hdu 5667 Sequence 矩阵快速幂+费马小定理

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,abfcn−1fn−2,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.

    1≤T≤10,1≤n≤1018,1≤a,b,c≤109,p is a prime number,and p≤109+7.

 

Output

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

 

Sample Input

1 5 3 3 3 233

 

Sample Output

190

 

Source

BestCoder Round #80

思路：就是f(n)=f(n-1)*c+f(n-2)  +b;

　　　求a^f(n)%p=a^(f(n)%(p-1))%p;//费马小

　　　矩阵构造博客：http://www.cnblogs.com/frog112111/archive/2013/05/19/3087648.html

　　　还有个坑点a%p==0

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
ll a,b,c,n,p;
struct is
{
    ll a[10][10];
};
is juzhenmul(is a,is b,ll hang ,ll lie)
{
    int i,t,j;
    is ans;
    memset(ans.a,0,sizeof(ans.a));
    for(i=1;i<=hang;i++)
    for(t=1;t<=lie;t++)
    for(j=1;j<=lie;j++)
    {
        ans.a[i][t]+=(a.a[i][j]*b.a[j][t]);
        ans.a[i][t]%=(p-1);
    }
    return ans;
}
is quickpow(is ans,is a,ll x)
{
    while(x)
    {
        if(x&1)  ans=juzhenmul(ans,a,3,3);
        a=juzhenmul(a,a,3,3);
        x>>=1;
    }
    return ans;
}
ll quickpow1(ll a,ll n)
{
    ll mul=1;
    if(a%p==0)
    return 0;
    while(n)
    {
        if(n&1) mul*=a,mul%=p;
        n>>=1;
        a=a*a;
        a%=p;
    }
    return mul;
}
int main()
{
    is ans,base,gg;
    int gggg;
    scanf("%d",&gggg);
    while(gggg--)
    {
        ll fn;
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&a,&b,&c,&p);
        if(n>2)
        {
            memset(ans.a,0,sizeof(ans.a));
            ans.a[1][1]=1;
            ans.a[2][2]=1;
            ans.a[3][3]=1;
            memset(base.a,0,sizeof(base.a));
            base.a[1][1]=c;
            base.a[1][2]=1;
            base.a[2][1]=1;
            base.a[3][1]=1;
            base.a[3][3]=1;
            ans=quickpow(ans,base,n-2);
            gg.a[1][1]=b;
            gg.a[1][2]=0;
            gg.a[1][3]=b;
            ans=juzhenmul(gg,ans,1,3);
            fn=ans.a[1][1];
        }
        else
        {
            if(n==1)
            fn=0;
            else
            fn=b;
        }
        printf("%I64d
",quickpow1(a,fn));
    }
    return 0;
}