Abol reaches heightA2 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:

 

Meaning: h1 = (h1 * x1 + y1) % m; h2 = (h2 * x2 + y2) % m; after t changes, ask if you can make h1 = a1; h2 = a2; if you can, output t; otherwise output-1

Idea: Consider loop section. Case 1: a1 or a2 cannot be reached no matter how many changes occur. Case 2: Yes, but either a1 or a2 is absent, or neither of them is in the circular section. Case 3: In the circular section, we use the same remainder equation to extend the Euclidean solution. <this question has a big pitfall, that is, when we use Euclidean to solve the problem, the value of x may overflow during operation. Special processing is required.>

For details, see the code:

#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;}

 

 

 

Great God code:

 

#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;}