首页 > ACM题库 > HDU-杭电 > HDU 4794-Arnold[解题报告]HOJ
2015
09-18

HDU 4794-Arnold[解题报告]HOJ

Arnold

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 182    Accepted Submission(s): 33



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
11 29 41
 


Sample Output
5 7 20

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。


思路:

先不看%N,假设现在的位置是Arnold,他会变成Arnold,
其中Arnold

所以其实就是斐波那契数列…… 然后要求的应该是每个斐波那契数列循环节的lcm….但是我打了一下表,发现100多以内的答案都等于0,1开头的斐波那契数列循环节…..所以问题转换为了求0,1开头的斐波那契数列的循环节,一道模板题,下面要做的都是模板,我就不多说了。

最后关于为什么要除以2,因为我求的是斐波那契序列的循环节,假设循环节长度是n, 那么斐波那契序列就是 f1, f2, f3 ….. fn, f1,f2….., 但是题目里面的变换是这样的 (f1,f2) -> (f3, f4) …. -> (f(n-1), fn) 这里n是偶数,所以要除以2, 如果n是奇数那么变换是这样的
(f1,f2)->(f3,f4)->……->(fn,f1)->(f2,f3)->……->(f(n-1),fn) 所以奇数就不用除以2


代码:

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <stdio.h>
#include <cmath>
#include <math.h>
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,b,a) for(int i = (b); i >= (a); --i)
#define clr(a,x) memset(a,(x),sizeof(a))
#define LL unsigned long long
#define eps 1e-8
using namespace std;
LL n;

LL gcd(LL a, LL b)
{
    while (a && b) {
        if (a > b) a %= b;
        else b %= a;
    }
    return a + b;
}

LL lcm(LL a,LL b)
{
    return a * b / gcd(a,b);
}

void mul(LL A[2][2],LL B[2][2],LL ret[2][2],LL mod)
{
    LL C[2][2] = { 0 };
    rep(i,0,2) rep(j,0,2) rep(k,0,2)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod) % mod;
    rep(i,0,2) rep(j,0,2)
        ret[i][j] = C[i][j];
}

void qpow(LL base[2][2],LL p,LL dest[2][2],LL mod)
{
    LL ret[2][2] = { 0 };
    ret[0][0] = ret[1][1] = 1;
    while (p > 0) {
        if (p & 1) mul(ret,base,ret,mod);
        mul(base,base,base,mod);
        p >>= 1;
    }
    rep(i,0,2) rep(j,0,2) dest[i][j] = ret[i][j];
}

LL qpow(LL base,LL p,LL mod)
{
    LL ret = 1;
    while (p) {
        if (p & 1) ret = ret * base % mod;
        base = base * base % mod;
        p >>= 1;
    }
    return ret;
}

LL S[10000],c;
LL f1,f2;
void F(LL p,LL mod)
{
    LL A[2][2] = { 0 };
    A[0][0] = 1; A[0][1] = 1;
    A[1][0] = 1; A[1][1] = 0;
    qpow(A,p-1,A,mod);
    f1 = (A[1][0]+A[1][1]) % mod; f2 = (A[0][0] + A[0][1]) % mod;
}

LL loop(LL mod)
{
    if (mod == 2) return 3;
    else if (mod == 3) return 8;
    else if (mod == 5) return 20;
    LL p;
    if (qpow(5,(mod-1)>>1,mod) == 1) p = mod - 1;
    else p = 2*(mod + 1);
    c = 0;
    for(LL i = 1; i * i <= p; ++i) if (p % i == 0) {
        LL x = i , y = p / i;
        F(x,mod);
        if (f1 == 0 && f2 == 1) return x;
        if (y != x) S[c++] = y;
    }
    while (c > 0) {
        F(S[--c],mod);
        if (f1 == 0 && f2 == 1) return S[c];
    }
    return 0;
}

int main()
{
   // cout << qpow(5,(mod - 1)/2,mod) << endl;
    #ifdef ACM
        freopen("in.txt", "r", stdin);
    #endif // ACM
    while (cin >> n) {
        LL x = n;
        LL ans = 1;
        for(LL i = 2; i * i <= x; ++i) if (x % i == 0) {
            LL len = loop(i);
            LL S = 1;
            do x /= i, S = S * i;
            while (x % i == 0);
            S /= i; S = S * len;
            ans = lcm(ans,S);
        }
        if (x > 1) {
            LL len = loop(x);
            ans = lcm(ans,len);
        }
        if (ans % 2 == 0) ans /= 2;
        printf("%I64u\n",ans);
    }
}




参考:http://blog.csdn.net/wsx1754175/article/details/41875117