HDU 1588 Gauss Fibonacci（矩阵高速幂+二分等比序列求和）

HDU 1588 Gauss Fibonacci（矩阵高速幂+二分等比序列求和）

ACM

题目地址：HDU 1588 Gauss Fibonacci

题意： 

g(i)=k*i+b;i为变量。 

给出k,b,n,M，问( f(g(0)) + f(g(1)) + ... + f(g(n)) ) % M的值。

分析： 

把斐波那契的矩阵带进去，会发现这个是个等比序列。

推倒：

S(g(i))
= F(b) + F(b+k) + F(b+2k) + .... + F(b+nk)
// 设 A = {1,1,0,1}, （花括号表示矩阵...）
// 也就是fib数的变化矩阵，F(x) = (A^x) * {1,0}
= F(b) + (A^k)F(b) + (A^2k)F(b)+….+(A^nk)F(b)
// 提取公因式 F(b)
= F(b) [ E +A^k + A^2k + ….+ A^nk] // (E表示的是单位矩阵)
// 令 K = A^k 后
E +A^k + A^2k + ….+ A^nk 变成 K^0+K^1+K^2+…+K^n

然后等比数列是能够二分求和的：数论_等比数列二分求和

代码：

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  Blog:        http://blog.csdn.net/hcbbt
*  File:        1588.cpp
*  Create Date: 2014-08-04 16:13:51
*  Descripton:  Matrix
*/
 
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;
 
const int N = 20;
const int SIZE = 2;        // max size of the matrix
 
ll MOD;
ll k, b, n;
 
struct Mat{
    int n;
    ll v[SIZE][SIZE];    // value of matrix
 
    Mat(int _n = SIZE) {
        n = _n;
    }
 
    void init(ll _v = 0) {
        memset(v, 0, sizeof(v));
        if (_v)
            repf (i, 0, n - 1)
                v[i][i] = _v;
    }
 
    void output() {
        repf (i, 0, n - 1) {
            repf (j, 0, n - 1)
                printf("%lld ", v[i][j]);
            puts("");
        }
        puts("");
    }
} a, B, C;
 
Mat operator * (Mat a, Mat b) {
    Mat c(a.n);
    repf (i, 0, a.n - 1) {
        repf (j, 0, a.n - 1) {
            c.v[i][j] = 0;
            repf (k, 0, a.n - 1) {
                c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;
                c.v[i][j] %= MOD;
            }
        }
    }
    return c;
}
 
Mat operator ^ (Mat a, ll k) {
    Mat c(a.n);
    c.init(1);
    while (k) {
        if (k&1) c = a * c;
        a = a * a;
        k >>= 1;
    }
    return c;
}
 
Mat operator + (Mat a, Mat b) {
    Mat c(a.n);
    repf (i, 0, a.n - 1)
        repf (j, 0, a.n - 1)
            c.v[i][j] = (b.v[i][j] + a.v[i][j]) % MOD;
    return c;
}
 
Mat operator + (Mat a, ll b) {
    Mat c = a;
    repf (i, 0, a.n - 1)
        c.v[i][i] = (a.v[i][i] + b) % MOD;
    return c;
}
 
// 二分求和1..n
Mat calc(Mat a, int n) {
    if (n == 1)
        return a;
    if (n&1)
        return (a^n) + calc(a, n - 1);
    else
        return calc(a, n/2) * ((a^(n/2)) + 1);
}
 
int main() {
    a.init();
    a.v[0][0] = a.v[0][1] = a.v[1][0] = 1;
    while (~scanf("%lld%lld%lld%lld", &k, &b, &n, &MOD)) {
        B = (a^k);
        C = calc(B, n - 1) + (a^0);
        C = (a^b) * C;
        printf("%lld\n", C.v[0][1]);
    }
    return 0;
}