hdu-5667 Sequence(矩阵快速幂+费马小定理+快速幂)

题目链接:

Sequence

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)


Problem Description
Holion August will eat every thing he has found.

Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.

fn=⎧⎩⎨⎪⎪1,ab,abfcn1fn2,n=1n=2otherwise

He gives you 5 numbers n,a,b,c,p,and he will eat fn foods.But there are only p foods,so you should tell him fn mod p.
 

 

Input
 
The first line has a number,T,means testcase.

Each testcase has 5 numbers,including n,a,b,c,p in a line.

1T10,1n10^18,1a,b,c10^9,p is a prime number,and p10^9+7.
 

 

Output
 
Output one number for each case,which is fn mod p.
 

 

Sample Input
 
1
5 3 3 3 233
 

 

Sample Output
 
190
 
题意:
 
问f(n)对p取模的结果;
 
思路:
 
(ab)p[n]= a* ((ab)p[n-1])c * ((ab)p[n-2]);递推式子可以这样写;
合并后变为(ab)p[n]=(ab)(c*p[n-1]+p[n-2]+1);
可以得到p[n]=c*p[n-1]+p[n-2]+1;
这样的递推式可以用矩阵乘法得到第n项;这是矩阵乘法的一个应用,给matrix67大神的博客地址可以学习,点这里
构造矩阵乘法:
 
p[n]    c    1    1    p[n-1]
p[n-1]   =  1   0    0   *   p[n-2]
1       0   0    1      1
 
然后这中间还有一个问题,就是取模的问题;ab*p[n]%mod=ab*p[n]%(mod-1)%mod;
这是根据费马小定理得到的;a(p-1)Ξ1%p;这里给出地址
 
ab*p[n]%mod=ab*p[n]/(mod-1)*(mod-1)+b*p[n]%(mod-1)%mod;
 
令x=b*p[n]/(mod-1)则ab*p[n]%mod=ax*(mod-1)+b*p[n]%(mod-1)%mod=ab*p[n]%(mod-1)%mod;
这就是取模的结果啦;
然后就是快速幂算答案啦;
还有要注意的就是a%p==0的情况;
 
 
AC代码:
 
 
/*5667    0MS    1584K    1835B    G++    2014300227*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e4+5;
typedef long long ll;
const int mod=1e9+7;
ll n,a,b,c,p;
struct matrix
{
    ll a[3][3];
};
matrix A;
void Iint(ll x)//矩阵初始化;
{
    A.a[0][0]=x;
    A.a[0][1]=1;
    A.a[0][2]=1;
    A.a[1][0]=1;
    A.a[1][1]=A.a[1][2]=0;
    A.a[2][0]=A.a[2][1]=0;
    A.a[2][2]=1;
}
matrix mul(matrix x,matrix y)//矩阵相乘
{
    matrix ans;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            ans.a[i][j]=0;
            for(int k=0;k<3;k++)
            {
                ans.a[i][j]+=(x.a[i][k]*y.a[k][j])%(p-1);
                ans.a[i][j]%=(p-1);
            }
        }
    }
    return ans;
}
ll fast_fun(matrix temp,ll num)//矩阵快速幂;
{
    matrix s,base;
    for(int i=0;i<3;i++)//s初始化为单位矩阵
    {
        for(int j=0;j<3;j++)
        {
            s.a[i][j]=0;
            base.a[i][j]=temp.a[i][j];
        }
    }
    s.a[0][0]=s.a[1][1]=s.a[2][2]=1;
    while(num)
    {
        if(num&1)
        {
            s=mul(s,base);
        }
        base=mul(base,base);
        num=(num>>1);
    }
    return (s.a[0][0]+s.a[0][2])%(p-1);
}
ll fastpow(ll fx,ll fy)//快速幂求结果;
{
    ll s=1,base=fx;
    while(fy)
    {
        if(fy&1)
        {
            s*=base;
            s%=p;
        }
        base*=base;
        base%=p;
        fy=(fy>>1);
    }
    return s;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);
        Iint(c);
        if(n==1)printf("1\n");
        else if(n==2)printf("%lld\n",fastpow(a,b));
        else
        {
            if(a%p==0)printf("0\n");
            else {
            ll gg=fast_fun(A,n-2)*b%(p-1);
            printf("%lld\n",fastpow(a,gg));
            }
        }
    }
    return 0;
}

 

 
posted @ 2016-04-17 16:15  LittlePointer  阅读(864)  评论(0编辑  收藏  举报