HDU-4565 So Easy! 公式化简

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;

LL a, b, n, m;

struct Matrix {
int r, c;
int a[2][2];
void unit() {
memset(a, 0, sizeof (a));
for (int i = 0; i < r; ++i) {
a[i][i] = 1;
}
}
void init() {
memset(a, 0, sizeof (a));
}
Matrix operator * (const Matrix & ot) const {
Matrix ret;
ret.r = r, ret.c = ot.c;
ret.init();
for (int i = 0; i < ret.r; ++i) {
for (int j = 0; j < ret.c; ++j) {
for (int k = 0; k < c; ++k) {
ret.a[i][j] += a[i][k] * ot.a[k][j];
ret.a[i][j] %= m;
}
}
}
return ret;
}
void show() {
puts("");
for (int i = 0; i < r; ++i) {
for (int j = 0; j < c; ++j) {
printf("%I64d ", a[i][j]);
}
puts("");
}
puts("");
}
};

Matrix _pow(Matrix a, LL b) {
Matrix ret;
ret.r = ret.c = 2;
ret.unit();
while (b) {
if (b & 1) {
ret = ret * a;
}
b >>= 1;
a = a * a;
}
return ret;
}

int main() {
while (scanf("%I64d %I64d %I64d %I64d", &a, &b, &n, &m) != EOF) {
Matrix A, B;
A.init();
A.r = 2, A.c = 1;
A.a[0][0] = (2*a*a+2*b) % m, A.a[1][0] = (2*a) % m;
if (n == 1) {
printf("%I64d\n", A.a[1][0]);
continue;
} else if (n == 2) {
printf("%I64d\n", A.a[0][0]);
continue;
}
B.r = B.c = 2;
B.a[0][0] = (2*a)%m, B.a[0][1] = (((b-a*a)%m)+m)%m;
B.a[1][0] = 1, B.a[1][1] = 0;
A = _pow(B, n-2) * A;
printf("%I64d\n", A.a[0][0]);
}
return 0;
}