hdu 3221 Brute-force Algorithm (矩阵快速幂+指数循环节)

  1/* ***********************************************
  2Author :111qqz
  3Created Time :Sun 30 Oct 2016 11:46:33 PM CST
  4File Name :code/hdu/3221.cpp
  5************************************************ */
  6
  7#include <cstdio>
  8#include <cstring>
  9#include <iostream>
 10#include <algorithm>
 11#include <vector>
 12#include <queue>
 13#include <set>
 14#include <map>
 15#include <string>
 16#include <cmath>
 17#include <cstdlib>
 18#include <ctime>
 19#define fst first
 20#define sec second
 21#define lson l,m,rt<<1
 22#define rson m+1,r,rt<<1|1
 23#define ms(a,x) memset(a,x,sizeof(a))
 24typedef long long LL;
 25#define pi pair < int ,int >
 26#define MP make_pair
 27
 28using namespace std;
 29const double eps = 1E-8;
 30const int dx4[4]={1,0,0,-1};
 31const int dy4[4]={0,-1,1,0};
 32const int inf = 0x3f3f3f3f;
 33LL a,b,p,n;
 34LL mod;
 35LL euler( LL x)
 36{
 37    LL ret = 1;
 38    for ( LL i = 2 ; i*i <= x;  i++)
 39    {
 40	if (x%i==0)
 41	{
 42	    x/=i;
 43	    ret*=(i-1);
 44	    while (x%i==0)
 45	    {
 46		x/=i;
 47		ret*=i;
 48	    }
 49	}
 50    }
 51    if (x>1) ret*=(x-1);
 52    return ret;
 53}
 54
 55struct Mat
 56{
 57    LL mat[2][2];
 58    void clear()
 59    {
 60	ms(mat,0);
 61    }
 62}M,M1;
 63
 64Mat operator * (Mat a,Mat b)
 65{
 66    Mat res;
 67    res.clear();
 68    for ( int i = 0 ; i < 2 ; i++)
 69	for ( int j = 0 ; j < 2 ; j++)
 70	    for ( int k = 0 ; k < 2 ; k++)
 71	    {
 72		res.mat[i][j] = (res.mat[i][j] + a.mat[i][k]*b.mat[k][j]);
 73		if (res.mat[i][j]>=mod)
 74		    res.mat[i][j] = res.mat[i][j] % mod + mod;
 75	    }
 76    return res;
 77}
 78Mat operator ^ (Mat a,LL b)
 79{
 80    Mat res;
 81    res.clear();
 82    for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
 83    while (b>0)
 84    {
 85	if (b&1) res = res * a;
 86	b = b >> 1LL;
 87	a = a * a;
 88    }
 89    return res;
 90}
 91void init()
 92{
 93    M.clear();
 94    M.mat[0][1] =  M.mat[1][0] = M.mat[1][1] = 1;
 95    M1.clear();
 96    M1.mat[0][0] = 0 ;
 97    M1.mat[1][0] = 1;
 98}
 99LL ksm( LL a,LL b,LL k)
100{
101    LL res = 1LL;
102    while (b>0)
103    {
104	if (b&1) res = (res * a) % k;
105	b = b >> 1LL;
106	a = ( a * a ) % k;
107    }
108    return res;
109}
110int main()
111{
112	#ifndef  ONLINE_JUDGE 
113	freopen("code/in.txt","r",stdin);
114  #endif
115	int T;
116	int cas = 0 ;
117	cin>>T;
118	while (T--)
119	{
120	    scanf("%lld%lld%lld%lld",&a,&b,&p,&n);
121	    printf("Case #%d: ",++cas);
122	    if (n==1)
123	    {
124		printf("%lld\n",a%p);
125		continue;
126	    }
127	    if (n==2)
128	    {
129		printf("%lld\n",b%p);
130		continue;
131	    }
132	    init();
133	    mod = euler(p);
134	    Mat ans;
135	    ans.clear();
136	    ans = (M^(n-2))*M1;
137	    LL x = ans.mat[0][0];
138	    LL y = ans.mat[1][0];
139	    LL ret = ksm(a,x,p)*ksm(b,y,p)%p;
140	    printf("%lld\n",ret);
141	   // printf("Case #%d: %lld\n",++cas,ret);
142	}
143
144  #ifndef ONLINE_JUDGE  
145  fclose(stdin);
146  #endif
147    return 0;
148}