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Codeforces Round #305 (Div. 2)C---Mike and Frog(扩欧+乱搞)

时间:2015-05-27 22:56:45      阅读:167      评论:0      收藏:0      [点我收藏+]

标签:数论

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 test(s)
Input

5
4 2
1 1
0 1
2 3

Output

3

Input

1023
1 2
1 0
1 2
1 1

Output

-1

Note

In the first sample, heights sequences are following:

Xaniar:

Abol:

m次以内一定有循环节,否则没周期
所以可以搞出循环节,之后可以得到一个二元一次方程,扩欧求解
找一个最小解即可

/*************************************************************************
    > File Name: C.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年05月27日 星期三 16时15分01秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int vis[1001010];

void ex_gcd(int a, int b, int &x, int &y) {
    if (!b) {
        x = 1;
        y = 0;
    }
    else {
        ex_gcd(b, a % b, y, x);
        y -= x * (a / b);
    }
}

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

int main() {
    int m;
    while (~scanf("%d", &m)) {
        int h1, a1, x1, y1;
        int h2, a2, x2, y2;
        scanf("%d%d%d%d", &h1, &a1, &x1, &y1);
        scanf("%d%d%d%d", &h2, &a2, &x2, &y2);
        for (int i = 0; i <= m; ++i) {
            vis[i] = -1;
        }
        int s1 = -1;
        vis[h1] = 0;
        LL cur = h1;
        int len1 = 0;
        for (int i = 1; i <= m; ++i) {
            cur = (cur * x1 + y1) % m;
            if (vis[cur] == -1) {
                vis[cur] = i;
                if (cur == a1) {
                    s1 = i;
                }
            }
            else {
                len1 = i - vis[cur];
                break;
            }
        }
        if (s1 == -1) {
            printf("-1\n");
            continue;
        }
        cur = a1;
        bool flag1 = 0;
        for (int i = s1 + 1; i <= len1 + s1; ++i) {
            cur = (cur * x1 + y1) % m;
        }
        if (cur == a1) {
            flag1 = 1;
        }
        cur = h2;
        int len2 = 0;
        int s2 = -1;
        for (int i = 0; i <= m; ++i) {
            vis[i] = -1;
        }
        vis[h2] = 0;
        for (int i = 1; i <= m;++i) {
            cur = (cur * x2 + y2) % m;
            if (vis[cur] == -1) {
                vis[cur] = i;
                if (cur == a2) {
                    s2 = i;
                }
            }
            else {
                len2 = i - vis[cur];
                break;
            }   
        }
        if (s2 == -1) {
            printf("-1\n");
            continue;
        }
        bool flag2 = 0;
        cur = a2;
        for (int i = s2 + 1; i <= s2 + len2; ++i) {
            cur = (cur * x2 + y2) % m;
        }
        if (cur == a2) {
            flag2 = 1;
        }
        if (!flag1 && !flag2) {
            if (s1 == s2) {
                printf("%d\n", s1);
            }
            else {
                printf("-1\n");
            }
            continue;
        }
        if (!flag1 && flag2) {
            if (s1 < s2) {
                printf("-1\n");
            }
            else {
                if ((s1 - s2) % len2) {
                    printf("-1\n");
                }
                else {
                    printf("%d\n", s1);
                }
            }
            continue;
        }
        if (flag1 && !flag2) {
            if (s1 > s2) {
                printf("-1\n");
            }
            else {
                if ((s2 - s1) % len1) {
                    printf("-1\n");
                }
                else {
                    printf("%d\n", s2);
                }
            }
            continue;
        }
        int GCD = gcd(len1, len2);
        if ((s2 - s1) % GCD) {
            printf("-1\n");
            continue;
        }
        int x0, y0;
        ex_gcd(len1, len2, x0, y0);
        x0 = x0 * (s2 - s1) / GCD;
        y0 *= -1;
        y0 = y0 * (s2 - s1) / GCD;
        int b = len2 / GCD;
        int a = len1 / GCD;
        if (x0 < 0) {
            LL k = ceil(-1.0 * x0 / b);
            x0 += k * b;
            y0 += k * a;
        }
        if (y0 < 0) {
            LL k = ceil(-1.0 * y0 / a);
            x0 += k * b;
            y0 += k * a;
        }
        LL ans = (LL)y0 * len2 + s2;
        while (1) {
            if (y0 - a >= 0) {
                y0 -= a;
                x0 -= b;
                ans = (LL)y0 * len2 + s2;
            }
            else {
                break;
            }
        }
        cout << ans << endl;
    }
    return 0;
}

Codeforces Round #305 (Div. 2)C---Mike and Frog(扩欧+乱搞)

标签:数论

原文地址:http://blog.csdn.net/guard_mine/article/details/46051379

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