hdu 4565 So Easy!

最新推荐文章于 2020-02-11 21:47:40 发布

moyan_min

最新推荐文章于 2020-02-11 21:47:40 发布

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公式推导：
( a + b ^ 0.5 ) ^ n +  ( a - b ^ 0.5 ) ^ n = Ck, Ck * ( ( a + b ^ 0.5 )  + ( a - b ^ 0.5 ) ) = ( ( a + b ^ 0.5 )  + ( a - b ^ 0.5 ) ) * ( a + b ^ 0.5 ) ^ n +  ( a - b ^ 0.5 ) ^ n 
=> 2 * a * Ck = Ck+1 + (a * a - b) * Ck-1   => Ck+1 + Ck = Ck * (2 * a + 1 ) + (a * a - b) * Ck-1  
==>  Ck + 1      =   2 * a       b - a * a      *       Ck
         Ck                  1             0                           Ck -1
==>  Ck + 1      =   2 * a       b - a * a     ^  n         *           2 * a
         Ck                  1             0                                            2

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#define LL long long
LL s[2], m;
struct node {
	LL arr[2][2];
};

node power(node p, int pos){
	if(pos == 1) return p;
	node v = power(p, pos / 2);
	node q;
	q.arr[0][0] = (v.arr[0][0] * v.arr[0][0] + v.arr[0][1] * v.arr[1][0]) % m;
	q.arr[0][1] = (v.arr[0][0] * v.arr[0][1] + v.arr[0][1] * v.arr[1][1]) % m;
	q.arr[1][0] = (v.arr[1][0] * v.arr[0][0] + v.arr[1][1] * v.arr[1][0]) % m;
	q.arr[1][1] = (v.arr[1][0] * v.arr[0][1] + v.arr[1][1] * v.arr[1][1]) % m;
	if(pos % 2 == 0) return q;
	v.arr[0][0] = (q.arr[0][0] * p.arr[0][0] + q.arr[0][1] * p.arr[1][0]) % m;
	v.arr[0][1] = (q.arr[0][0] * p.arr[0][1] + q.arr[0][1] * p.arr[1][1]) % m;
	v.arr[1][0] = (q.arr[1][0] * p.arr[0][0] + q.arr[1][1] * p.arr[1][0]) % m;
	v.arr[1][1] = (q.arr[1][0] * p.arr[0][1] + q.arr[1][1] * p.arr[1][1]) % m;
	return v;
}

int main()
{
   // freopen("in.txt", "r", stdin);
    LL a, b;
    int n;
    while (scanf("%I64d %I64d %d %I64d", &a, &b, &n, &m) != EOF){
    	if(n == 1){
    		printf("%lld\n", a * 2 % m);
    		continue;
    	}
        s[0] = 2 * a % m, s[1] = 2;
        node p;
        p.arr[0][0] = a * 2 % m;
        p.arr[0][1] = (b - a * a) % m;
        p.arr[1][0] = 1, p.arr[1][1] = 0;
        node q = p;
        q = power(p, n - 1);
        printf("%I64d\n", (q.arr[0][0] * s[0] % m + q.arr[0][1] * s[1] % m + m * 2) % m);
    }
    return 0;
}