Codeforces Round #305 (Div. 1) A Mike and Frog 循环节，EXCRT

 此题主要学到了处理循环节的问题。

对于每个生物而言，有可能有无数能到a1，有可能仅1次到a1，也可能一次也到不了a1

对于一次也到不了，直接输出-1

对于仅1次跑到a1，for m次，如果没有就没有了

对于无数次，联立同余方程，可以用EXGCD或者EXCRT求解。

#pragma warning(disable:4996)

#include<iostream>
#include<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_map>
#include<fstream>
#include<iomanip>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define inf 0x7FFFFFFF
#define MOD 1000000007
#define moD 1000000003
#define pii pair<int,string>
#define eps 1e-8
#define equals(a,b) (fabs(a-b)<eps)
#define bug puts("bug")
#define re  register
#define fi first
#define se second
const int maxn = 1e6 + 5;
const double Inf = 10000.0;
const double PI = acos(-1.0);
typedef  long long ll;
typedef unsigned long long ull;
using namespace std;

int f1, f2;
int n;
vector<ll> v1, v2;



ll ai[maxn], bi[maxn];

ll mul(ll a, ll b, ll mod)
{
    ll res = 0;
    while (b > 0)
    {
        if (b & 1) res = (res + a) % mod;
        a = (a + a) % mod;
        b >>= 1;
    }
    return res;
}

ll exgcd(ll a, ll b, ll& x, ll& y)
{
    if (b == 0) { x = 1; y = 0; return a; }
    ll gcd = exgcd(b, a % b, x, y);
    ll tp = x;
    x = y; y = tp - a / b * y;
    return gcd;
}

ll excrt()
{
    ll x, y, k;
    ll M = bi[1], ans = ai[1];
    for (int i = 2; i <= n; i++)
    {
        ll a = M, b = bi[i], c = (ai[i] - ans % b + b) % b;
        ll gcd = exgcd(a, b, x, y), bg = b / gcd;
        if (c % gcd != 0) return -1;

        x = mul(x, c / gcd, bg);
        ans += x * M;
        M *= bg;
        ans = (ans % M + M) % M;
    }
    return (ans % M + M) % M;
}


int main() {
    ll m, h1, a1, h2, a2, x1, y1, x2, y2;
    int pos1, pos2;
    scanf("%lld", &m);
    ll tmp, s1, s2, p1, p2;
    scanf("%lld %lld %lld %lld %lld %lld %lld %lld", &h1, &a1, &x1, &y1, &h2, &a2, &x2, &y2);
    tmp = h1;
    v1.push_back(tmp);
    for (int i = 1; i < 2 * m; i++) {
        tmp = (x1 * tmp + y1) % m;
        v1.push_back(tmp);
    }
    for (int i = 0; i < 2 * m; i++) {
        if (v1[i] == a1) {
            f1++;
            if (f1 == 1) s1 = i;
            if (f1 == 2) {
                p1 = i - s1;
                break;
            }
        }
    }
    if (f1 != 2) p1 = 0;
    tmp = h2;
    v2.push_back(tmp);
    for (int i = 1; i < 2 * m; i++) {
        tmp = (x2 * tmp + y2) % m;
        v2.push_back(tmp);
    }
    for (int i = 0; i < 2 * m; i++) {
        if (v2[i] == a2) {
            f2++;
            if (f2 == 1) s2 = i;
            if (f2 == 2) {
                p2 = i - s2;
                break;
            }
        }
    }
    if (f2 != 2) p2 = 0;
    if (!p2 || !p2) {
        for (int i = 0; i < m; i++) {
            if (v1[i] == a1 && v2[i] == a2) {
                printf("%d", i);
                return 0;
            }
        }
        puts("-1");
        return 0;
    }
    if (!f1 || !f2) {
        puts("-1");
        return 0;
    }
    for (int i = 0; i < m; i++) {
        if (v1[i] == a1 && v2[i] == a2) {
            printf("%d", i);
            return 0;
        }
    }
    n = 2;
    bi[1] = p1 , bi[2] = p2;
    ai[1] = s1 , ai[2] = s2;
    ll res = excrt();
    if (res == -1) puts("-1");
    else printf("%lld", res);
}