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

Posted by 111qqz on Sunday, October 30, 2016

TOC

题目链接

题意:给出了一段伪代码。分析得知其实就是f[1]= a,f[2] = b,f[n]=f[n-1] * f[n-2]

思路:一眼题,和hdu4549很类似hdu4549解题报告

不同的是这道题中p不一定是质数(其实不是也无所谓啊…hdu4549只不过是因为1E9+7是指数,又用费马小定理化简了一下,这道理%phi(p)即可)

还有这道题让我知道了

括号里的话是错误的。只有当x<phi(c)的时候,这个公式才成立。

这道题就是反例,不加判断会wa。

/* ***********************************************
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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