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

Posted by 111qqz on Sunday, October 30, 2016

## TOC

``````/* ***********************************************
Author :111qqz
Created Time :Sun 30 Oct 2016 11:46:33 PM CST
File Name :code/hdu/3221.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
LL a,b,p,n;
LL mod;
LL euler( LL x)
{
LL ret = 1;
for ( LL i = 2 ; i*i <= x;  i++)
{
if (x%i==0)
{
x/=i;
ret*=(i-1);
while (x%i==0)
{
x/=i;
ret*=i;
}
}
}
if (x>1) ret*=(x-1);
return ret;
}

struct Mat
{
LL mat[2][2];
void clear()
{
ms(mat,0);
}
}M,M1;

Mat operator * (Mat a,Mat b)
{
Mat res;
res.clear();
for ( int i = 0 ; i < 2 ; i++)
for ( int j = 0 ; j < 2 ; j++)
for ( int k = 0 ; k < 2 ; k++)
{
res.mat[i][j] = (res.mat[i][j] + a.mat[i][k]*b.mat[k][j]);
if (res.mat[i][j]>=mod)
res.mat[i][j] = res.mat[i][j] % mod + mod;
}
return res;
}
Mat operator ^ (Mat a,LL b)
{
Mat res;
res.clear();
for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
while (b>0)
{
if (b&1) res = res * a;
b = b >> 1LL;
a = a * a;
}
return res;
}
void init()
{
M.clear();
M.mat[0][1] =  M.mat[1][0] = M.mat[1][1] = 1;
M1.clear();
M1.mat[0][0] = 0 ;
M1.mat[1][0] = 1;
}
LL ksm( LL a,LL b,LL k)
{
LL res = 1LL;
while (b>0)
{
if (b&1) res = (res * a) % k;
b = b >> 1LL;
a = ( a * a ) % k;
}
return res;
}
int main()
{
#ifndef  ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif
int T;
int cas = 0 ;
cin>>T;
while (T--)
{
scanf("%lld%lld%lld%lld",&a,&b,&p,&n);
printf("Case #%d: ",++cas);
if (n==1)
{
printf("%lld\n",a%p);
continue;
}
if (n==2)
{
printf("%lld\n",b%p);
continue;
}
init();
mod = euler(p);
Mat ans;
ans.clear();
ans = (M^(n-2))*M1;
LL x = ans.mat[0][0];
LL y = ans.mat[1][0];
LL ret = ksm(a,x,p)*ksm(b,y,p)%p;
printf("%lld\n",ret);
// printf("Case #%d: %lld\n",++cas,ret);
}

#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}
``````

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