hdoj 5667 Sequence 【矩阵快速幂 + 费马小定理】

题目链接：hdoj 5667 Sequence

Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1030 Accepted Submission(s): 338

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

f[i] = 0, i = 1
f[i] = a^b, i = 2
f[i] = a^b * f[i-1]^c * f[i-2], i = 3

对a取log，
g[i] = b + c * g[i-1] + g[i-2]。
发现a^g[i] = f[i] (%p)由费马小定理-> a^(g[i]%(p-1)) = f[i] (%p)。
特判a % p == 0的情况。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <iostream>
#include <cmath>
#include <queue>
#include <stack>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e3 + 10;
LL p;
void add(LL &x, LL y) { x += y; x %= (p-1); }
struct Ma {
    LL a[3][3];
};
Ma multi(Ma x, Ma y) {
    Ma z; CLR(z.a, 0);
    for(int i = 0; i < 3; i++) {
        for(int k = 0; k < 3; k++) {
            if(x.a[i][k] == 0) continue;
            for(int j = 0; j < 3; j++) {
                add(z.a[i][j], x.a[i][k]*y.a[k][j]%(p-1));
            }
        }
    }
    return z;
}
Ma Solve(LL n, Ma x) {
    Ma y; CLR(y.a, 0);
    for(int i = 0; i < 3; i++) {
        y.a[i][i] = 1;
    }
    while(n > 0) {
        if(n & 1LL) {
            y = multi(x, y);
        }
        x = multi(x, x);
        n >>= 1;
    }
    return y;
}
LL pow_mod(LL a, LL n) {
    LL ans = 1LL;
    while(n) {
        if(n & 1LL) {
            ans = ans * a % p;
        }
        a = a * a % p;
        n >>= 1LL;
    }
    return ans;
}
int main()
{
    int t; scanf("%d", &t);
    while(t--) {
        LL a, b, c, n;
        scanf("%lld%lld%lld%lld%lld", &n, &a, &b, &c, &p);
        if(a % p == 0) {
            printf("0\n");
            continue;
        }
        if(n == 1) {
            printf("1\n");
            continue;
        }
        Ma ori, res; CLR(ori.a, 0);
        ori.a[0][1] = ori.a[1][0] = ori.a[2][1] = ori.a[2][2] = 1;; ori.a[1][1] = c;
        res = Solve(n-2, ori);
        LL f[3], g[3]; f[0] = 0; f[1] = f[2] = b;
        for(int i = 0; i < 3; i++) {
            g[i] = 0;
            for(int j = 0; j < 3; j++) {
                add(g[i], f[j] * res.a[j][i] % (p-1));
            }
        }
        //cout << g[1] << endl;
        printf("%lld\n", pow_mod(a%p, g[1]));
    }
    return 0;
}