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

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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可以用矩阵快速幂求解
之后可以设B = A^k
所以式子可以转化为sum = A^b(1 + B + B^2..+ B^(n - 1))
sum就可以使用等比数列二分求和来解决了。

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[N][N];    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;}

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