HDU1588-Gauss Fibonacci(矩阵高速幂+等比数列二分求和)

题目链接

题意：g(x) = k * x + b。f(x) 为Fibonacci数列。求f(g(x))，从x = 1到n的数字之和sum。并对m取模。

思路：

设A = |（1， 1），（1， 0）|

sum = f(b) + f(k + b) + f(2k + b)...+f((n-1)k + b) (f(x) 为Fibonacci数列)

sum = A^b + A^(k + b) + A^(2k + b)...+ A^((n-1)k + b)

sum = A^b(1 + A^k + A^2k...+A^(n-1)k)

所以A^b与A^k能够用矩阵高速幂求解

之后能够设B = A^k

所以式子能够转化为sum = A^b(1 + B + B^2..+ B^(n - 1))

sum就能够使用等比数列二分求和来攻克了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

//typedef long long ll;
typedef __int64 ll;

const int N = 2;

struct mat{
ll s

;
mat(ll a = 0, ll b = 0, ll c = 0, ll d = 0) {
s[0][0] = a;
s[0][1] = b;
s[1][0] = c;
s[1][1] = d;
}
mat operator * (const mat& c) {
mat ans;
memset(ans.s, 0, sizeof(ans.s));
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
ans.s[i][j] = (s[i][0] * c.s[0][j] + s[i][1] * c.s[1][j]);
return ans;
}
mat operator % (int mod) {
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
s[i][j] %= mod;
return *this;
}
mat operator + (const mat& c) {
mat ans;
memset(ans.s, 0, sizeof(ans.s));
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
ans.s[i][j] = s[i][j] + c.s[i][j];
return ans;
}
void put() {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf("%I64d ", s[i][j]);
printf("\n");
}
}
}c(1, 1, 1, 0), tmp(1, 0, 0, 1);

ll k, b, n, M;

mat pow_mod(int n, mat c) {
if (n == 0)
return tmp;
if (n == 1)
return c;
mat a = pow_mod(n / 2, c);
mat ans = a * a % M;
if (n % 2)
ans = ans * c % M;
return ans;
}

mat sum(int n, mat a) {
if (n == 1)
return a;
if (n & 1)
return (pow_mod(n, a) + sum(n - 1, a)) % M;
else
return (((pow_mod(n / 2, a) + tmp) % M) * sum(n / 2, a) % M);
}

int main() {
while (scanf("%I64d%I64d%I64d%I64d", &k, &b, &n, &M) != EOF) {
mat A = pow_mod(b, c);
mat B = pow_mod(k, c);
mat C = sum(n - 1, B) + tmp;
C = C * A;
printf("%I64d\n", C.s[0][1] % M);
}
return 0;
}