HDU 4794 Arnold （斐波那契数模 n 的应用）

Arnold

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Do you know Vladimir Arnold? He's a mathematician who demonstrated an image transformation method called arnold transformation, which could shuffle all pixels in an image, and after a serials of this transformation, the image would be transformed to its original form.
The transformation method is quite simple. For a given image with N × N pixels (Width and height are both equal to N), a pixel at location (x, y) will be shuffled to location ( (x + y) % N , (x + 2 × y) % N ) (0 ≤ x < N, 0 ≤ y < N). In one step of transformation, all N × N pixels will be shuffled to new corresponding location, making the image a chaotic one. You can do the transformation as many times as you can.
The arnold transformation is very interesting. For every image of size N × N, after finite steps of transformation, the image will become exact the same as the original one. The minimum number of steps which make every possible image become the same as origin will be called as period of arnold transformation. For a given N, can you calculate the period?

 

Input

There will be about 200 test cases. For each test case, there will be an integer N in one line. Here N (2 ≤ N ≤ 4000000000) equals to width and height of images.

 

Output

For each test case, please calculate the period of arnold transformation and output it in one line.

 

Sample Input

112941

 

Sample Output

5720

题意：对一个N*N的矩阵进行若干次转换，每一次转换是矩阵的每一个像素(x,y) 会转移到((x+y)%N,(x+2*y)%N), 经过若干次转换会变回原来的矩阵，问最少要转换多少次才会变回原来的矩阵。

 

题解：http://pan.baidu.com/s/1dE0Tv5R  密码：usdy

AC代码：

#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#include <functional>using namespace std;typedef unsigned long long LL;const int M = 2;struct Matrix //矩阵{        LL m[M][M];};Matrix A;Matrix I = {1, 0, 0, 1};Matrix multi(Matrix a, Matrix b, LL MOD) //矩阵的乘法{        Matrix c;        for (int i = 0; i < M; i++)        {                for (int j = 0; j < M; j++)                {                        c.m[i][j] = 0;                        for (int k = 0; k < M; k++)                        {                                c.m[i][j] = (c.m[i][j] % MOD + (a.m[i][k] % MOD) * (b.m[k][j] % MOD) % MOD) % MOD;                        }                        c.m[i][j] %= MOD;                }        }        return c;}Matrix power(Matrix a, LL k, LL MOD)  //矩阵的快速幂{        Matrix ans = I, p = a;        while (k)        {                if (k & 1)                {                        ans = multi(ans, p, MOD);                        k--;                }                k >>= 1;                p = multi(p, p, MOD);        }        return ans;}LL gcd(LL a, LL b){        return b ? gcd(b, a % b) : a;}const int N = 400005;const int NN = 5005;LL num[NN], pri[NN];LL fac[NN];int cnt, c;bool prime[N]; //判断是否是素数int p[N];      //素数表int k;         //素数个数void isprime() //打素数表{        k = 0;        memset(prime, true, sizeof(prime));        for (int i = 2; i < N; i++)        {                if (prime[i])                {                        p[k++] = i;                        for (int j = i + i; j < N; j += i)                        {                                prime[j] = false;                        }                }        }}LL quick_mod(LL a, LL b, LL m) //快速幂的模运算{        LL ans = 1;        a %= m;        while (b)        {                if (b & 1)                {                        ans = ans * a % m;                        b--;                }                b >>= 1;                a = a * a % m;        }        return ans;}LL legendre(LL a, LL p) //赖让德公式  判断a是不是模P的二次剩余{        if (quick_mod(a, (p - 1) >> 1, p) == 1)        {                return 1;        }        else        {                return -1;        }}void Solve(LL n, LL pri[], LL num[]) //将n质因数分解 pri存底数数，num存幂{        cnt = 0;        LL t = (LL)sqrt(1.0 * n);        for (int i = 0; p[i] <= t; i++)        {                if (n % p[i] == 0)                {                        int a = 0;                        pri[cnt] = p[i];                        while (n % p[i] == 0)                        {                                a++;                                n /= p[i];                        }                        num[cnt] = a;                        cnt++;                }        }        if (n > 1) //有余数        {                pri[cnt] = n;                num[cnt] = 1;                cnt++;        }}void Work(LL n) //n的所有因子{        c = 0;        LL t = (LL)sqrt(1.0 * n);        for (int i = 1; i <= t; i++)        {                if (n % i == 0)                {                        if (i * i == n)                        {                                fac[c++] = i;                        }                        else                        {                                fac[c++] = i;                                fac[c++] = n / i;                        }                }        }}LL find_loop(LL n)  //找循环节{        Solve(n, pri, num);        LL ans = 1;        for (int i = 0; i < cnt; i++)        {                LL record = 1;                if (pri[i] == 2)                {                        record = 3;                }                else if (pri[i] == 3)                {                        record = 8;                }                else if (pri[i] == 5)                {                        record = 20;                }                else                {                        if (legendre(5, pri[i]) == 1)                        {                                Work(pri[i] - 1);                        }                        else                        {                                Work(2 * (pri[i] + 1));                        }                        sort(fac, fac + c);                        for (int k = 0; k < c; k++) //判断因子中哪一个是循环节的长度                        {                                Matrix a = power(A, fac[k] - 1, pri[i]);                                LL x = (a.m[0][0] % pri[i] + a.m[0][1] % pri[i]) % pri[i];                                LL y = (a.m[1][0] % pri[i] + a.m[1][1] % pri[i]) % pri[i];                                if (x == 1 && y == 0)                                {                                        record = fac[k];                                        break;                                }                        }                }                for (int k = 1; k < num[i]; k++)                {                        record *= pri[i];                }                ans = ans / gcd(ans, record) * record;        }        return ans;}void Init(){        A.m[0][0] = 1;        A.m[0][1] = 1;        A.m[1][0] = 1;        A.m[1][1] = 0;}int main(){        LL n;        Init();        isprime();        while (cin >> n)        {                LL ans = find_loop(n);                if (ans % 2 == 0)                {                        ans /= 2;                }                cout << ans << endl;        }        return 0;}